# NCERT Solutions Class 10 Maths Chapter 5

## NCERT Solutions For Class 10 Mathematics Chapter 5 Arithmetic Progressions

Mathematics is one subject that most students find difficult. The only way to actually score good marks in Mathematics is to practice solving as many problems as one can. That is why students find NCERT Solutions For Class 10 Mathematics Chapter 5 by Extramarks really useful in their preparation. If students ever get stuck on any of the exercise questions, they can always refer to these solutions. These are also great resources for completing assignments and for last-minute revision.

### Access NCERT Solutions for Mathematics Chapter 5 – Arithmetic Progression

Chapter 5 of Class 10 Mathematics NCERT textbook introduces students to Arithmetic Progressions, which are nothing but lists of numbers where each term, except the first term, can be derived by adding or subtracting a fixed number to its preceding term. These are really useful to express patterns that we see in our daily lives and in nature. The chapter further discusses the different types of Arithmetic Progressions, the sum of Arithmetic Progressions, deriving a general formula for the nth term of an Arithmetic Progression, and much more.

### What is Arithmetic Progression?

An Arithmetic Progression is a series of numbers where each number can be derived from its predecessor by adding or subtracting a fixed number. For example, consider the series of numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. Here, each term can be derived from its predecessor by adding 1 to it.

The fixed difference between the terms of an AP is called the common difference. In the above example, the common difference is 1.

## NCERT Solutions for Class 10 Mathematics

Extramarks provides detailed NCERT Solutions for Class 10 Mathematics. Since the only way to get good at Mathematics is to solve problems, these solutions can be very useful for students. Extramarks provides detailed solutions to the questions given in all of the NCERT Mathematics textbooks chapters listed below:

• Chapter 1 – Real Numbers
• Chapter 2 – Polynomials
• Chapter 3 – Pair of Linear Equations in Two Variables
• Chapter 4 – Quadratic Equations
• Chapter 5 – Arithmetic Progressions
• Chapter 6 – Triangles
• Chapter 7 – Coordinate Geometry
• Chapter 8 – Introduction to Trigonometry
• Chapter 9 – Some Applications of Trigonometry
• Chapter 10 – Circles
• Chapter 11 – Constructions
• Chapter 12 – Areas Related to Circles
• Chapter 13 – Surface Areas and Volumes
• Chapter 14 – Statistics
• Chapter 15–Probability

The General Form of an Arithmetic Progression

The general form of an AP can be expressed as follows:

A, A + d, A + 2d, A + 3d,….

Here, A is the first term and d is the common difference.

 Position Of Terms Representation of terms Value of Terms 1 a1 A=a+(1-1) d 2 a2 A+d=a+(2-1) d 3 a3 A+2d=a+(3-1) d 4 a4 A+3d=a+(4-1) d . . . . N An A+(n-1)d

The Formulas

There are some essential formulae that students will need to solve problems related to Arithmetic Progressions. These formulae have been listed below:

• Arithmetic Progression’s nth Term (AP)

Students will face many problems related to this chapter where they will be expected to calculate the nth term of an Arithmetic Progression. The formula to calculate the nth term of an AP is shown below:

An = a + (n – 1) d

Here,

a= The initial term

d= The difference value

n= the number of terms

An= the nth term

Consider the following problem. Calculate the 15th term in the AP:  1, 2, 3, 4, 5 …

Here, we are supposed to calculate the 15th term. So, the value of n is 15. The first term, a, is 1 and the common difference is also 1. Using the formula to calculate the nth term of an AP, the 15th term can be calculated as follows:

A15 = 1 + (15 – 1) 1 = 1 + 14 = 15.

• The Sum of the First n Terms

Another common type of problem that students will face related to Arithmetic Progressions is to calculate the sum of the first n terms of a given AP. The formula to do so is shown below:

S = (n / 2)*(2a+(n−1)d)

Here,

n is the number of terms

a is the first term in the AP

d is the common difference

Here, using the formula the nth term of an AP, we can further simplify this expression to the one shown below:

S = (n/2)*(a+An)/2

Here,

An is the nth term of the AP

Also, if the nth term is actually the last term of the AP, then the formula is further simplified to the following:

S = n / 2 (first term + last term)

We’ve also included all of the crucial formulas in this chapter in a table for easy reference.

 General Form of AP A, a + d, a + 2d, a + 3d, a + 4d, …, a + nd The nth term of AP An = a + (n – 1) x d Sum of n terms in AP S= (n/2)*(2a + (n – 1)d) Sum of all terms in a finite AP with the last term as I (n/2)*(a + I)

### Why should students refer to NCERT Solutions from Extramarks?

Extramarks is committed to supporting students in every way. . We work to ensure that students can manage their academic and personal responsibilities more efficiently. The NCERT Solutions provides solutions for the Class 10 Chapter by Extramarks will help students cross-check their answers or even get the right answers to the textbook questions. Additionally, students will also get an idea of how to attempt a question in the right manner.

### The Benefits of Referring to NCERT Solutions Mathematics

NCERT Solutions for Class 10 Mathematics Chapter 5 from Extramarks is a useful resource that will definitely help students to step up their preparation. Here are some of the benefits:

• All the answers are written following the CBSE guidelines. When students study from it, they will get an edge over their peers.
• Subject matter experts write these solutions giving utmost importance to accuracy so that students are able to understand every concept and answer any question easily.
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Q.1 Write first four terms of the AP, when the first terma and the common difference d are given as follows:(i) a=10, d=10 (ii) a=2, d=0(iii) a=4, d=3 (iv) a=1, d=12(v) a=1.25, d=0.25

Ans.

$\begin{array}{l}\text{(i)}\\ \text{We have,}\\ \text{First term}=\text{a}=10\text{}\\ \text{and}\\ \text{Common difference}=\text{d}=10\\ \text{Now,}\\ \text{Second term}=\text{a}+\text{d}=\text{10}+10=20,\\ \text{Third term}=\text{a}+2\text{d}=\text{10}+2×10=30,\\ \text{Fourth term}=\text{a}+3\text{d}=\text{10}+3×10=40,\\ \text{Therefore, first four terms of the AP are:}\\ \text{10, 20, 30, 40}\\ \text{(ii)}\\ \text{We have,}\\ \text{First term}=\text{a}=-2\\ \text{and}\\ \text{Common difference}=\text{d}=0\\ \text{Now,}\\ \text{Second term}=\text{a}+\text{d}=-2+0=-2,\\ \text{Third term}=\text{a}+2\text{d}=-2+2×0=-2,\\ \text{Fourth term}=\text{a}+3\text{d}=-2+3×0=-2,\\ \text{Therefore, first four terms of the AP are:}\\ -2\text{,}-2\text{,}-2\text{,}-2\\ \text{(iii)}\\ \text{We have,}\\ \text{First term}=\text{a}=4\\ \text{and}\\ \text{Common difference}=\text{d}=-3\\ \text{Now,}\\ \text{Second term}=\text{a}+\text{d}=4-3=1,\\ \text{Third term}=\text{a}+2\text{d}=4+2×\left(-3\right)=-2,\\ \text{Fourth term}=\text{a}+3\text{d}=4+3×\left(-3\right)=-5,\\ \text{Therefore, first four terms of the AP are:}\\ 4\text{,}1\text{,\hspace{0.17em}\hspace{0.17em}}-2\text{,\hspace{0.17em}\hspace{0.17em}}-5\\ \\ \text{(iv)}\\ \text{We have,}\\ \text{First term}=\text{a}=-1\\ \text{and}\\ \text{Common difference}=\text{d}=\frac{1}{2}\\ \text{Now,}\\ \text{Second term}=\text{a}+\text{d}=-1+\frac{1}{2}=-\frac{1}{2}\\ \text{Third term}=\text{a}+2\text{d}=-1+2×\frac{1}{2}=0\\ \text{Fourth term}=\text{a}+3\text{d}=-1+3×\frac{1}{2}=\frac{1}{2}\\ \text{Therefore, first four terms of the AP are:}\\ -\text{1,}-\frac{1}{2},\text{0,}\frac{1}{2}\\ \text{(v)}\\ \text{We have,}\\ \text{First term}=\text{a}=-1.25\\ \text{and}\\ \text{Common difference}=\text{d}=-0.25\\ \text{Now,}\\ \text{Second term}=\text{a}+\text{d}=-1.25-0.25=-1.50,\\ \text{Third term}=\text{a}+2\text{d}=-1.25+2×\left(-0.25\right)=-1.75,\\ \text{Fourth term}=\text{a}+3\text{d}=-1.25+3×\left(-0.25\right)=-2,\\ \text{Therefore, first four terms of the AP are:}\\ -1.25\text{,}-1.50\text{,\hspace{0.17em}\hspace{0.17em}}-1.75\text{,\hspace{0.17em}\hspace{0.17em}}-2\end{array}$

Q.2 For the following APs, write the first term and thecommon difference:(i) 3,  1,1,3, . . . (ii)5,  1, 3, 7, . . .(iii) 13,  53, 93,  133,. . . (iv) 0.6, 1.7, 2.8, 3.9, . . .

Ans.

$\begin{array}{l}\text{(i)}\\ \text{Following is the given AP.}\\ \text{}3,\text{\hspace{0.17em}\hspace{0.17em}}1,\text{\hspace{0.17em}}-1,\text{\hspace{0.17em}}-3,\text{}.\text{}.\text{}.\text{}\\ \text{Here, first term}=3\\ \text{and}\\ \text{common difference}=\text{difference between a term and}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} its preceding term}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=1-3=-2\\ \text{(ii)}\\ \text{Following is the given AP.}\\ \text{}-5,\text{\hspace{0.17em}\hspace{0.17em}}-1,\text{\hspace{0.17em}}3,\text{\hspace{0.17em}}7,\text{}.\text{}.\text{}.\text{}\\ \text{Here, first term}=-5\\ \text{and}\\ \text{common difference}=\text{difference between a term and}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} its preceding term}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=-1-5=-6\\ \\ \text{(iii)}\\ \text{Following is the given AP.}\\ \text{}\frac{1}{3},\text{\hspace{0.17em}\hspace{0.17em}}\frac{5}{3},\text{}\frac{9}{3},\text{}\frac{13}{9},\text{}.\text{}.\text{}.\text{}\\ \text{Here, first term}=\frac{1}{3}\\ \text{and}\\ \text{common difference}=\text{difference between a term and}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} its preceding term}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{5}{3}-\frac{1}{3}=\frac{4}{3}\\ \\ \text{(iv)}\\ \text{Following is the given AP.}\\ \text{0.6, 1.7, 2.8, 3.9,}.\text{}.\text{}.\text{}\\ \text{Here, first term}=0.6\\ \text{and}\\ \text{common difference}=\text{difference between a term and}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} its preceding term}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=1.7-0.6=\text{1.1}\end{array}$

Q.3 Which of the following are APs? If they form an AP,find the common difference d and write three moreterms.(i) 2, 4, 8, 16, . . .(ii) 2, 52, 3, 72, . . .(iii) 1.2, 3.2, 5.2, 7.2, . . .(iv) 10, 6,2, 2, . . .(v) 3, 3+2,  3+22,  3+32,  . . .(vi) 0.2, 0.22, 0.222, 0.2222,  . . .(vii) 0, 4, 8, 12, . . .(viii) 12, 12, 12, 12, . . .(ix) 1, 3, 9, 27, ...(x) a, 2a, 3a, 4a, ...(xi) a, a2, a3, a4, ...(xii) 2, 8, 18, 32, ...(xiii) 3, 6, 9, 12, ...(xiv) 12, 32, 52, 72, ...(xv) 12, 52, 72, 73, ...

Ans.

$\begin{array}{l}\text{(i) 2, 4, 8, 16, . . .}\\ \text{Here, 4}-2=2\ne 8-4\text{}\\ \text{Therefore, the given list of numbers is not an AP.}\\ \text{(ii) 2,}\frac{\text{5}}{2}\text{, 3,}\frac{\text{7}}{2}\text{, . . .}\\ \text{We have,}\\ {\text{a}}_{2}-{\text{a}}_{1}=\frac{\text{5}}{2}-2=\frac{1}{2}\\ {\text{a}}_{3}-{\text{a}}_{2}=\text{3}-\frac{\text{5}}{2}=\frac{1}{2}\\ {\text{a}}_{4}-{\text{a}}_{3}=\frac{\text{7}}{2}-3=\frac{1}{2}\\ {\text{i.e., a}}_{\mathrm{k}+1}-{\text{a}}_{\mathrm{k}}\text{is the same every time.}\\ \text{Therefore, the given list of numbers is an AP with the common}\\ \text{difference}\mathrm{d}=\frac{1}{2}.\\ \text{The next three terms are:}\frac{\text{7}}{2}+\frac{1}{2}=4,\text{\hspace{0.17em}\hspace{0.17em}}4+\frac{1}{2}=\frac{9}{2}\text{and}\frac{9}{2}+\frac{1}{2}=5.\\ \text{(iii)}-\text{1.2,}-\text{3.2,}-\text{5.2,}-\text{7.2, . . .}\\ \text{We have,}\\ {\text{a}}_{2}-{\text{a}}_{1}=-\text{3.2}-\left(-\text{1.2}\right)=-\text{3.2}+\text{1.2}=-2\\ {\text{a}}_{3}-{\text{a}}_{2}=-\text{5.2}-\left(-3.\text{2}\right)=-\text{5.2}+3.\text{2}=-2\\ {\text{a}}_{4}-{\text{a}}_{3}=-7.\text{2}-\left(-5.\text{2}\right)=-7.\text{2}+5.\text{2}=-2\\ {\text{i.e., a}}_{\mathrm{k}+1}-{\text{a}}_{\mathrm{k}}\text{is the same every time.}\\ \text{Therefore, the given list of numbers is an AP with the common}\\ \text{difference}\mathrm{d}=-2.\\ \text{The next three terms are:}-\text{7.2}-2=-9.\text{2},\text{\hspace{0.17em}\hspace{0.17em}}-9.\text{2}-2=-11.\text{2}\\ \text{and}-11.\text{2}-2=-13.\text{2}.\\ \text{(iv)}-\text{10,}-\text{6,}-\text{2, 2, . . .}\\ \text{We have,}\\ {\text{a}}_{2}-{\text{a}}_{1}=-\text{6}-\left(-\text{10}\right)=-\text{6}+\text{10}=4\\ {\text{a}}_{3}-{\text{a}}_{2}=-\text{2}-\left(-6\right)=-\text{2}+6=4\\ {\text{a}}_{4}-{\text{a}}_{3}=\text{2}-\left(-\text{2}\right)=\text{2}+\text{2}=4\\ {\text{i.e., a}}_{\mathrm{k}+1}-{\text{a}}_{\mathrm{k}}\text{is the same every time.}\\ \text{Therefore, the given list of numbers is an AP with the common}\\ \text{difference}\mathrm{d}=4.\\ \text{The next three terms are: 2}+4=6,\text{\hspace{0.17em}\hspace{0.17em}}6+4=10\text{and}10+4=14.\\ \text{(v) 3,}3+\sqrt{2},\text{\hspace{0.17em}\hspace{0.17em}}3+2\sqrt{2},\text{\hspace{0.17em}\hspace{0.17em}}3+3\sqrt{2},\text{\hspace{0.17em}\hspace{0.17em}}.\text{}.\text{}.\\ \text{We have,}\\ {\text{a}}_{2}-{\text{a}}_{1}=3+\sqrt{2}-3=\sqrt{2}\\ {\text{a}}_{3}-{\text{a}}_{2}=3+2\sqrt{2}-\left(3+\sqrt{2}\right)=\sqrt{2}\\ {\text{a}}_{4}-{\text{a}}_{3}=3+3\sqrt{2}-\left(3+2\sqrt{2}\right)=\sqrt{2}\\ {\text{i.e., a}}_{\mathrm{k}+1}-{\text{a}}_{\mathrm{k}}\text{is the same every time.}\\ \text{Therefore, the given list of numbers is an AP with the common}\\ \text{difference}\mathrm{d}=\sqrt{2}.\\ \text{The next three terms are:}3+3\sqrt{2}+\sqrt{2}=3+4\sqrt{2},\text{\hspace{0.17em}}\\ \text{\hspace{0.17em}}3+4\sqrt{2}+\sqrt{2}=3+5\sqrt{2}\text{and}3+5\sqrt{2}+\sqrt{2}=3+6\sqrt{2}.\\ \text{(vi)}0.2,\text{}0.22,\text{}0.222,\text{}0.2222,\text{\hspace{0.17em}\hspace{0.17em}}.\text{}.\text{}.\\ \text{We have,}\\ {\text{a}}_{2}-{\text{a}}_{1}=0.22-0.2=0.02\\ {\text{a}}_{3}-{\text{a}}_{2}=0.222-0.22=0.002\\ {\text{So, a}}_{2}-{\text{a}}_{1}\ne {\text{a}}_{3}-{\text{a}}_{2}\\ \text{Therefore, the given list of numbers is not an AP.}\\ \text{(vii) 0,}-\text{4,}-\text{8,}-\text{12, . . .}\\ \text{We have,}\\ {\text{a}}_{2}-{\text{a}}_{1}=-4-\text{0}=-4\\ {\text{a}}_{3}-{\text{a}}_{2}=-8-\left(-4\right)=-8+4=-4\\ {\text{a}}_{4}-{\text{a}}_{3}=-1\text{2}-\left(-8\right)=-1\text{2}+\text{8}=-4\\ {\text{i.e., a}}_{\mathrm{k}+1}-{\text{a}}_{\mathrm{k}}\text{is the same every time.}\\ \text{Therefore, the given list of numbers is an AP with the common}\\ \text{difference}\mathrm{d}=-4.\\ \text{The next three terms are:}-1\text{2}-4=-16,\text{\hspace{0.17em}\hspace{0.17em}}-16-4=-20\\ \text{and}-20-4=-24.\\ \text{(viii)}-\frac{\text{1}}{2}\text{,}-\frac{\text{1}}{2}\text{,}-\frac{\text{1}}{2}\text{,}-\frac{\text{1}}{2}\text{,\hspace{0.17em}\hspace{0.17em}. . .}\\ {\text{Obviously, a}}_{\mathrm{k}+1}-{\text{a}}_{\mathrm{k}}\text{is the same every time.}\\ \text{Therefore, the given list of numbers is an AP with the common}\\ \text{difference}\mathrm{d}=0.\\ \text{The next three terms are:}-\frac{\text{1}}{2}\text{,}-\frac{\text{1}}{2}\text{,}-\frac{\text{1}}{2}.\\ \text{(ix) 1, 3, 9, 27,}...\\ \text{We have,}\\ {\text{a}}_{2}-{\text{a}}_{1}=3-1=2\\ {\text{a}}_{3}-{\text{a}}_{2}=9-3=6\\ {\text{So, a}}_{2}-{\text{a}}_{1}\ne {\text{a}}_{3}-{\text{a}}_{2}\\ \text{Therefore, the given list of numbers is not an AP.}\\ \text{(x) a, 2a, 3a, 4a,}...\\ \text{We have,}\\ {\text{a}}_{2}-{\text{a}}_{1}=2\text{a}-\text{a}=\text{a}\\ {\text{a}}_{3}-{\text{a}}_{2}=\text{3a}-2\text{a}=\text{a}\\ {\text{a}}_{4}-{\text{a}}_{3}=\text{4a}-\text{3a}=\text{a}\\ {\text{i.e., a}}_{\mathrm{k}+1}-{\text{a}}_{\mathrm{k}}\text{is the same every time.}\\ \text{Therefore, the given list of numbers is an AP with the common}\\ \text{difference}\mathrm{d}=\text{a}.\\ \text{The next three terms are:}4\text{a}+\text{a}=5\text{a},\text{\hspace{0.17em}\hspace{0.17em}}5\text{a}+\text{a}=6\text{a}\\ \text{and 6a}+\text{a}=7\text{a}.\\ {\text{(xi) a, a}}^{2},{\text{a}}^{3},{\text{a}}^{4},\text{}...\\ {\text{Obviously, a}}_{\mathrm{k}+1}-{\text{a}}_{\mathrm{k}}\text{is not the same every time.}\\ \text{Therefore, the given list of numbers is not an AP.}\\ \text{(xii)}\sqrt{2},\text{}\sqrt{8},\text{}\sqrt{18},\text{}\sqrt{32},\text{}...\\ \text{We have,}\\ {\text{a}}_{2}-{\text{a}}_{1}=\sqrt{8}-\sqrt{2}=2\sqrt{2}-\sqrt{2}=\sqrt{2}\\ {\text{a}}_{3}-{\text{a}}_{2}=\sqrt{18}-\sqrt{8}=3\sqrt{2}-2\sqrt{2}=\sqrt{2}\\ {\text{a}}_{4}-{\text{a}}_{3}=\sqrt{32}-\sqrt{18}=4\sqrt{2}-3\sqrt{2}=\sqrt{2}\\ {\text{i.e., a}}_{\mathrm{k}+1}-{\text{a}}_{\mathrm{k}}\text{is the same every time.}\\ \text{Therefore, the given list of numbers is an AP with the common}\\ \text{difference}\mathrm{d}=\sqrt{2}.\\ \text{The next three terms are:}\sqrt{32}+\sqrt{2}=4\sqrt{2}+\sqrt{2}=5\sqrt{2}=\sqrt{50},\text{\hspace{0.17em}}\\ \text{\hspace{0.17em}}\sqrt{50}+\sqrt{2}=5\sqrt{2}+\sqrt{2}=6\sqrt{2}=\sqrt{72}\text{}\\ \text{and}\sqrt{72}+\sqrt{2}=6\sqrt{2}+\sqrt{2}=7\sqrt{2}=\sqrt{98}.\\ \text{(xiii)}\sqrt{3},\text{}\sqrt{6},\text{}\sqrt{9},\text{}\sqrt{12},\text{}...\\ \text{We have,}\\ {\text{a}}_{2}-{\text{a}}_{1}=\sqrt{6}-\sqrt{3}=\sqrt{2}×\sqrt{3}-\sqrt{3}=\sqrt{3}\left(\sqrt{2}-1\right)\\ {\text{a}}_{3}-{\text{a}}_{2}=\sqrt{9}-\sqrt{6}=\sqrt{3}×\sqrt{3}-\sqrt{2}×\sqrt{3}=\sqrt{3}\left(\sqrt{3}-\sqrt{2}\right)\\ {\text{So, a}}_{2}-{\text{a}}_{1}\ne {\text{a}}_{3}-{\text{a}}_{2}\\ \text{Therefore, the given list of numbers is not an AP.}\\ {\text{(xiv) 1}}^{2}{\text{, 3}}^{2},{\text{5}}^{2},{\text{7}}^{2},\text{}...\\ \text{We have,}\\ {\text{a}}_{2}-{\text{a}}_{1}={\text{3}}^{2}-{\text{1}}^{2}=9-1=8\\ {\text{a}}_{3}-{\text{a}}_{2}={\text{5}}^{2}-{\text{3}}^{2}=25-9=16\\ {\text{So, a}}_{2}-{\text{a}}_{1}\ne {\text{a}}_{3}-{\text{a}}_{2}\\ \text{Therefore, the given list of numbers is not an AP.}\\ {\text{(xv) 1}}^{2}{\text{, 5}}^{2},{\text{7}}^{2},\text{73},\text{}...\\ \text{We have,}\\ {\text{a}}_{2}-{\text{a}}_{1}={5}^{2}-{\text{1}}^{2}=25-1=24\\ {\text{a}}_{3}-{\text{a}}_{2}={\text{7}}^{2}-{\text{5}}^{2}=49-25=24\\ {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}a}}_{4}-{\text{a}}_{3}=\text{73}-{\text{7}}^{2}=73-49=24\\ {\text{i.e., a}}_{\mathrm{k}+1}-{\text{a}}_{\mathrm{k}}\text{is the same every time.}\\ \text{Therefore, the given list of numbers is an AP with the common}\\ \text{difference}\mathrm{d}=24.\\ \text{The next three terms are: 73}+24=97,\text{\hspace{0.17em}}97+24=121\\ \text{and}121+24=145.\end{array}$

Q.4 Fill in the blanks in the following table, given that a is the first term, d the common difference and an the nth term of the AP: Ans.

$\begin{array}{l}\text{(i)}\\ \text{We have}\\ \mathrm{a}=7,\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{d}=3,\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{n}=8\\ \text{We know that}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{a}}_{\mathrm{n}}=\mathrm{a}+\left(\mathrm{n}-1\right)\mathrm{d}\\ ⇒{\mathrm{a}}_{\mathrm{n}}=7+\left(8-1\right)3=28\\ \text{(ii)}\\ \text{We have}\\ \mathrm{a}=-18,\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{n}=10,\text{}{\mathrm{a}}_{\mathrm{n}}=0\\ \text{We know that}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{a}}_{\mathrm{n}}=\mathrm{a}+\left(\mathrm{n}-1\right)\mathrm{d}\\ ⇒0=-18+\left(10-1\right)\mathrm{d}\\ ⇒\mathrm{d}=\frac{18}{9}=2\\ \text{(iii)}\\ \text{We have}\\ \mathrm{d}=-3,\text{}\mathrm{n}=18,\text{}{\mathrm{a}}_{\mathrm{n}}=-5\\ \text{We know that}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{a}}_{\mathrm{n}}=\mathrm{a}+\left(\mathrm{n}-1\right)\mathrm{d}\\ ⇒-5=\mathrm{a}+\left(18-1\right)\left(-3\right)\\ ⇒\mathrm{a}=51-5=46\\ \text{(iv)}\\ \text{We have}\\ \mathrm{a}=-18.9,\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{d}=2.5,\text{}{\mathrm{a}}_{\mathrm{n}}=3.6\\ \text{We know that}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{a}}_{\mathrm{n}}=\mathrm{a}+\left(\mathrm{n}-1\right)\mathrm{d}\\ ⇒3.6=-18.9+\left(\mathrm{n}-1\right)\left(2.5\right)\\ ⇒\left(\mathrm{n}-1\right)\left(2.5\right)=3.6+18.9=22.5\\ ⇒\mathrm{n}=\frac{22.5}{2.5}+1=10\\ \text{(v)}\\ \text{We have}\\ \mathrm{a}=3.5,\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{d}=0,\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{n}=105\\ \text{We know that}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{a}}_{\mathrm{n}}=\mathrm{a}+\left(\mathrm{n}-1\right)\mathrm{d}\\ ⇒{\mathrm{a}}_{\mathrm{n}}=3.5+\left(105-1\right)×0=3.5\end{array}$

Q.5 Choose the correct choice in the following and justify:(i) 30th term of the AP: 10, 7, 4, . . . , is (A) 97 (B) 77 (C) 77 (D) 87(ii) 11th term of the AP: 3, 12, 2, . . ., is (A) 28 (B) 22 (C) 38 (D) 4812

Ans.

$\begin{array}{l}\text{(i)}\\ \text{The given AP is:}\\ \text{10, 7, 4, . . .}\\ \text{Here,}\mathrm{a}=10,\text{}\mathrm{d}=7-10=-3,\text{}\mathrm{n}=30\\ \text{We know that}\\ \text{}{\mathrm{a}}_{\mathrm{n}}=\mathrm{a}+\left(\mathrm{n}-1\right)\mathrm{d}\\ ⇒\text{}{\mathrm{a}}_{30}=10+\left(30-1\right)\left(-3\right)=10-87=-77\\ \text{Hence, the correct answer is (C).}\\ \text{(ii)}\\ \text{The given AP is:}\\ \text{}-\text{3,}\frac{-\text{1}}{2}\text{, 2, . . .}\\ \text{Here,}\mathrm{a}=-3,\text{}\mathrm{d}=2-\frac{-1}{2}=\frac{5}{2},\text{}\mathrm{n}=11\\ \text{We know that}\\ \text{}{\mathrm{a}}_{\mathrm{n}}=\mathrm{a}+\left(\mathrm{n}-1\right)\mathrm{d}\\ ⇒\text{}{\mathrm{a}}_{11}=-3+\left(11-1\right)×\frac{5}{2}=-3+25=22\\ \text{Hence, the correct answer is (B).}\end{array}$

Q.6 InthefollowingAPs,findthemissingtermsintheboxes:i 2,,26ii ,13,,3iii 5, , ,912iv4,, ,, ,6v ,38, , , ,22

Ans.

$\begin{array}{l}\left(\mathrm{i}\right)\\ \mathrm{The}\text{\hspace{0.17em}}\mathrm{given}\text{\hspace{0.17em}}\mathrm{AP}\text{\hspace{0.17em}}\mathrm{is}:\\ 2,\overline{)},\text{\hspace{0.17em}}26\\ \mathrm{Let}\text{\hspace{0.17em}}\mathrm{the}\text{\hspace{0.17em}}\mathrm{missing}\text{\hspace{0.17em}}\mathrm{term}\text{\hspace{0.17em}}\mathrm{is}\text{\hspace{0.17em}}‘\mathrm{t}‘.\text{\hspace{0.17em}}\mathrm{Then},\text{\hspace{0.17em}}\mathrm{we}\text{\hspace{0.17em}}\mathrm{have}\\ \mathrm{t}-2=26-\mathrm{t}\\ ⇒2\mathrm{t}=26+2=28\\ ⇒\mathrm{t}=\frac{28}{2}=14\\ \left(\mathrm{ii}\right)\\ \mathrm{The}\text{\hspace{0.17em}}\mathrm{given}\text{\hspace{0.17em}}\mathrm{AP}\text{\hspace{0.17em}}\mathrm{is}:\\ \overline{)},\text{\hspace{0.17em}}13,\overline{)},\text{\hspace{0.17em}}3\\ \mathrm{Let}\text{\hspace{0.17em}}\mathrm{the}\text{\hspace{0.17em}}\mathrm{common}\text{\hspace{0.17em}}\mathrm{difference}\text{\hspace{0.17em}}\mathrm{is}\text{\hspace{0.17em}}‘\mathrm{d}‘.\text{\hspace{0.17em}}\mathrm{Then},\text{\hspace{0.17em}}\mathrm{we}\text{\hspace{0.17em}}\mathrm{have}\\ {\mathrm{a}}_{4}-{\mathrm{a}}_{2}=2\mathrm{d}=3-13\\ ⇒2\mathrm{d}=-10\\ ⇒\mathrm{d}=\frac{-10}{2}=-5\\ \mathrm{Therefore},\\ \mathrm{first}\text{\hspace{0.17em}}\mathrm{term}={\mathrm{a}}_{1}={\mathrm{a}}_{2}-\mathrm{d}=13-\left(-5\right)=13+5=18\\ \mathrm{third}\text{\hspace{0.17em}}\mathrm{term}={\mathrm{a}}_{3}={\mathrm{a}}_{4}-\mathrm{d}=3-\left(-5\right)=3+5=8\\ \mathrm{Hence},\text{\hspace{0.17em}}\mathrm{the}\text{\hspace{0.17em}}\mathrm{given}\text{\hspace{0.17em}}\mathrm{APis}:\\ \overline{)18},13,\overline{)8},3\\ \left(\mathrm{iii}\right)\\ \mathrm{The}\text{\hspace{0.17em}}\mathrm{given}\text{\hspace{0.17em}}\mathrm{AP}\text{\hspace{0.17em}}\mathrm{is}:\\ 5,\text{\hspace{0.17em}}\overline{)},\text{\hspace{0.17em}}\overline{)},\text{\hspace{0.17em}}9\frac{1}{2}\\ \mathrm{Let}\text{\hspace{0.17em}}\mathrm{the}\text{\hspace{0.17em}}\mathrm{common}\text{\hspace{0.17em}}\mathrm{difference}\text{\hspace{0.17em}}\mathrm{is}\text{\hspace{0.17em}}‘\mathrm{d}‘.\text{\hspace{0.17em}}\mathrm{Then},\text{\hspace{0.17em}}\mathrm{we}\text{\hspace{0.17em}}\mathrm{have}\\ {\mathrm{a}}_{2}={\mathrm{a}}_{1}+\mathrm{d},\\ {\mathrm{a}}_{3}={\mathrm{a}}_{2}+\mathrm{d}={\mathrm{a}}_{1}+\mathrm{d}+\mathrm{d}={\mathrm{a}}_{1}+2\mathrm{d},\\ {\mathrm{a}}_{4}={\mathrm{a}}_{3}+\mathrm{d}={\mathrm{a}}_{1}+2\mathrm{d}+\mathrm{d}={\mathrm{a}}_{1}+3\mathrm{d}\\ ⇒\frac{19}{2}=5+3\mathrm{d}\\ ⇒3\mathrm{d}=\frac{19}{2}-5=\frac{9}{2}\\ ⇒\mathrm{d}=\frac{9}{2}×\frac{1}{3}=\frac{3}{2}\\ \mathrm{Therefore},\text{\hspace{0.17em}}\mathrm{missing}\text{\hspace{0.17em}}\mathrm{terms}\text{\hspace{0.17em}}\mathrm{of}\text{\hspace{0.17em}}\mathrm{the}\text{\hspace{0.17em}}\mathrm{given}\text{\hspace{0.17em}}\mathrm{AP}\text{\hspace{0.17em}}\mathrm{are}:\\ {\mathrm{a}}_{2}={\mathrm{a}}_{1}+\mathrm{d}=5+\frac{3}{2}=\frac{13}{2},\\ {\mathrm{a}}_{3}={\mathrm{a}}_{2}+\mathrm{d}=\frac{13}{2}+\frac{3}{2}=\frac{16}{2}=8\\ \mathrm{Hence},\text{\hspace{0.17em}}\mathrm{the}\text{\hspace{0.17em}}\mathrm{given}\text{\hspace{0.17em}}\mathrm{AP}\text{\hspace{0.17em}}\mathrm{is}:\\ 5,\overline{)\frac{13}{2}},\overline{)8},9\frac{1}{2}\\ \left(\mathrm{iv}\right)\\ \mathrm{The}\text{\hspace{0.17em}}\mathrm{given}\text{\hspace{0.17em}}\mathrm{AP}\text{\hspace{0.17em}}\mathrm{is}:\\ -4,\text{\hspace{0.17em}}\overline{)},\text{\hspace{0.17em}}\overline{)},\text{\hspace{0.17em}}\overline{)},\text{\hspace{0.17em}}\overline{)},\text{\hspace{0.17em}}6\\ \mathrm{Let}\text{\hspace{0.17em}}\mathrm{the}\text{\hspace{0.17em}}\mathrm{common}\text{\hspace{0.17em}}\mathrm{difference}\text{\hspace{0.17em}}\mathrm{is}\text{\hspace{0.17em}}‘\mathrm{d}‘.\text{\hspace{0.17em}}\mathrm{Then},\text{\hspace{0.17em}}\mathrm{we}\text{\hspace{0.17em}}\mathrm{have}\\ {\mathrm{a}}_{6}={\mathrm{a}}_{1}+5\mathrm{d}\\ ⇒6=-4+5\mathrm{d}\\ ⇒5\mathrm{d}=6+4=10\\ ⇒\mathrm{d}=\frac{10}{5}=2\\ \mathrm{Therefore},\text{\hspace{0.17em}}\mathrm{missing}\text{\hspace{0.17em}}\mathrm{terms}\text{\hspace{0.17em}}\mathrm{of}\text{\hspace{0.17em}}\mathrm{the}\text{\hspace{0.17em}}\mathrm{given}\text{\hspace{0.17em}}\mathrm{AP}\text{\hspace{0.17em}}\mathrm{are}:\\ {\mathrm{a}}_{2}={\mathrm{a}}_{1}+\mathrm{d}=-4+2=-2\\ {\mathrm{a}}_{3}={\mathrm{a}}_{1}+2\mathrm{d}=-4+2×2=0,\\ {\mathrm{a}}_{4}={\mathrm{a}}_{1}+3\mathrm{d}=-4+3×2=2,\\ {\mathrm{a}}_{5}={\mathrm{a}}_{1}+4\mathrm{d}=-4+4×2=4\\ \mathrm{Hence},\text{\hspace{0.17em}}\mathrm{the}\text{\hspace{0.17em}}\mathrm{given}\text{\hspace{0.17em}}\mathrm{AP}\text{\hspace{0.17em}}\mathrm{is}:\\ -4,\overline{)-2},\overline{)0},\overline{)2},\overline{)4},6\\ \left(\mathrm{v}\right)\\ \mathrm{The}\text{\hspace{0.17em}}\mathrm{given}\text{\hspace{0.17em}}\mathrm{AP}\text{\hspace{0.17em}}\mathrm{is}:\\ \overline{)},\text{\hspace{0.17em}}38,\overline{)},\text{\hspace{0.17em}}\overline{)},\text{\hspace{0.17em}}\overline{)},\text{\hspace{0.17em}}-22\\ \mathrm{Let}\text{\hspace{0.17em}}\mathrm{the}\text{\hspace{0.17em}}\mathrm{common}\text{\hspace{0.17em}}\mathrm{difference}\text{\hspace{0.17em}}\mathrm{is}\text{\hspace{0.17em}}‘\mathrm{d}‘.\text{\hspace{0.17em}}\mathrm{Then},\text{\hspace{0.17em}}\mathrm{we}\text{\hspace{0.17em}}\mathrm{have}\\ {\mathrm{a}}_{6}={\mathrm{a}}_{1}+5\mathrm{d}\\ ⇒{\mathrm{a}}_{6}={\mathrm{a}}_{1}+\mathrm{d}+4\mathrm{d}={\mathrm{a}}_{2}+4\mathrm{d}\\ ⇒-22=38+4\mathrm{d}\\ ⇒4\mathrm{d}=-22-38=-60\\ ⇒\mathrm{d}=\frac{-60}{4}=-15\\ \mathrm{Therefore},\text{\hspace{0.17em}}\mathrm{missing}\text{\hspace{0.17em}}\mathrm{terms}\text{\hspace{0.17em}}\mathrm{of}\text{\hspace{0.17em}}\mathrm{the}\text{\hspace{0.17em}}\mathrm{given}\text{\hspace{0.17em}}\mathrm{AP}\text{\hspace{0.17em}}\mathrm{are}:\\ {\mathrm{a}}_{1}={\mathrm{a}}_{2}+\mathrm{d}=38-\left(-15\right)=38+15=53,\\ {\mathrm{a}}_{3}={\mathrm{a}}_{1}+2\mathrm{d}=53+2×\left(-15\right)=23,\\ {\mathrm{a}}_{4}={\mathrm{a}}_{1}+3\mathrm{d}=53+3×\left(-15\right)=8\\ {\mathrm{a}}_{5}={\mathrm{a}}_{1}+4\mathrm{d}=53+4×\left(-15\right)=-7\\ \mathrm{Hence},\text{\hspace{0.17em}}\mathrm{the}\text{\hspace{0.17em}}\mathrm{given}\text{\hspace{0.17em}}\mathrm{AP}\text{\hspace{0.17em}}\mathrm{is}:\\ \overline{)53},38,\overline{)23},\overline{)8},\overline{)-7},-22\end{array}$

Q.7 Which term of the AP: 3, 8, 13, 18, . . . , is 78?

Ans.

$\begin{array}{l}\text{The given AP is:}\\ \text{3, 8, 13, 18, . . .}\\ {\text{Let n}}^{\text{th}}\text{term of the given AP is 78.}\\ \text{We know that}\\ \text{}{\mathrm{a}}_{\mathrm{n}}=\mathrm{a}+\left(\mathrm{n}-1\right)\mathrm{d}\\ ⇒78=3+\left(\mathrm{n}-1\right)5\\ ⇒\left(\mathrm{n}-1\right)5=78-3=75\\ ⇒\mathrm{n}-1=\frac{75}{5}=15\\ ⇒\mathrm{n}=15+1=16\\ {\text{Therefore, 16}}^{\text{th}}\text{term of the given AP is 78.}\end{array}$

Q.8 Find the number of terms in each of the following APs:(i) 7, 13, 19, . . . , 205 (ii) 18, 1512, 13, . . . , 47

Ans.

$\begin{array}{l}\text{(i) 7, 13, 19, . . . , 205}\\ \text{Let the given AP has}\mathrm{n}\text{terms.}\\ \text{We know that}\\ \text{}{\mathrm{a}}_{\mathrm{n}}=\mathrm{a}+\left(\mathrm{n}-1\right)\mathrm{d}\\ ⇒\text{205}=7+\left(\mathrm{n}-1\right)6\\ ⇒\left(\mathrm{n}-1\right)6=205-7=198\\ ⇒\mathrm{n}-1=\frac{198}{6}=33\\ ⇒\mathrm{n}=33+1=34\\ \text{(ii) 18, 15}\frac{1}{2},\text{13, . . . ,}-\text{47}\\ \text{Let the given AP has}\mathrm{n}\text{terms.}\\ \text{We know that}\\ \text{}{\mathrm{a}}_{\mathrm{n}}=\mathrm{a}+\left(\mathrm{n}-1\right)\mathrm{d}\\ ⇒-\text{47}=18+\left(\mathrm{n}-1\right)\left(\frac{31}{2}-18\right)=18+\left(\mathrm{n}-1\right)\left(\frac{-5}{2}\right)\\ ⇒\left(\mathrm{n}-1\right)\left(\frac{-5}{2}\right)=-47-18=-65\\ ⇒\mathrm{n}-1=-65×\frac{-2}{5}=26\\ ⇒\mathrm{n}=26+1=27\end{array}$

Q.9 Check whether –150 is a term of the AP: 11, 8, 5, 2, . . .

Ans.

$\begin{array}{l}\text{The given AP is \hspace{0.17em}}\mathbf{11},\text{}\mathbf{8},\text{}\mathbf{5},\text{}\mathbf{2},\text{}.\text{}.\text{}.\\ {\text{Let n}}^{\mathrm{th}}\text{term of the given AP is}-\text{150.}\\ \text{We know that}\\ \text{}{\mathrm{a}}_{\mathrm{n}}=\mathrm{a}+\left(\mathrm{n}-1\right)\mathrm{d}\\ ⇒-\text{150}=11+\left(\mathrm{n}-1\right)\left(8-11\right)=11+\left(\mathrm{n}-1\right)\left(-3\right)\\ ⇒\left(\mathrm{n}-1\right)\left(-3\right)=-150-11=-161\\ ⇒\mathrm{n}-1=-161×\frac{-1}{3}=53\frac{2}{3}\\ ⇒\mathrm{n}=53\frac{2}{3}+1=54\frac{2}{3}\\ \text{But}54\frac{2}{3}\text{is not a natural number. Therefore,}-150\text{is not}\\ \text{a term of the given AP.}\end{array}$

Q.10 Find the 31st term of an AP whose 11th term is 38 and the 16th term is 73.

Ans.

$\begin{array}{l}\text{We have}\\ {\text{a}}_{11}=38\text{and}{\mathrm{a}}_{16}=73\\ {\text{Or a}}_{11}={\mathrm{a}}_{1}+10\mathrm{d}=38\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\text{(1)}\\ \text{and\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{a}}_{16}={\mathrm{a}}_{1}+15\mathrm{d}=73\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\text{(2)}\\ \text{Subtracting (1) from (2), we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}5\mathrm{d}=35\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{d}=7\\ \text{On putting this value of}\mathrm{d}\text{in (1), we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{a}}_{1}+10\mathrm{d}=38\\ ⇒\text{}{\mathrm{a}}_{1}+10×7=38\\ ⇒\text{}{\mathrm{a}}_{1}=38-70=-32\\ \text{Now,}\\ {\mathrm{a}}_{31}={\mathrm{a}}_{1}+30\mathrm{d}=-32+30×7=-32+210=178\end{array}$

Q.11 An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.

Ans.

$\begin{array}{l}\text{We have}\\ {\text{a}}_{3}=12\text{and}{\mathrm{a}}_{50}=106\\ {\text{Or a}}_{3}={\mathrm{a}}_{1}+2\mathrm{d}=12\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\text{(1)}\\ \text{and\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{a}}_{50}={\mathrm{a}}_{1}+49\mathrm{d}=106\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\text{(2)}\\ \text{Subtracting (1) from (2), we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}47\mathrm{d}=94\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{d}=\frac{94}{47}=2\\ \text{On putting this value of}\mathrm{d}\text{in (1), we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{a}}_{1}+2×2=12\\ ⇒\text{}{\mathrm{a}}_{1}=12-4=8\\ \text{Now,}\\ {\mathrm{a}}_{29}={\mathrm{a}}_{1}+28\mathrm{d}=8+28×2=8+56=64\end{array}$

Q.12 If the 3rd and the 9th terms of an AP are 4 and –8 respectively, which term of this AP is zero?

Ans.

$\begin{array}{l}\text{We have}\\ {\text{a}}_{3}=4\text{and}{\mathrm{a}}_{9}=-8\\ {\text{Or a}}_{3}={\mathrm{a}}_{1}+2\mathrm{d}=4\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\text{(1)}\\ \text{and\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{a}}_{9}={\mathrm{a}}_{1}+8\mathrm{d}=-8\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\text{(2)}\\ \text{Subtracting (1) from (2), we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}6\mathrm{d}=-12\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{d}=\frac{-12}{6}=-2\\ \text{On putting this value of}\mathrm{d}\text{in (1), we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{a}}_{1}+2×\left(-2\right)=4\\ ⇒\text{}{\mathrm{a}}_{1}=4+4=8\\ {\text{Let n}}^{\text{th}}\text{term of the given AP is zero. Then, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{a}}_{\mathrm{n}}={\mathrm{a}}_{1}+\left(\mathrm{n}-1\right)\mathrm{d}\\ ⇒0=8+\left(\mathrm{n}-1\right)\left(-2\right)\\ ⇒\mathrm{n}-1=\frac{-8}{-2}=4\\ ⇒\mathrm{n}=4+1=5\\ \text{Hence, 5th term of the given AP is zero.}\end{array}$

Q.13 The 17th term of an AP exceeds its 10th term by 7. Find the common difference.

Ans.

$\begin{array}{l}\text{We have}\\ {\text{a}}_{17}={\text{a}}_{10}+7\text{}\\ ⇒{\text{a}}_{17}-{\mathrm{a}}_{10}=7\\ ⇒{\mathrm{a}}_{1}+16\mathrm{d}-\left({\mathrm{a}}_{1}+9\mathrm{d}\right)=7\text{[}{\mathrm{a}}_{\mathrm{n}}=\mathrm{a}+\left(\mathrm{n}-1\right)\mathrm{d}\right]\\ ⇒7\mathrm{d}=7\\ ⇒\mathrm{d}=1\\ \text{Therefore, the common difference is 1.}\end{array}$

Q.14 Which term of the AP: 3, 15, 27, 39, . . . will be 132 more than its 54th term?

Ans.

$\begin{array}{l}\text{Let}{\mathrm{n}}^{\mathrm{th}}{\text{term be 132 more than 54}}^{\mathrm{th}}\text{term of the given AP.}\\ \text{We have}\\ {\text{a}}_{\mathrm{n}}={\text{a}}_{54}+132\text{}\\ ⇒{\text{a}}_{\mathrm{n}}-{\mathrm{a}}_{54}=132\\ ⇒{\mathrm{a}}_{1}+\left(\mathrm{n}-1\right)\mathrm{d}-\left({\mathrm{a}}_{1}+53\mathrm{d}\right)=132\text{[}{\mathrm{a}}_{\mathrm{n}}=\mathrm{a}+\left(\mathrm{n}-1\right)\mathrm{d}\right]\\ ⇒\left(\mathrm{n}-54\right)\mathrm{d}=132\\ ⇒\mathrm{n}-54=\frac{132}{\mathrm{d}}=\frac{132}{12}\text{[The given AP is: 3, 15, 27, 39,}...\text{]}\\ ⇒\mathrm{n}=11+54=65\\ \text{Therefore,}{65}^{\mathrm{th}}{\text{term is 132 more than 54}}^{\mathrm{th}}\text{term of the given AP.}\end{array}$

Q.15 Two APs have the same common difference. The difference between their 100th terms is 100, what is the difference between their 1000th terms?

Ans.

$\begin{array}{l}\text{It is given that two APs have the same common difference.}\\ \text{Let}{\mathrm{a}}_{\mathrm{n}}\text{and}{\mathrm{b}}_{\mathrm{n}}\text{represent the terms of the first and second}\\ \text{AP respectively for all natural numbers}\mathrm{n}\text{.}\\ \text{We have}\\ \text{}{\mathrm{a}}_{100}-{\mathrm{b}}_{100}=100\\ ⇒{\text{a}}_{1}+99\mathrm{d}-{\mathrm{b}}_{1}-99\mathrm{d}=100\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}[}{\mathrm{a}}_{\mathrm{n}}=\mathrm{a}+\left(\mathrm{n}-1\right)\mathrm{d}\right]\\ ⇒{\mathrm{a}}_{1}-{\mathrm{b}}_{1}=100\text{}\\ \text{Now,}\\ {\mathrm{a}}_{1000}-{\mathrm{b}}_{1000}={\mathrm{a}}_{1}+999\mathrm{d}-{\mathrm{b}}_{1}-999\mathrm{d}={\mathrm{a}}_{1}-{\mathrm{b}}_{1}=100\text{}\\ \text{Therefore, the difference between 1000th terms of the two}\\ \text{APs is 100.}\end{array}$

Q.16 How many three-digit numbers are divisible by 7?

Ans.

$\begin{array}{l}\text{The smallest three-digit number divisible by 7 is 105.}\\ \text{The largest three-digit number divisible by 7 is 994.}\\ \text{Three-digit numbers divisible by 7 are:}\\ \text{105, 112, 119,}...\text{,994}\\ \text{These multiples of 7 form an AP with common difference 7,}\\ \text{first term 105 and last term 994. Thus, we have}\\ {\mathrm{a}}_{1}=105,\text{}{\mathrm{a}}_{\mathrm{n}}=994\text{and}\mathrm{d}=7.\\ \text{We know that}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{a}}_{\mathrm{n}}={\mathrm{a}}_{1}+\left(\mathrm{n}-1\right)\mathrm{d}\\ ⇒994=105+\left(\mathrm{n}-1\right)7\\ ⇒\left(\mathrm{n}-1\right)7=994-105=889\text{}\\ ⇒\mathrm{n}-1=\frac{889}{7}=127\\ ⇒\mathrm{n}=127+1=128\text{}\\ \text{Therefore, 128 three-digit numbers are divisible by 7.}\end{array}$

Q.17 How many multiples of 4 lie between 10 and 250?

Ans.

$\begin{array}{l}\text{Multiples of 4 between 10 and 250 are:}\\ \text{12, 16, 20,}...\text{,248}\\ \text{These multiples of 4 form an AP with common difference 4,}\\ \text{first term 12 and last term 248. Thus, we have}\\ {\mathrm{a}}_{1}=12,\text{}{\mathrm{a}}_{\mathrm{n}}=248\text{and}\mathrm{d}=4.\\ \text{We know that}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{a}}_{\mathrm{n}}={\mathrm{a}}_{1}+\left(\mathrm{n}-1\right)\mathrm{d}\\ ⇒248=12+\left(\mathrm{n}-1\right)4\\ ⇒\left(\mathrm{n}-1\right)4=248-12=236\\ ⇒\mathrm{n}-1=\frac{236}{4}=59\\ ⇒\mathrm{n}=59+1=60\\ \text{Therefore, 60}\mathbf{multiples}\text{}\mathbf{of}\text{}\mathbf{4}\text{}\mathbf{lie}\text{}\mathbf{between}\text{}\mathbf{10}\text{}\mathbf{and}\text{}\mathbf{250}.\end{array}$

Q.18 For what value of n, are the nth terms of two APs: 63, 65, 67, . . . and 3, 10, 17, . . . equal?

Ans.

$\begin{array}{l}\text{Let}\mathrm{n}\text{th term of the given APs are equal.}\\ \text{Let}{\mathrm{a}}_{\mathrm{n}}\text{and}{\mathrm{b}}_{\mathrm{n}}\text{represent the terms of the first and second}\\ \text{AP respectively for all natural numbers}\mathrm{n}\text{. Also, let}{\mathrm{d}}_{1}\text{and}{\mathrm{d}}_{2}\text{are}\\ \text{common diffrences of first and second AP respectively.}\\ \text{The given two APs are:}\\ \mathbf{63},\text{}\mathbf{65},\text{}\mathbf{67},\text{}.\text{}.\text{}.\text{}\mathbf{and}\text{}\mathbf{3},\text{}\mathbf{10},\text{}\mathbf{17},\text{}.\text{}.\text{}.\\ \text{We have}\\ \text{}{\mathrm{a}}_{\mathrm{n}}={\mathrm{b}}_{\mathrm{n}}\\ ⇒\text{}{\mathrm{a}}_{1}+\left(\mathrm{n}-1\right){\mathrm{d}}_{1}={\mathrm{b}}_{1}+\left(\mathrm{n}-1\right){\mathrm{d}}_{2}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}[}{\mathrm{a}}_{\mathrm{n}}=\mathrm{a}+\left(\mathrm{n}-1\right)\mathrm{d}\right]\\ ⇒63+\left(\mathrm{n}-1\right)2=3+\left(\mathrm{n}-1\right)7\\ ⇒\left(\mathrm{n}-1\right)2-\left(\mathrm{n}-1\right)7=3-63\\ ⇒-5\left(\mathrm{n}-1\right)=-60\\ ⇒\mathrm{n}-1=\frac{-60}{-5}=12\\ ⇒\mathrm{n}=12+1=13\\ \text{Therefore, 13th terms of the given APs are equal.}\end{array}$

Q.19 Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12.

Ans.

$\begin{array}{l}\text{We have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{a}}_{3}=16\text{and}{\mathrm{a}}_{7}={\mathrm{a}}_{5}+12\\ ⇒{\mathrm{a}}_{1}+2\mathrm{d}=16\text{and}{\mathrm{a}}_{7}-{\mathrm{a}}_{5}=12\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}[}{\mathrm{a}}_{\mathrm{n}}=\mathrm{a}+\left(\mathrm{n}-1\right)\mathrm{d}\right]\\ ⇒{\mathrm{a}}_{1}+2\mathrm{d}=16\text{and}{\mathrm{a}}_{1}+6\mathrm{d}-{\mathrm{a}}_{1}-4\mathrm{d}=12\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\\ ⇒{\mathrm{a}}_{1}+2\mathrm{d}=16\text{and}2\mathrm{d}=12\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\\ ⇒{\mathrm{a}}_{1}+2\mathrm{d}=16\text{and}\mathrm{d}=6\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\\ ⇒{\mathrm{a}}_{1}+2×6=16\text{and}\mathrm{d}=6\text{\hspace{0.17em}}\\ ⇒{\mathrm{a}}_{1}=4\text{and}\mathrm{d}=6\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}\\ \text{Therefore, the required AP is: 4, 10, 16, 22,}...\end{array}$

Q.20 Find the 20th term from the last term of the AP: 3, 8, 13, . . ., 253.

Ans.

$\begin{array}{l}\text{Here,}\mathrm{a}=3,\text{}\mathrm{d}=8-3=5,\text{}\mathrm{l}=253\text{where}\mathrm{l}=\mathrm{a}+\left(\mathrm{n}-1\right)\mathrm{d}\\ \text{To find the 20th term from the last term, we will find the}\\ \text{total number of terms in the AP.}\\ \text{So},\text{}253=3+\left(\mathrm{n}-1\right)5\\ \text{i.e.,}253-3=\left(\mathrm{n}-1\right)5\\ \text{i.e.,}\mathrm{n}-1=\frac{250}{5}=50\\ \text{or}\mathrm{n}=50+1=51\\ \text{So, there are 51 terms in the given AP.}\\ \text{The 20th term from the last term will be the 32nd term(51-20+1) from the beginning.}\\ {\text{So, a}}_{32}=3+31×5=3+155=158\\ \text{i.e., the 20th term from the last term is 158.}\end{array}$

Q.21 The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the AP.

Ans.

$\begin{array}{l}\text{Here,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{a}}_{4}+{\mathrm{a}}_{8}=24\\ ⇒\mathrm{a}+3\mathrm{d}+\mathrm{a}+7\mathrm{d}=24\\ ⇒2\mathrm{a}+10\mathrm{d}=24\\ ⇒\mathrm{a}+5\mathrm{d}=12\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(1\right)\\ \text{Also,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{a}}_{6}+{\mathrm{a}}_{10}=44\\ ⇒\mathrm{a}+5\mathrm{d}+\mathrm{a}+9\mathrm{d}=44\\ ⇒2\mathrm{a}+14\mathrm{d}=44\\ ⇒\mathrm{a}+7\mathrm{d}=22\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(2\right)\\ \text{We subtract (2) from (1) and get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}2\mathrm{d}=10\\ ⇒\mathrm{d}=5\\ \text{On putting this value of d in (1), we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{a}+5×5=12\\ ⇒\mathrm{a}=12-25=-13\\ \text{First three terms of the AP are}\mathrm{a}\text{,}\mathrm{a}+\mathrm{d}\text{and}\mathrm{a}+2\mathrm{d}\text{i.e.,}\\ -\text{13,}-1\text{3}+\text{5 and}-1\text{3}+\text{2}×\text{5 i.e.,}-1\text{3,}-\text{8 and}-\text{3.}\end{array}$

Q.22 Subba Rao started work in 1995 at an annual salary of ₹ 5000 and received an increment of ₹ 200 each year. In which year did his income reach ₹ 7000?

Ans.

$\begin{array}{l}\text{Subba Rao’s annual salary in 1995 was ₹ 5000.}\\ \text{He received an increment of ₹ 200 each year.}\\ \text{So, Subba Rao’s annual salaries forms an AP with}\\ \text{a=5000 and d=200.}\\ {\text{Let 7000 is n}}^{\text{th}}\text{term of this AP.}\\ \text{We know that}\\ {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}a}}_{\text{n}}\text{=a+(n-1)d}\\ \mathrm{⇒}\text{7000=5000+(n-1)200}\\ \mathrm{⇒}\text{(n-1)200=7000-5000}\\ \mathrm{⇒}\text{n-1=}\frac{\text{2000}}{\text{200}}\text{=10}\\ \mathrm{⇒}\text{n=10+1=11}\\ {\text{Therefore, in 11}}^{\text{th}}\text{year, Subba Rao’s salary reach ₹ 7000.}\end{array}$

Q.23 Ramkali saved ₹ 5 in the first week of a year and then increased her weekly savings by ₹ 1.75. If in the nth week, her weekly savings become ₹ 20.75, find n.

Ans.

$\begin{array}{l}\text{}₹\text{ }\\ \text{}₹\text{}..\\ \text{So,}‘\text{s forms an AP with}\\ \mathrm{a}=5\text{and}\mathrm{d}=.\text{.}\\ {\text{It is given that in n}}^{\text{th}}\text{week her weekly savings become}\\ ‘\text{}.\text{.}\\ \therefore {\mathrm{a}}_{\mathrm{n}}=.\\ \text{We know that}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{a}}_{\mathrm{n}}=\mathrm{a}+\left(\mathrm{n}-1\right)\mathrm{d}\\ ⇒.=5+\left(\mathrm{n}-1\right)×.\\ ⇒\left(\mathrm{n}-1\right)×.=.-5\\ ⇒\mathrm{n}-1=\frac{15.75}{.}=9\\ ⇒\mathrm{n}=9+1=10\end{array}$

Q.24

$\begin{array}{l}\text{Find the sum of the following APs:}\\ \text{(i) 2, 7, 12, . . ., to 10 terms.}\\ \text{(ii)}-\text{37,}-\text{33,}-\text{29, . . ., to 12 terms.}\\ \text{(iii) 0.6, 1.7, 2.8, . . ., to 100 terms.}\\ \text{(iv)}\frac{\text{1}}{15}\text{,}\frac{\text{1}}{12}\text{,}\frac{\text{1}}{10}\text{, . . ., to 11 terms.}\end{array}$

Ans.

$\begin{array}{l}\text{(i)}\\ \text{Here,}\mathrm{a}=2,\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{d}={\mathrm{a}}_{2}-{\mathrm{a}}_{1}=7-2=5,\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{n}=10\\ \text{We know that sum of first\hspace{0.17em}\hspace{0.17em}n terms of an AP is given by}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{S}=\frac{\mathrm{n}}{2}\left[2\mathrm{a}+\left(\mathrm{n}-1\right)\mathrm{d}\right]\\ \text{So, \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{S}=\frac{10}{2}\left[2×2+\left(10-1\right)5\right]=5\left[4+45\right]\\ \text{or}\mathrm{S}=245\\ \text{So, the sum of the first 10 terms of the given AP is 245.}\\ \text{(ii)}\\ \text{Here,}\mathrm{a}=-\text{37},\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{d}={\mathrm{a}}_{2}-{\mathrm{a}}_{1}=-\text{33}-\left(-\text{37}\right)=4,\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{n}=12\\ \text{We know that sum of first\hspace{0.17em}\hspace{0.17em}n terms of an AP is given by}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{S}=\frac{\mathrm{n}}{2}\left[2\mathrm{a}+\left(\mathrm{n}-1\right)\mathrm{d}\right]\\ \text{So, \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{S}=\frac{12}{2}\left[2×\left(-\text{37}\right)+\left(12-1\right)4\right]=6\left[-74+44\right]\\ \text{or}\mathrm{S}=-180\\ \text{So, the sum of the first 12 terms of the given AP is}-180.\\ \text{(iii)}\\ \text{Here,}\mathrm{a}=\text{0.6},\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{d}={\mathrm{a}}_{2}-{\mathrm{a}}_{1}=\text{1.7}-\text{0.6}=1.1,\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{n}=100\\ \text{We know that sum of first\hspace{0.17em}\hspace{0.17em}n terms of an AP is given by}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{S}=\frac{\mathrm{n}}{2}\left[2\mathrm{a}+\left(\mathrm{n}-1\right)\mathrm{d}\right]\\ \text{So, \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{S}=\frac{100}{2}\left[2×\text{0.6}+\left(100-1\right)×1.1\right]=50\left[1.2+108.9\right]\\ \text{or}\mathrm{S}=5505\\ \text{So, the sum of the first 100 terms of the given AP is}5505.\\ \text{(iv)}\\ \text{Here,}\mathrm{a}=\text{}\frac{\text{1}}{15},\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{d}={\mathrm{a}}_{2}-{\mathrm{a}}_{1}=\frac{\text{1}}{12}-\frac{\text{1}}{15}=\frac{1}{60},\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{n}=11\\ \text{We know that sum of first\hspace{0.17em}\hspace{0.17em}n terms of an AP is given by}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{S}=\frac{\mathrm{n}}{2}\left[2\mathrm{a}+\left(\mathrm{n}-1\right)\mathrm{d}\right]\\ \text{So, \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{S}=\frac{11}{2}\left[2×\frac{\text{1}}{15}+\left(11-1\right)×\frac{1}{60}\right]=\frac{11}{2}\left[\frac{2}{15}+\frac{1}{6}\right]\\ \text{or}\mathrm{S}=\frac{33}{20}\\ \text{So, the sum of the first 11 terms of the given AP is}\frac{33}{20}\text{.}\end{array}$

Q.25

$\begin{array}{l}\text{Find the sums given below :}\\ \text{(i)}7+10\frac{1}{2}+14+\text{}.\text{}.\text{}.\text{}+84\text{}\\ \text{(ii)}34+32+30+\text{}.\text{}.\text{}.\text{}+10\\ \text{(iii)}-5+\left(-8\right)+\left(-11\right)+.\text{}.\text{}.+\left(-230\right)\end{array}$

Ans.

$\begin{array}{l}\text{(i)}7+10\frac{1}{2}+14+\text{}.\text{}.\text{}.\text{}+84\text{}\\ \text{Here,}\mathrm{a}=7,\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{d}={\mathrm{a}}_{2}-{\mathrm{a}}_{1}=10\frac{1}{2}-7=3\frac{1}{2}=\frac{7}{2},\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{l}=84\\ \text{We know that}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{l}=\mathrm{a}+\left(\mathrm{n}-1\right)\mathrm{d}\\ \text{So,\hspace{0.17em}\hspace{0.17em}}84=7+\left(\mathrm{n}-1\right)\frac{7}{2}\\ \text{or}84-7=\left(\mathrm{n}-1\right)\frac{7}{2}\\ \text{or 77}×\frac{2}{7}+1=\mathrm{n}\\ \text{or}\mathrm{n}=23\\ \text{We know that sum of first\hspace{0.17em}\hspace{0.17em}n terms of an AP is given by}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{n}}{2}\left(\mathrm{a}+\mathrm{l}\right)\\ \text{So, \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{S}}_{23}=\frac{23}{2}\left(7+84\right)=\frac{23}{2}×91\\ \text{or}{\mathrm{S}}_{23}=1046.5\\ \text{So, the requird sum is}1046.5.\\ \text{(ii)}34+32+30+\text{}.\text{}.\text{}.\text{}+10\\ \text{Here,}\mathrm{a}=34,\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{d}={\mathrm{a}}_{2}-{\mathrm{a}}_{1}=32-34=-2,\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{l}=10\\ \text{We know that}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{l}=\mathrm{a}+\left(\mathrm{n}-1\right)\mathrm{d}\\ \text{So,\hspace{0.17em}\hspace{0.17em}}10=34+\left(\mathrm{n}-1\right)\left(-2\right)\\ \text{or}10-34=\left(\mathrm{n}-1\right)\left(-2\right)\\ \text{or}\frac{-24}{-2}=\mathrm{n}-1\\ \text{or}\mathrm{n}=13\\ \text{We know that sum of first\hspace{0.17em}\hspace{0.17em}n terms of an AP is given by}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{n}}{2}\left(\mathrm{a}+\mathrm{l}\right)\\ \text{So, \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{S}}_{13}=\frac{13}{2}\left(34+10\right)=\frac{13}{2}×44\\ \text{or}{\mathrm{S}}_{13}=286\\ \text{So, the requird sum is}286.\\ \text{(iii)}-5+\left(-8\right)+\left(-11\right)+.\text{}.\text{}.+\left(-230\right)\\ \text{Here,}\mathrm{a}=-5,\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{d}={\mathrm{a}}_{2}-{\mathrm{a}}_{1}=-8-\left(-5\right)=-3,\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{l}=-230\\ \text{We know that}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{l}=\mathrm{a}+\left(\mathrm{n}-1\right)\mathrm{d}\\ \text{So,\hspace{0.17em}\hspace{0.17em}}-230=-5+\left(\mathrm{n}-1\right)\left(-3\right)\\ \text{or}-230+5=\left(\mathrm{n}-1\right)\left(-3\right)\\ \text{or}\frac{-225}{-3}+1=\mathrm{n}\\ \text{or}\mathrm{n}=76\\ \text{We know that sum of first\hspace{0.17em}\hspace{0.17em}n terms of an AP is given by}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{n}}{2}\left(\mathrm{a}+\mathrm{l}\right)\\ \text{So, \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{S}}_{75}=\frac{76}{2}\left(-5-230\right)=38×\left(-235\right)\\ \text{or}{\mathrm{S}}_{75}=-8930\\ \text{So, the requird sum is}-8930.\end{array}$

Q.26

$\begin{array}{l}\text{In an AP:}\\ \text{(i) given}\mathrm{a}=5,\text{}\mathrm{d}=3,\text{}{\mathrm{a}}_{\mathrm{n}}=50\text{, find n and}{\mathrm{S}}_{\mathrm{n}}\text{.}\\ \text{(ii) given}\mathrm{a}=7,\text{}{\mathrm{a}}_{13}=35\text{, find d and}{\mathrm{S}}_{13}\text{.}\\ \text{(iii) given}{\mathrm{a}}_{12}=37,\text{}\mathrm{d}=3\text{, find a and}{\mathrm{S}}_{12}\text{.}\\ \text{(iv) given}{\mathrm{a}}_{3}=15,\text{}{\mathrm{S}}_{10}=125,\text{find}\mathrm{d}\text{and}{\mathrm{a}}_{10}\text{.}\\ \text{(v) given}\mathrm{d}=5,\text{}{\mathrm{S}}_{9}=75,\text{find}\mathrm{a}\text{and}{\mathrm{a}}_{9}\text{.}\\ \text{(vi) given}\mathrm{a}=2,\text{}\mathrm{d}=8,\text{}{\mathrm{S}}_{\mathrm{n}}=90,\text{find}\mathrm{n}\text{and}{\mathrm{a}}_{\mathrm{n}}\text{.}\\ \text{(vii) given}\mathrm{a}=8,\text{}{\mathrm{a}}_{\mathrm{n}}=62,\text{}{\mathrm{S}}_{\mathrm{n}}=210,\text{find}\mathrm{n}\text{and}\mathrm{d}\text{.}\\ \text{(viii) given}{\mathrm{a}}_{\mathrm{n}}=4,\text{}\mathrm{d}=2,\text{}{\mathrm{S}}_{\mathrm{n}}=-14,\text{find}\mathrm{n}\text{and}\mathrm{a}\text{.}\\ \text{(ix) given}\mathrm{a}=3,\text{}\mathrm{n}=8,\text{}\mathrm{S}=192,\text{find}\mathrm{d}\text{.}\\ \text{(x) given}\mathrm{l}=28,\text{}\mathrm{S}=144,\text{and there are total 9 terms.}\\ \text{Find }\mathrm{a}\text{.}\end{array}$

Ans.

$\begin{array}{l}\text{(i)}\\ \text{Here,}\mathrm{a}=5,\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{d}=3,\text{\hspace{0.17em}\hspace{0.17em}}{\mathrm{a}}_{\mathrm{n}}=50\\ \text{We have to find n and}{\mathrm{S}}_{\mathrm{n}}\text{.}\\ \text{We know that}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{a}}_{\mathrm{n}}=\mathrm{a}+\left(\mathrm{n}-1\right)\mathrm{d}\\ \text{So,\hspace{0.17em}\hspace{0.17em}}50=5+\left(\mathrm{n}-1\right)3\\ \text{or}50-5=\left(\mathrm{n}-1\right)3\\ \text{or}\frac{45}{3}+1=\mathrm{n}\\ \text{or}\mathrm{n}=16\\ \text{We know that sum of first\hspace{0.17em}\hspace{0.17em}n terms of an AP is given by}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{n}}{2}\left(\mathrm{a}+{\mathrm{a}}_{\mathrm{n}}\right)\\ \text{So, \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{S}}_{16}=\frac{16}{2}\left(5+50\right)=8×55\\ \text{or}{\mathrm{S}}_{16}=440\\ \text{(ii)}\\ \text{Here,}\mathrm{a}=7,\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{a}}_{13}=35\\ \text{We have to find}\mathrm{d}\text{and}{\mathrm{S}}_{13}\text{.}\\ \text{We know that}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{a}}_{\mathrm{n}}=\mathrm{a}+\left(\mathrm{n}-1\right)\mathrm{d}\\ \text{So,\hspace{0.17em}\hspace{0.17em}}{\mathrm{a}}_{13}=7+\left(13-1\right)\mathrm{d}\\ \text{or}35-7=12\mathrm{d}\\ \text{or}\frac{28}{12}=\mathrm{d}\\ \text{or}\mathrm{d}=\frac{7}{3}\\ \text{We know that sum of first\hspace{0.17em}\hspace{0.17em}n terms of an AP is given by}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{n}}{2}\left(\mathrm{a}+{\mathrm{a}}_{\mathrm{n}}\right)\\ \text{So, \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{S}}_{13}=\frac{13}{2}\left(7+35\right)=\frac{13}{2}×42\\ \text{or}{\mathrm{S}}_{13}=273\\ \text{(iii)}\\ \text{Here,\hspace{0.17em}\hspace{0.17em}}{\mathrm{a}}_{12}=37,\text{}\mathrm{d}=3\\ \text{We have to find}\mathrm{a}\text{and}{\mathrm{S}}_{12}\text{.}\\ \text{We know that}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{a}}_{\mathrm{n}}=\mathrm{a}+\left(\mathrm{n}-1\right)\mathrm{d}\\ \text{So,\hspace{0.17em}\hspace{0.17em}}{\mathrm{a}}_{12}=\mathrm{a}+\left(12-1\right)3\\ \text{or}37=\mathrm{a}+33\\ \text{or \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{a}=37-33=4\\ \text{We know that sum of first\hspace{0.17em}\hspace{0.17em}n terms of an AP is given by}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{n}}{2}\left(\mathrm{a}+{\mathrm{a}}_{\mathrm{n}}\right)\\ \text{So, \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{S}}_{12}=\frac{12}{2}\left(\mathrm{a}+{\mathrm{a}}_{12}\right)=6\left(4+37\right)\\ \text{or}{\mathrm{S}}_{12}=246\\ \text{(iv)}\\ \text{Here,\hspace{0.17em}\hspace{0.17em}}{\mathrm{a}}_{3}=15,\text{}{\mathrm{S}}_{10}=125\\ \text{We have to find}\mathrm{d}\text{and}{\mathrm{a}}_{10}\text{.}\\ \text{We know that}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{a}}_{\mathrm{n}}=\mathrm{a}+\left(\mathrm{n}-1\right)\mathrm{d}\\ \text{So,\hspace{0.17em}\hspace{0.17em}}{\mathrm{a}}_{3}=\mathrm{a}+\left(3-1\right)\mathrm{d}\\ \text{or \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{a}+2\mathrm{d}=15\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(1\right)\\ \text{We know that sum of first\hspace{0.17em}\hspace{0.17em}n terms of an AP is given by}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{n}}{2}\left[2\mathrm{a}+\left(\mathrm{n}-1\right)\mathrm{d}\right]\\ \text{So, \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{S}}_{10}=\frac{10}{2}\left[2\mathrm{a}+\left(10-1\right)\mathrm{d}\right]=5\left[2\mathrm{a}+9\mathrm{d}\right]\\ \text{or}125=5\left[2\mathrm{a}+9\mathrm{d}\right]\\ \text{or}2\mathrm{a}+9\mathrm{d}=25\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}2\left(15-2\mathrm{d}\right)+9\mathrm{d}=25\text{[From (1),}\mathrm{a}=15-2\mathrm{d}\right]\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}5\mathrm{d}=25-30=-5\\ \text{or}\mathrm{d}=-1\\ \text{putting this value of}\mathrm{d}\text{in (1), we get}\mathrm{a}=17.\\ \text{Now,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{a}}_{10}=\mathrm{a}+\left(10-1\right)\mathrm{d}=17+9×\left(-1\right)=8\\ \text{(v)}\\ \text{Here,\hspace{0.17em}\hspace{0.17em}}\mathrm{d}=5,\text{}{\mathrm{S}}_{9}=75\\ \text{We have to find}\mathrm{a}\text{and}{\mathrm{a}}_{9}\text{.}\\ \text{We know that}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{a}}_{\mathrm{n}}=\mathrm{a}+\left(\mathrm{n}-1\right)\mathrm{d}\\ \text{So,\hspace{0.17em}\hspace{0.17em}}{\mathrm{a}}_{9}=\mathrm{a}+\left(9-1\right)5\\ \text{or \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{a}}_{9}-\mathrm{a}=40\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(1\right)\\ \text{We know that sum of first\hspace{0.17em}\hspace{0.17em}n terms of an AP is given by}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{n}}{2}\left[2\mathrm{a}+\left(\mathrm{n}-1\right)\mathrm{d}\right]\\ \text{So, \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{S}}_{9}=\frac{9}{2}\left[2\mathrm{a}+\left(9-1\right)5\right]=\frac{9}{2}\left[2\mathrm{a}+40\right]\\ \text{or}75=9\left[\mathrm{a}+20\right]\\ \text{or}\mathrm{a}=\frac{75}{9}-20=\frac{75-180}{9}=\frac{-105}{9}=\frac{-35}{3}\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{a}=\frac{-35}{3}\\ \text{On putting this value of}\mathrm{a}\text{in (1), we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{a}}_{9}-\frac{-35}{3}=40\\ ⇒\text{}{\mathrm{a}}_{9}=\text{\hspace{0.17em}\hspace{0.17em}}40-\frac{35}{3}=\frac{120-35}{3}=\frac{85}{3}\\ \text{(vi)}\\ \text{Here,\hspace{0.17em}\hspace{0.17em}}\mathrm{a}=2,\text{}\mathrm{d}=8,\text{}{\mathrm{S}}_{\mathrm{n}}=90\\ \text{We have to find}\mathrm{n}\text{and}{\mathrm{a}}_{\mathrm{n}}\text{.}\\ \text{We know that}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{a}}_{\mathrm{n}}=\mathrm{a}+\left(\mathrm{n}-1\right)\mathrm{d}\\ \text{So,\hspace{0.17em}\hspace{0.17em}}{\mathrm{a}}_{\mathrm{n}}=2+\left(\mathrm{n}-1\right)8\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(1\right)\\ \text{We know that sum of first\hspace{0.17em}\hspace{0.17em}n terms of an AP is given by}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{n}}{2}\left[2\mathrm{a}+\left(\mathrm{n}-1\right)\mathrm{d}\right]\\ \text{So, \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{n}}{2}\left[2×2+\left(\mathrm{n}-1\right)8\right]\\ \text{or}90=2\mathrm{n}+4\mathrm{n}\left(\mathrm{n}-1\right)=2\mathrm{n}+4{\mathrm{n}}^{2}-4\mathrm{n}\\ \text{or}90=4{\mathrm{n}}^{2}-2\mathrm{n}\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}2{\mathrm{n}}^{2}-\mathrm{n}-45=0\\ \text{or}2{\mathrm{n}}^{2}-10\mathrm{n}+9\mathrm{n}-45=0\\ \text{or}2\mathrm{n}\left(\mathrm{n}-5\right)+9\left(\mathrm{n}-5\right)=0\\ \text{or}\left(2\mathrm{n}+9\right)\left(\mathrm{n}-5\right)=0\\ \text{or}\mathrm{n}=5\\ \text{On putting this value of}\mathrm{n}\text{in (1), we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{a}}_{5}=2+\left(5-1\right)8=34\text{\hspace{0.17em}}\\ \text{(vii)}\\ \text{Here,\hspace{0.17em}\hspace{0.17em}}\mathrm{a}=8,\text{}{\mathrm{a}}_{\mathrm{n}}=62,\text{}{\mathrm{S}}_{\mathrm{n}}=210\\ \text{We have to find}\mathrm{n}\text{and}\mathrm{d}\text{.}\\ \text{We know that}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{a}}_{\mathrm{n}}=\mathrm{a}+\left(\mathrm{n}-1\right)\mathrm{d}\\ \text{So,\hspace{0.17em}\hspace{0.17em}}62=8+\left(\mathrm{n}-1\right)\mathrm{d}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(1\right)\\ \text{We know that sum of first\hspace{0.17em}\hspace{0.17em}n terms of an AP is given by}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{n}}{2}\left(\mathrm{a}+{\mathrm{a}}_{\mathrm{n}}\right)\\ \text{So, \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}210=\frac{\mathrm{n}}{2}\left(8+62\right)\\ \text{or}210=35\mathrm{n}\\ \text{or}\mathrm{n}=\frac{210}{35}=6\\ \text{On putting this value of}\mathrm{n}\text{in (1), we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}62=8+\left(\mathrm{n}-1\right)\mathrm{d}\text{\hspace{0.17em}}=8+\left(6-1\right)\mathrm{d}=8+5\mathrm{d}\\ \text{or}\mathrm{d}=\frac{62-8}{5}\text{\hspace{0.17em}}=\frac{54}{5}\\ \text{(viii)}\\ \text{Here,\hspace{0.17em}\hspace{0.17em}}{\mathrm{a}}_{\mathrm{n}}=4,\text{}\mathrm{d}=2,\text{}{\mathrm{S}}_{\mathrm{n}}=-14\\ \text{We have to find}\mathrm{n}\text{and}\mathrm{a}\text{.}\\ \text{We know that}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{a}}_{\mathrm{n}}=\mathrm{a}+\left(\mathrm{n}-1\right)\mathrm{d}\\ \text{So,\hspace{0.17em}\hspace{0.17em}}4=\mathrm{a}+2\left(\mathrm{n}-1\right)=\mathrm{a}+2\mathrm{n}-2\text{\hspace{0.17em}\hspace{0.17em}}\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{a}+2\mathrm{n}=6\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{a}=6-2\mathrm{n}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(1\right)\\ \text{We know that sum of first\hspace{0.17em}\hspace{0.17em}n terms of an AP is given by}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{n}}{2}\left(\mathrm{a}+{\mathrm{a}}_{\mathrm{n}}\right)\\ \text{So,}-14=\frac{\mathrm{n}}{2}\left(\mathrm{a}+4\right)\\ \text{or}\mathrm{n}\left(\mathrm{a}+4\right)=-28\\ \text{or}\mathrm{n}\left(6-2\mathrm{n}+4\right)=-28\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\text{From (1)]}\\ \text{or}10\mathrm{n}-2{\mathrm{n}}^{2}+28=0\\ \text{or}{\mathrm{n}}^{2}-5\mathrm{n}-14=0\\ \text{or}{\mathrm{n}}^{2}-7\mathrm{n}+2\mathrm{n}-14=0\\ \text{or}\mathrm{n}\left(\mathrm{n}-7\right)+2\left(\mathrm{n}-7\right)=0\\ \text{or}\left(\mathrm{n}-7\right)\left(\mathrm{n}+2\right)=0\\ \text{or}\mathrm{n}-7=0\\ \text{or}\mathrm{n}=7\text{[}\mathrm{n}\text{is a natural number]}\\ \text{On putting this value of}\mathrm{n}\text{in (1), we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{a}=6-2\mathrm{n}\text{\hspace{0.17em}\hspace{0.17em}}=6-2×7=-8\\ \text{(ix)}\\ \text{Here,\hspace{0.17em}\hspace{0.17em}}\mathrm{a}=3,\text{}\mathrm{n}=8,\text{}\mathrm{S}=192\\ \text{We know that sum of first\hspace{0.17em}\hspace{0.17em}}\mathrm{n}\text{terms of an AP is given by}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{S}=\frac{\mathrm{n}}{2}\left[2\mathrm{a}+\left(\mathrm{n}-1\right)\mathrm{d}\right]\\ \text{So, \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}192=\frac{8}{2}\left[2×3+\left(8-1\right)\mathrm{d}\right]\\ \text{or}192=4\left[6+7\mathrm{d}\right]\\ \text{or}6+7\mathrm{d}=\frac{192}{4}=48\\ \text{or}7\mathrm{d}=48-6=42\\ \text{or}\mathrm{d}=\frac{42}{7}=6\\ \text{(x)}\\ \text{Here,\hspace{0.17em}\hspace{0.17em}}\mathrm{l}=28,\text{}\mathrm{S}=144,\text{}\mathrm{n}=9\\ \text{We have to find}\mathrm{a}\text{.}\\ \text{We know that sum of first\hspace{0.17em}\hspace{0.17em}n terms of an AP is given by}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{S}=\frac{\mathrm{n}}{2}\left(\mathrm{a}+\mathrm{l}\right)\\ \text{So, \hspace{0.17em}\hspace{0.17em}}144=\frac{9}{2}\left(\mathrm{a}+28\right)\\ \text{or}\mathrm{a}+28=\text{\hspace{0.17em}}144×\frac{2}{9}=16×2=32\\ \text{or}\mathrm{a}=32-28=4\text{\hspace{0.17em}\hspace{0.17em}}\end{array}$

Q.27

$\text{How many terms of the AP : 9, 17, 25, . . . must be taken to give a sum of 636?}$

Ans.

$\begin{array}{l}\text{Here,\hspace{0.17em}\hspace{0.17em}}\mathrm{a}=9,\text{}\mathrm{d}={\mathrm{a}}_{2}-{\mathrm{a}}_{1}=17-9=8,\text{}{\mathrm{S}}_{\mathrm{n}}=636\\ \text{We know that sum of first\hspace{0.17em}\hspace{0.17em}}\mathrm{n}\text{terms of an AP is given by}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{S}=\frac{\mathrm{n}}{2}\left[2\mathrm{a}+\left(\mathrm{n}-1\right)\mathrm{d}\right]\\ \text{So, \hspace{0.17em}\hspace{0.17em}}636=\frac{\mathrm{n}}{2}\left[2×9+\left(\mathrm{n}-1\right)8\right]\\ \text{or}636=\frac{\mathrm{n}}{2}×2\left[9+4\mathrm{n}-4\right]=4{\mathrm{n}}^{2}+5\mathrm{n}\\ \text{or}4{\mathrm{n}}^{2}+5\mathrm{n}-636=0\\ \text{or}4{\mathrm{n}}^{2}+53\mathrm{n}-48\mathrm{n}-636=0\\ \text{or}\mathrm{n}\left(4\mathrm{n}+53\right)-12\left(4\mathrm{n}+53\right)=0\\ \text{or}\left(4\mathrm{n}+53\right)\left(\mathrm{n}-12\right)=0\\ \text{or}\mathrm{n}=12\text{[}\mathrm{n}\text{is a natural number]}\\ \text{Therefore, 12 terms of the given AP must be taken to give a}\\ \text{sum of 636.}\end{array}$

Q.28 The first term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.

Ans.

$\begin{array}{l}\text{Here,\hspace{0.17em}\hspace{0.17em}}\mathrm{a}=5,\text{}{\mathrm{a}}_{\mathrm{n}}=\mathrm{l}=45,\text{}{\mathrm{S}}_{\mathrm{n}}=400\\ \text{We have to find}\mathrm{n}\text{and}\mathrm{d}\text{.}\\ \text{We know that}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{a}}_{\mathrm{n}}=\mathrm{a}+\left(\mathrm{n}-1\right)\mathrm{d}\\ \text{So,\hspace{0.17em}\hspace{0.17em}}45=5+\left(\mathrm{n}-1\right)\mathrm{d}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(1\right)\\ \text{We know that sum of first\hspace{0.17em}\hspace{0.17em}n terms of an AP is given by}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{n}}{2}\left(\mathrm{a}+{\mathrm{a}}_{\mathrm{n}}\right)\\ \text{So, \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}400=\frac{\mathrm{n}}{2}\left(5+45\right)\\ \text{or}400=25\mathrm{n}\\ \text{or}\mathrm{n}=\frac{400}{25}=16\\ \text{On putting this value of}\mathrm{n}\text{in (1), we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}45=5+\left(\mathrm{n}-1\right)\mathrm{d}\text{\hspace{0.17em}}=5+\left(16-1\right)\mathrm{d}=5+15\mathrm{d}\\ \text{or}\mathrm{d}=\frac{45-5}{15}\text{\hspace{0.17em}}=\frac{40}{15}=\frac{8}{3}\end{array}$

Q.29 The first and the last terms of an AP are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum?

Ans.

$\begin{array}{l}\text{Here,\hspace{0.17em}\hspace{0.17em}}\mathrm{a}=17,\text{}{\mathrm{a}}_{\mathrm{n}}=\mathrm{l}=350,\text{}\mathrm{d}=9\\ \text{We have to find}\mathrm{n}\text{and}{\mathrm{S}}_{\mathrm{n}}\text{.}\\ \text{We know that}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{a}}_{\mathrm{n}}=\mathrm{a}+\left(\mathrm{n}-1\right)\mathrm{d}\\ \text{So,\hspace{0.17em}\hspace{0.17em}}350=17+\left(\mathrm{n}-1\right)9\text{\hspace{0.17em}}\\ \text{or}\mathrm{n}-1=\frac{350-17}{9}=\frac{333}{9}=37\\ \text{or}\mathrm{n}=37+1=38\\ \text{We know that sum of first\hspace{0.17em}\hspace{0.17em}n terms of an AP is given by}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{n}}{2}\left(\mathrm{a}+{\mathrm{a}}_{\mathrm{n}}\right)\\ \text{So, \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{S}}_{38}=\frac{38}{2}\left(17+350\right)=19×367\\ \text{or}{\mathrm{S}}_{38}=6973\end{array}$

Q.30 Find the sum of first 22 terms of an AP in which d = 7 and 22nd term is 149.

Ans.

$\begin{array}{l}\text{Here,\hspace{0.17em}\hspace{0.17em}}{\mathrm{a}}_{22}=149,\text{}\mathrm{d}=7\\ \text{We have to find}{\mathrm{S}}_{22}\text{.}\\ \text{We know that}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{a}}_{\mathrm{n}}=\mathrm{a}+\left(\mathrm{n}-1\right)\mathrm{d}\\ \text{So,\hspace{0.17em}\hspace{0.17em}}{\mathrm{a}}_{22}=\mathrm{a}+\left(22-1\right)\mathrm{d}\\ \text{or}149=\mathrm{a}+21×7\text{\hspace{0.17em}}=\mathrm{a}+147\\ \text{or}\mathrm{a}=149-147=2\\ \text{We know that sum of first n terms of an AP is given by}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{n}}{2}\left(\mathrm{a}+{\mathrm{a}}_{\mathrm{n}}\right)\\ \text{So, \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{S}}_{22}=\frac{22}{2}\left(\mathrm{a}+{\mathrm{a}}_{22}\right)=11\left(2+149\right)=11×151\\ \text{or}{\mathrm{S}}_{22}=1661\end{array}$

Q.31 Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.

Ans.

$\begin{array}{l}\text{Here,\hspace{0.17em}\hspace{0.17em}}{\mathrm{a}}_{2}=14,\text{}{\mathrm{a}}_{3}=18\\ \text{Therefore,}\mathrm{d}={\mathrm{a}}_{3}-{\mathrm{a}}_{2}=18-14=4\\ \text{We know that}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{a}}_{\mathrm{n}}=\mathrm{a}+\left(\mathrm{n}-1\right)\mathrm{d}\\ \text{So,\hspace{0.17em}\hspace{0.17em}}{\mathrm{a}}_{2}=\mathrm{a}+\left(2-1\right)\mathrm{d}\\ \text{or}14=\mathrm{a}+1×4\text{\hspace{0.17em}}=\mathrm{a}+4\\ \text{or}\mathrm{a}=14-4=10\\ \text{We know that sum of first n terms of an AP is given by}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{n}}{2}\left[2\mathrm{a}+\left(\mathrm{n}-1\right)\mathrm{d}\right]\\ \text{So, \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{S}}_{51}=\frac{51}{2}\left[2×10+\left(51-1\right)4\right]=\frac{51}{2}\left(20+200\right)=51×110\\ \text{or}{\mathrm{S}}_{51}=5610\end{array}$

Q.32 If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first n terms.

Ans.

$\begin{array}{l}\text{Here,\hspace{0.17em}\hspace{0.17em}}{\mathrm{S}}_{7}=49,\text{}{\mathrm{S}}_{17}=289\\ \text{We know that sum of first n terms of an AP is given by}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{n}}{2}\left[2\mathrm{a}+\left(\mathrm{n}-1\right)\mathrm{d}\right]\\ \text{So, \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{S}}_{7}=\frac{7}{2}\left[2\mathrm{a}+\left(7-1\right)\mathrm{d}\right]\\ \text{or}49=\frac{7}{2}\left(2\mathrm{a}+6\mathrm{d}\right)\\ \text{or}\frac{49}{7}=\frac{2}{2}\left(\mathrm{a}+3\mathrm{d}\right)\\ \text{or}\mathrm{a}+3\mathrm{d}=7\text{}...\text{(1)}\\ \text{Also,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{S}}_{17}=\frac{17}{2}\left[2\mathrm{a}+\left(17-1\right)\mathrm{d}\right]\\ \text{or}289=\frac{17}{2}\left(2\mathrm{a}+16\mathrm{d}\right)=\frac{17}{2}×2\left(\mathrm{a}+8\mathrm{d}\right)\\ \text{or}\mathrm{a}+8\mathrm{d}=17\\ \text{or}7-3\mathrm{d}+8\mathrm{d}=17\text{[From (1),}\mathrm{a}=7-3\mathrm{d}\right]\\ \text{or}5\mathrm{d}=17-7=10\\ \text{or}\mathrm{d}=\frac{10}{5}=2\\ \text{On putting this value of}\mathrm{d}\text{in (1), we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{a}=7-3\mathrm{d}=7-3×2=1\\ \text{Therefore, sum of first n terms of the AP with a}=1\text{and}\mathrm{d}=2\text{is}\\ \text{given by}\\ \text{}{\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{n}}{2}\left[2×1+\left(\mathrm{n}-1\right)2\right]=\mathrm{n}\left(1+\mathrm{n}-1\right)={\mathrm{n}}^{2}\end{array}$

Q.33 Show that a1, a2, . . ., an, . . . form an AP where an is defined as below :
(i) an = 3 + 4n (ii) an = 9 – 5n
Also find the sum of the first 15 terms in each case.

Ans.

$\begin{array}{l}\text{(i)}\\ \text{Here,\hspace{0.17em}\hspace{0.17em}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathbf{a}}_{n}=\mathbf{3}+\mathbf{4}\mathbf{n}\\ ⇒{\mathbf{a}}_{\mathbf{n}+1}=\mathbf{3}+\mathbf{4}\left(\mathbf{n}+1\right)=3+4\mathrm{n}+4=7+4\mathrm{n}\\ \text{Now,}\\ {\mathbf{a}}_{\mathbf{n}+1}-{\mathbf{a}}_{n}=7+4\mathrm{n}-\mathbf{3}-\mathbf{4}\mathbf{n}=4\\ \text{i.e., difference between a term and its preceding term is always 4.}\\ \text{Therefore,}{\mathbf{a}}_{1},\text{}{\mathbf{a}}_{2},\text{}.\text{}.\text{}.,\text{}{\mathbf{a}}_{n},\text{}.\text{}.\text{}.\text{form an AP where}{\mathbf{a}}_{n}=\mathbf{3}+\mathbf{4}\mathbf{n}.\\ \text{Now, we have}\\ \mathrm{d}={\mathbf{a}}_{\mathbf{n}+1}-{\mathbf{a}}_{n}=4\\ \text{and}\\ \mathrm{a}={\mathrm{a}}_{1}=3+4×1=7\text{}\\ \text{We know that sum of first n terms of an AP is given by}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{n}}{2}\left[2\mathrm{a}+\left(\mathrm{n}-1\right)\mathrm{d}\right]\\ \text{So, \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{S}}_{15}=\frac{15}{2}\left[2×7+\left(15-1\right)4\right]\\ \text{or}{\mathrm{S}}_{15}=\frac{15}{2}×2\left[7+14×2\right]=15×35=525\\ \text{(ii)}\\ \text{Here,\hspace{0.17em}\hspace{0.17em}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathbf{a}}_{n}=\mathbf{9}-\mathbf{5}\mathbf{n}\\ ⇒{\mathbf{a}}_{\mathbf{n}+1}=9-5\left(\mathbf{n}+1\right)=9-5\mathrm{n}-5=4-5\mathrm{n}\\ \text{Now,}\\ {\mathbf{a}}_{\mathbf{n}+1}-{\mathbf{a}}_{n}=4-5\mathrm{n}-\left(\mathbf{9}-\mathbf{5}\mathbf{n}\right)=4-5\mathrm{n}-\mathbf{9}+\mathbf{5}\mathbf{n}=-5\\ \text{i.e., difference between a term and its preceding term is}\\ \text{always}-\text{5.}\\ \text{Therefore,}{\mathbf{a}}_{1},\text{}{\mathbf{a}}_{2},\text{}.\text{}.\text{}.,\text{}{\mathbf{a}}_{n},\text{}.\text{}.\text{}.\text{form an AP where}{\mathbf{a}}_{n}=\mathbf{9}-\mathbf{5}\mathbf{n}.\\ \text{Now, we have}\\ \mathrm{d}={\mathbf{a}}_{\mathbf{n}+1}-{\mathbf{a}}_{n}=-5\\ \text{and}\\ \mathrm{a}={\mathrm{a}}_{1}=9-5×1=4\\ \text{We know that sum of first n terms of an AP is given by}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{n}}{2}\left[2\mathrm{a}+\left(\mathrm{n}-1\right)\mathrm{d}\right]\\ \text{So, \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{S}}_{15}=\frac{15}{2}\left[2×4+\left(15-1\right)\left(-5\right)\right]\\ \text{or}{\mathrm{S}}_{15}=\frac{15}{2}×2\left[4-5×7\right]=-31×15=-465\end{array}$

Q.34 If the sum of the first n terms of an AP is 4n – n2, what is the first term (that is S1)? What is the sum of first two terms? What is the second term? Similarly, find the 3rd, the 10th and the nth terms.

Ans.

$\begin{array}{l}\text{Here,\hspace{0.17em}\hspace{0.17em}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{S}}_{}=-{\mathrm{n}}^{2}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{S}}_{1}=×1-{1}^{2}=3\\ \text{i.e.,}\mathrm{a}={\mathrm{S}}_{1}=3\\ {\mathrm{S}}_{2}=×2-{2}^{2}=4\\ \text{Now,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{a}}_{2}=\text{\hspace{0.17em}}{\mathrm{S}}_{2}-{\mathrm{S}}_{1}=\left(×2-{2}^{2}\right)-3=\left(8-4\right)-3=1\\ \text{Similarly,}\\ {\mathrm{a}}_{3}=\text{\hspace{0.17em}}{\mathrm{S}}_{3}-{\mathrm{S}}_{2}=\left(×3-{3}^{2}\right)-\left(×2-{2}^{2}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\left(12-9\right)-\left(8-4\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=3-4=-1\\ {\mathrm{a}}_{10}=\text{\hspace{0.17em}}{\mathrm{S}}_{10}-{\mathrm{S}}_{9}=\left(×10-{10}^{2}\right)-\left(×9-{9}^{2}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\left(40-100\right)-\left(36-81\right)=-60-\left(-45\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=-60+45=-15\\ {\mathrm{a}}_{\mathrm{n}}=\text{\hspace{0.17em}}{\mathrm{S}}_{\mathrm{n}}-{\mathrm{S}}_{\mathrm{n}-1}=\left(×\mathrm{n}-{\mathrm{n}}^{2}\right)-\left(×\left(\mathrm{n}-1\right)-{\left(\mathrm{n}-1\right)}^{2}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{n}\left(4-\mathrm{n}\right)-\left(\mathrm{n}-1\right)\left(4-\mathrm{n}+1\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=\mathrm{n}\left(4-\mathrm{n}\right)-\left(\mathrm{n}-1\right)\left(5-\mathrm{n}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=4\mathrm{n}-{\mathrm{n}}^{2}-5\mathrm{n}+{\mathrm{n}}^{2}+5-\mathrm{n}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=-2\mathrm{n}+5\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=5-2\mathrm{n}\end{array}$

Q.35 Find the sum of the first 40 positive integers divisible by 6.

Ans.

$\begin{array}{l}\text{First 40 positive integers divisible by 6 forms an AP with}\\ \mathrm{a}=6\text{and}\mathrm{d}=6.\\ \text{We know that sum of first n terms of an AP is given by}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{n}}{2}\left[2\mathrm{a}+\left(\mathrm{n}-1\right)\mathrm{d}\right]\\ \text{So,}{\mathrm{S}}_{40}=\frac{40}{2}\left[2×6+\left(40-1\right)6\right]=20\left[12+39×6\right]\\ \text{or}{\mathrm{S}}_{40}=20\left[12+234\right]=20×246=4920\end{array}$

Q.36 Find the sum of the first 15 multiples of 8.

Ans.

$\begin{array}{l}\text{First 15 multiples of 8 forms an AP with}\\ \mathrm{a}=8\text{and}\mathrm{d}=8.\\ \text{We know that sum of first n terms of an AP is given by}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{n}}{2}\left[2\mathrm{a}+\left(\mathrm{n}-1\right)\mathrm{d}\right]\\ \text{So,}{\mathrm{S}}_{15}=\frac{15}{2}\left[2×8+\left(15-1\right)8\right]=\frac{15}{2}×8\left[2+14\right]\\ \text{or}{\mathrm{S}}_{15}=60×16=960\end{array}$

Q.37 Find the sum of the odd numbers between 0 and 50.

Ans.

$\begin{array}{l}\text{Odd numbers between 0 and 50 forms an AP with}\\ \mathrm{a}=1\text{and}\mathrm{d}=2.\\ \text{Also, last term of this AP is 49.}\\ \text{Therefore,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{l}={\mathrm{a}}_{\mathrm{n}}=49\\ ⇒\mathrm{a}+\left(\mathrm{n}-1\right)\mathrm{d}=49\\ ⇒1+\left(\mathrm{n}-1\right)2=49\\ ⇒\mathrm{n}-1=\frac{49-1}{2}=24\\ ⇒\mathrm{n}=24+1=25\\ \text{We know that sum of first n terms of an AP is given by}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{n}}{2}\left[2\mathrm{a}+\left(\mathrm{n}-1\right)\mathrm{d}\right]\\ \text{So,}{\mathrm{S}}_{25}=\frac{25}{2}\left[2×1+\left(25-1\right)2\right]=\frac{25}{2}×2\left[1+24\right]\\ \text{or}{\mathrm{S}}_{25}=25×25=625\end{array}$

Q.38 A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows: ₹ 200 for the first day, ₹ 250 for the second day, ₹300 for the third day, etc., the penalty for each succeeding day being ₹ 50 more than for the preceding day. How much money the contractor has to pay as penalty, if he has delayed the work by 30 days?

Ans.

$\begin{array}{l}\text{The given 30 penalties forms an AP with first term as 200}\\ \text{and common difference as 50.}\\ \text{So, we have}\\ \mathrm{a}=200,\text{}\mathrm{d}=50\text{and}\mathrm{n}=30.\\ \text{We know that sum of first n terms of an AP is given by}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{n}}{2}\left[2\mathrm{a}+\left(\mathrm{n}-1\right)\mathrm{d}\right]\\ \text{So,}{\mathrm{S}}_{30}=\frac{30}{2}\left[2×200+\left(30-1\right)50\right]=15\left[400+29×50\right]\\ \text{or}{\mathrm{S}}_{30}=15\left[400+1450\right]=1850×15=27,750\\ \text{Therefore, the contractor has to pay}₹\text{\hspace{0.17em}}27,750\text{as penalty.}\end{array}$

Q.39 A sum of ₹ 700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is ₹ 20 less than its preceding prize, find the value of each of the prizes.

Ans.

$\begin{array}{l}\text{The given 7 cash prizes forms an AP with common difference}\\ \text{as}-\text{20 and sum of all the 7 prizes as 700.}\\ \text{So, we have}\\ {\mathrm{S}}_{7}=700,\text{}\mathrm{d}=-20\text{and}\mathrm{n}=7.\\ \text{We know that sum of first n terms of an AP is given by}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{n}}{2}\left[2\mathrm{a}+\left(\mathrm{n}-1\right)\mathrm{d}\right]\\ \text{So,}{\mathrm{S}}_{7}=\frac{7}{2}\left[2\mathrm{a}+\left(7-1\right)\left(-20\right)\right]=\frac{7}{2}×2\left[\mathrm{a}-60\right]\\ \text{or 7}00=7\left[\mathrm{a}-60\right]\\ \text{or}\mathrm{a}-60=\frac{700}{7}=100\\ \text{or}\mathrm{a}=100+60=160\\ \text{Therefore, the values of all the 7 prizes are}₹\text{\hspace{0.17em}}160\text{,}₹\text{\hspace{0.17em}}140,\text{}₹\text{\hspace{0.17em}}120,\\ ₹\text{\hspace{0.17em}}100,\text{}₹\text{\hspace{0.17em}}80,\text{}₹\text{\hspace{0.17em}}60\text{and}₹\text{\hspace{0.17em}40}.\end{array}$

Q.40 In a school, students thought of planting trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, will be the same as the class, in which they are studying, e.g., a section of Class I will plant 1 tree, a section of Class II will plant 2 trees and so on till Class XII. There are three sections of each class. How many trees will be planted by the students?

Ans.

$\begin{array}{l}\text{It is given that}\mathbf{a}\text{}\mathbf{section}\text{}\mathbf{of}\text{}\mathbf{Class}\text{}\mathbf{I}\text{}\mathbf{will}\text{}\mathbf{plant}\text{}\mathbf{1}\text{}\mathbf{tree},\text{}\mathbf{a}\text{}\mathbf{section}\\ \mathbf{of}\text{}\mathbf{Class}\text{}\mathbf{II}\text{}\mathbf{will}\text{}\mathbf{plant}\text{}\mathbf{2}\text{}\mathbf{trees}\text{}\mathbf{and}\text{}\mathbf{so}\text{}\mathbf{on}\text{}\mathbf{till}\text{}\mathbf{Class}\text{}\mathbf{XII}.\text{}\\ \text{The number of trees planted by a section of each class}\\ \text{forms an AP with}\mathrm{a}=1\text{and}\mathrm{d}=1.\\ \text{So, we have}\\ \mathrm{a}=1,\text{}\mathrm{d}=1\text{and}\mathrm{n}=12.\\ \text{We know that sum of first n terms of an AP is given by}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{n}}{2}\left[2\mathrm{a}+\left(\mathrm{n}-1\right)\mathrm{d}\right]\\ \text{So,}{\mathrm{S}}_{12}=\frac{12}{2}\left[2×1+\left(12-1\right)1\right]=6\left[2+11\right]\\ \text{or}{\mathrm{S}}_{12}=78\\ \text{Therefore,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Sum of trees planted by a section of each class}=78\\ ⇒\text{\hspace{0.17em}Sum of trees planted by 3 sections of each class}=78×3\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=234\end{array}$

Q.41 A spiral is made up of successive semicircles, with centres alternately at A and B, starting with centre at A, of radii 0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm, . . . as shown in the following figure. What is the total length of such a spiral made up of thirteen consecutive semicircles? (Take π =22/7) [Hint: Length of successive semicircles is l1, l2, l3, l4, . . . with centres at A, B, A, B, . . ., respectively.]

Ans.

$\begin{array}{l}\text{Here, successive semicircles are of raddi 0.5 cm, 1.0 cm,}\\ \text{1.5 cm, 2.0 cm,}.\dots \text{}\\ \text{Let}{\mathrm{l}}_{1},\text{}{\mathrm{l}}_{2},\text{}{\mathrm{l}}_{3},\text{}{\mathrm{l}}_{4},\text{}.\text{}.\text{}.\text{are the lengths of successive semicircles}\\ \text{in the spiral.}\\ \text{We know that length of semicircle is}\mathrm{\pi }\text{r.}\\ \text{So, lengths of successive semicircles are 0.5}\mathrm{\pi }\text{cm,}\mathrm{\pi }\text{cm,}\\ \text{1.5}\mathrm{\pi }\text{cm, 2}\mathrm{\pi }\text{cm,}.\dots \text{}\\ \text{Sum of lengths of 13 semicircles is}\mathrm{\pi }\left(\text{0.5}+1.0+1.5+2.0+\\ \text{}...\text{to 13th terms) cm.}\\ \text{Now,}\\ \text{List of numbers involved in sum 0.5}+1.0+1.5+2.0+\text{}...\\ \text{to 13th terms forms an AP with}\mathrm{a}=0.5\text{and}\mathrm{d}=0.5\\ \text{We know that sum of first n terms of an AP is given by}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{n}}{2}\left[2\mathrm{a}+\left(\mathrm{n}-1\right)\mathrm{d}\right]\\ \text{So,}{\mathrm{S}}_{13}=\frac{13}{2}\left[2×0.5+\left(13-1\right)\left(0.5\right)\right]=\frac{13}{2}×0.5\left[2+12\right]\\ \text{or}{\mathrm{S}}_{13}=\frac{13}{2}×0.5×14=45.5\\ \text{Therefore,}\\ \text{Sum of lengths of 13 semicircles}\\ =\mathrm{\pi }\left(\text{0.5}+1.0+1.5+2.0+...\text{to 13th terms) cm}\\ =45.5\mathrm{\pi }\text{cm}\\ =45.5×\frac{22}{7}\text{cm}\\ =143\text{cm}\\ \text{Therefore, total length of the spiral is}143\text{cm}.\end{array}$

Q.42 200 logs are stacked in the following manner: 20 logs in the bottom row, 19 in the next row, 18 in the row next to it and so on (see the following figure). In how many rows are the 200 logs placed and how many logs are in the top row? Ans.

$\begin{array}{l}\text{It is given that}\mathbf{200}\text{}\mathbf{logs}\text{}\mathbf{are}\text{}\mathbf{stacked}\text{}\mathbf{in}\text{such a way that}\\ \mathbf{20}\text{}\mathbf{logs}\text{are}\mathbf{in}\text{}\mathbf{the}\text{}\mathbf{bottom}\text{}\mathbf{row},\text{}\mathbf{19}\text{logs are}\mathbf{in}\text{}\mathbf{the}\text{}\mathbf{next}\text{}\mathbf{row},\\ \mathbf{18}\text{}\mathrm{log}\text{s are}\mathbf{in}\text{}\mathbf{the}\text{}\mathbf{row}\text{}\mathbf{next}\text{}\mathbf{to}\text{}\mathbf{it}\text{}\mathbf{and}\text{}\mathbf{so}\text{}\mathbf{on}.\\ \text{List of numbers 20, 19, 18,}...\text{involved in stacking of logs}\\ \text{form an AP with}\mathrm{a}=20,\text{}\mathrm{d}=-1\text{and}{\mathrm{S}}_{\mathrm{n}}=200.\\ \text{We know that sum of first n terms of an AP is given by}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{n}}{2}\left[2\mathrm{a}+\left(\mathrm{n}-1\right)\mathrm{d}\right]\\ \text{So,}200=\frac{\mathrm{n}}{2}\left[2×20+\left(\mathrm{n}-1\right)\left(-1\right)\right]=\frac{\mathrm{n}}{2}\left[40+-\mathrm{n}+1\right]\\ \text{or}200=\frac{\mathrm{n}}{2}\left[41-\mathrm{n}\right]\\ \text{or 400}=41\mathrm{n}-{\mathrm{n}}^{2}\\ \text{or}{\mathrm{n}}^{2}-41\mathrm{n}+400=0\\ \text{or}{\mathrm{n}}^{2}-16\mathrm{n}-25\mathrm{n}+400=0\\ \text{or}\mathrm{n}\left(\mathrm{n}-16\right)-25\left(\mathrm{n}-16\right)=0\\ \text{or}\left(\mathrm{n}-16\right)\left(\mathrm{n}-25\right)=0\\ \text{or}\mathrm{n}=16,\text{}\mathrm{n}=25\\ \text{Now,}\\ {\mathrm{a}}_{25}=\mathrm{a}+24\mathrm{d}=20+24×\left(-1\right)=-4\\ \text{But, number of logs can not be negative.}\\ \text{Therefore, number of rows of logs can not be 25 i.e.,\hspace{0.17em}}\mathrm{n}\ne 25.\\ \text{Also,}\\ {\mathrm{a}}_{16}=\mathrm{a}+15\mathrm{d}=20+15×\left(-1\right)=5\\ \text{Therefore, 200 logs are placed in 16 rows and 5 logs are}\\ \text{placed in the top row.}\end{array}$

Q.43 In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato, and the other potatoes are placed 3 m apart in a straight line. There are ten potatoes in the line(see the following figure) A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in, and she continues in the same way until all the potatoes are in the bucket. What is the total distance the competitor has to run?
[Hint: To pick up the first potato and the second potato, the total distance (in metres) run by a competitor is 2 × 5 + 2 × (5 + 3)]

Ans. $\begin{array}{l}\text{To pick up the first potato and the second potato, the total}\\ \text{distance}\left(\text{in metres}\right)\text{run by a competitor is 2×5+2×}\left(\text{5 + 3}\right)\text{.}\\ \text{Similarly, the total distance}\left(\text{in metres}\right)\text{run by a competitor}\\ \text{to pick up 10 potatoes is 2×5+2×}\left(\text{5+3}\right)\text{+2×}\left(\text{5+3+3}\right)\text{+}...\text{up to}\\ \text{10th terms.}\\ \text{Now,}\\ \text{Total distance = 2×5+2×}\left(\text{5+3}\right)\text{+2×}\left(\text{5+3+3}\right)\text{+}...\text{up to\hspace{0.17em}\hspace{0.17em}10th terms.}\\ \text{\hspace{0.17em}\hspace{0.17em} = 2}\left(\text{5+8+11+}...\text{up to 10th terms}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}= 2×}\frac{\text{10}}{\text{2}}\left[\text{2×5+(10-1)3}\right]\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[{\text{S}}_{\text{n}}\text{=}\frac{\text{n}}{\text{2}}\text{[2a+(n-1)d]}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}= 10}\left[\text{10+27}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}= 370 m}\end{array}$

Q.44 Which term of the AP: 121, 117, 113, . . ., is its first negative term?
[Hint: Find n for an < 0]

Ans.

$\begin{array}{l}\text{The given AP is:}121,\text{}117,\text{}113,\text{}.\text{}.\text{}.\\ \text{Here,}\mathrm{a}=121,\text{}\mathrm{d}=117-121=-4\\ \text{Now,}\\ {\mathrm{a}}_{\mathrm{n}}=\mathrm{a}+\left(\mathrm{n}-1\right)\text{}\mathrm{d}\\ \text{}=121+\left(\mathrm{n}-1\right)\left(-4\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\text{}121-4\mathrm{n}+4\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\text{}125-4\mathrm{n}\\ \text{We have to find the first negative term of this A.P.}\\ \mathrm{So},\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{a}}_{\mathrm{n}}<0\\ ⇒125-4\mathrm{n}<0\\ ⇒4\mathrm{n}>125\\ ⇒\mathrm{n}>\frac{125}{4}\\ ⇒\mathrm{n}>31.25\\ \therefore \mathrm{n}=32\\ \text{Hence,}{32}^{\text{nd}}\text{term of the given AP is first negative term.}\end{array}$

Q.45 The sum of the third and the seventh terms of an AP is 6 and their product is 8. Find the sum of first sixteen terms of the AP.

Ans.

$\begin{array}{l}\text{We have}\\ {\text{a}}_{3}+{\mathrm{a}}_{7}=6,{\text{a}}_{3}×{\mathrm{a}}_{7}=8\\ {\text{We know that n}}^{\text{th}}\text{terms of an AP is given by}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{a}}_{\mathrm{n}}=\mathrm{a}+\left(\mathrm{n}-1\right)\mathrm{d}\\ \text{So,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{a}}_{3}=\mathrm{a}+2\mathrm{d}\text{and}{\mathrm{a}}_{7}=\mathrm{a}+6\mathrm{d}\\ \text{Now,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{a}}_{3}+{\mathrm{a}}_{7}=6\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{a}+2\mathrm{d}+\mathrm{a}+6\mathrm{d}=6\\ ⇒\text{2}\mathrm{a}+8\mathrm{d}=6\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{a}+4\mathrm{d}=3\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{a}=3-4\mathrm{d}\text{}...\text{(1)}\\ \text{Again,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{a}}_{3}×{\mathrm{a}}_{7}=8\\ ⇒\text{}\left(\mathrm{a}+2\mathrm{d}\right)×\left(\mathrm{a}+6\mathrm{d}\right)=8\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}}\left(3-4\mathrm{d}+2\mathrm{d}\right)\left(3-4\mathrm{d}+6\mathrm{d}\right)=8\text{[From (1)]}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left(3-2\mathrm{d}\right)\left(3+2\mathrm{d}\right)=8\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}9-4{\mathrm{d}}^{2}=8\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}4{\mathrm{d}}^{2}=1\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{d}=\frac{1}{2}\text{or}-\frac{1}{2}\\ \text{On putting this value of}\mathrm{d}\text{in (1), we get}\\ \mathrm{a}=1,\text{when}\mathrm{d}=\frac{1}{2}\\ \text{and}\\ \mathrm{a}=5,\text{when}\mathrm{d}=-\frac{1}{2}\\ \text{When}\mathrm{a}=1\text{and}\mathrm{d}=\frac{1}{2}\text{then,}\\ {\mathrm{S}}_{16}=\frac{16}{2}\left[2×1+15×\frac{1}{2}\right]=76\\ \text{When}\mathrm{a}=5\text{and}\mathrm{d}=-\frac{1}{2}\text{then,}\\ {\mathrm{S}}_{16}=\frac{16}{2}\left[2×5+15×\left(-\frac{1}{2}\right)\right]=20\end{array}$

Q.46

$\begin{array}{l}\text{A ladder has rungs 25 cm apart.}\left(\text{see the following figure}\right)\text{.}\\ \text{The rungs decrease uniformly in length from 45 cm at the}\\ \text{bottom to 25 cm at the top. If the top and the bottom}\\ \text{rungs are 2}\frac{\text{1}}{\text{2}}\text{m apart, what is the length of the wood}\\ \text{required for the rungs?}\\ \left[\text{Hint:Number of rungs=}\frac{\text{250}}{\text{25}}\right]\end{array}$ Ans.

$\begin{array}{l}\text{It is given that the rungs are 25 cm apart and the top and}\\ \text{bottom rungs are}\frac{\text{1}}{\text{2}}\text{m apart.}\\ \therefore \text{Total number of rungs=}\frac{\text{2}\frac{\text{1}}{\text{2}}\text{×100}}{\text{25}}\text{+1=}\frac{\text{250}}{\text{25}}\text{+1=11}\\ {\text{Lengths of the rungs form an AP with a=45 and a}}_{\text{11}}\text{=l=25.}\\ {\text{S}}_{\text{n}}\text{=}\frac{\text{n}}{\text{2}}\left(\text{a+l}\right)\text{=}\frac{\text{11}}{\text{2}}\left(\text{45+25}\right)\text{=}\frac{\text{11}}{\text{2}}\text{×70=385}\\ \text{Therefore, the length of the wood required for the rungs is}\\ \text{385 cm.}\end{array}$

Q.47 The houses of a row are numbered consecutively from 1 to 49. Show that there is a value of x such that the sum of the numbers of the houses preceding the house numbered x is equal to the sum of the numbers of the houses following it. Find this value of x.
[Hint: Sx–1 = S49 – Sx]

Ans.

$\begin{array}{l}\text{The houses are numbered consecutively as 1, 2, 3,}...\text{, 49.}\\ \text{It can be observed that the number of houses are in an AP}\\ \text{having}\mathrm{a}\text{as 1 and}\mathrm{d}\text{also as 1.}\\ \text{Let there is a number x in this AP such that the sum of the}\\ \text{numbers of the houses preceding the house numbered x is}\\ \text{equal to the sum of the numbers of the houses following it.}\\ \text{We know that,}\\ \text{Sum of first n terms of an A.P is given by}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{n}}{2}\left[2\mathrm{a}+\left(\mathrm{n}-1\right)\mathrm{d}\right]\\ \therefore {\text{Sum of number of houses preceding x}}^{\text{th}}\text{house}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\mathrm{S}}_{\mathrm{x}-1}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{x}-1}{2}\left[2×1+\left(\mathrm{x}-1-1\right)1\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{x}-1}{2}\left[2+\left(\mathrm{x}-2\right)\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{x}\left(\mathrm{x}-1\right)}{2}\\ {\text{Sum of number of houses following x}}^{\text{th}}\text{house}\\ \text{}={\mathrm{S}}_{49}-{\mathrm{S}}_{\mathrm{x}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{49}{2}\left[2+\left(49-1\right)1\right]-\frac{\mathrm{x}}{2}\left[2+\left(\mathrm{x}-1\right)1\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=49×25-\frac{\mathrm{x}}{2}\left(1+\mathrm{x}\right)\\ \text{It is given that}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{S}}_{\mathrm{x}-1}={\mathrm{S}}_{49}-{\mathrm{S}}_{\mathrm{x}}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{x}\left(\mathrm{x}-1\right)}{2}=49×25-\frac{\mathrm{x}}{2}\left(1+\mathrm{x}\right)\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{x}}^{2}-\mathrm{x}=2450-\mathrm{x}-{\mathrm{x}}^{2}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}2{\mathrm{x}}^{2}=2450\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{x}}^{2}=1225\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}=±35\text{}\\ \text{But, number of houses can not be negative.}\\ \therefore \mathrm{x}=35\end{array}$

Q.48

$\begin{array}{l}\text{A small terrace at a football ground comprises of 15}\\ \text{steps each of which is 50 m long and\hspace{0.17em}\hspace{0.17em}built of solid}\\ \text{concrete. Each step has a rise of}\frac{\text{1}}{\text{4}}\text{m and a tread of}\\ \frac{\text{1}}{\text{2}}\text{m.}\left(\text{see the following figure}\right)\text{. Calculate the total}\\ \text{volume of concrete required to build the terrace.}\\ \text{[Hint: Volume of concrete required to build the first}\\ \text{step=}\frac{\text{1}}{\text{4}}\text{×}\frac{\text{1}}{\text{2}}{\text{×50 m}}^{\text{3}}\text{]}\end{array}$ Ans. $\begin{array}{l}\text{According to question, widths of successive steps in meter}\\ \text{from top are}\frac{1}{2},\text{\hspace{0.17em}}1,\text{\hspace{0.17em}}\frac{3}{2},\text{\hspace{0.17em}}2,\text{\hspace{0.17em}\hspace{0.17em}}...\text{which form an AP with}\mathrm{a}=\mathrm{d}=\frac{1}{2}\text{.}\\ \text{Each step is 50 m long and\hspace{0.17em}\hspace{0.17em}has a rise of}\frac{\text{1}}{\text{4}}\text{m}.\text{}\\ \text{Now,}\\ \text{Volume of concrete required to build the 1st step}\\ \text{}=\frac{\text{1}}{\text{4}}×\frac{\text{1}}{\text{2}}×{\text{50 m}}^{\text{3}}=\frac{25}{4}{\text{m}}^{\text{3}}\\ \text{Volume of concrete required to build the 2nd step}\\ \text{}=\frac{\text{1}}{\text{4}}×\text{1}×{\text{50 m}}^{\text{3}}=\frac{25}{2}{\text{m}}^{\text{3}}\\ \text{Volume of concrete required to build the 3rd step}\\ \text{}=\frac{\text{1}}{\text{4}}×\frac{3}{2}×{\text{50 m}}^{\text{3}}=\frac{75}{4}{\text{m}}^{\text{3}}\\ \text{List of numbers}\frac{25}{4},\text{\hspace{0.17em}}\frac{25}{2},\text{\hspace{0.17em}}\frac{75}{4},\text{\hspace{0.17em}\hspace{0.17em}}...\text{form an AP with}\mathrm{a}=\frac{25}{4}\text{and}\\ \mathrm{d}=\frac{25}{2}-\frac{25}{4}=\frac{25}{4}.\\ \text{We know that}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{n}}{2}\left[2\mathrm{a}+\left(\mathrm{n}-1\right)\mathrm{d}\right]\\ \mathrm{So},\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{S}}_{15}=\frac{15}{2}\left[2×\frac{25}{4}+\left(15-1\right)\frac{25}{4}\right]\\ \text{or}{\mathrm{S}}_{15}=\frac{15}{2}\left[\frac{25}{2}+14×\frac{25}{4}\right]=\frac{15}{2}\left[\frac{25}{2}+\frac{175}{2}\right]=\frac{15}{2}\left[100\right]=750\\ \text{Therefore,}\\ \text{Volume of concrete required to build the terrace is}750{\text{\hspace{0.17em}m}}^{3}.\end{array}$

Q.49

$\begin{array}{l}\text{In which of the following situations, does the list of}\\ \text{numbers involved make an arithmetic progression,}\\ \text{and why?}\\ \text{(i) The taxi fare after each km when the fare is}₹\text{15}\\ \text{for the first km and}₹\text{8 for each additional km.}\\ \text{(ii) The amount of air present in a cylinder when a}\\ \text{vacuum pump removes}\frac{1}{4}\text{of the air remaining}\\ \text{in the cylinder at a time.}\\ \text{(iii) The cost of digging a well after every metre of}\\ \text{digging, when it costs}₹\text{150 for the first metre}\\ \text{and rises by}₹\text{50 for each subsequent metre.}\\ \text{(iv) The amount of money in the account every year,}\\ \text{when}₹\text{10000 is deposited at compound interest}\\ \text{at 8 % per annum.}\end{array}$

Ans.

$\begin{array}{l}\text{(i)}\\ \text{Fare for the first km \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\hspace{0.17em}\hspace{0.17em}}₹\text{15}\\ \text{Fare for the second km =\hspace{0.17em}\hspace{0.17em}}₹\text{15 + \hspace{0.17em}}₹\text{8 = \hspace{0.17em}}₹\text{23}\\ \text{Fare for the third km =\hspace{0.17em}\hspace{0.17em}}₹\text{15 +}₹\text{8 + \hspace{0.17em}}₹\text{8 =\hspace{0.17em} }₹\text{31}\\ \text{and so on.}\\ \text{Therefore, fare for each successive kilometre in rupees}\\ \text{are:}\\ \text{15, 23, 31, 39,}...\\ \text{Difference between fares for a particular kilometre and its}\\ \text{preceding kilometre is}‘\text{8.}\\ \text{Therefore, 15, 23, 31, 39,}...\text{forms an AP.}\\ \\ \text{(ii)}\\ \text{Let the amount of air present in the cylinder initially is V.}\\ \text{The vaccum pump removes}\frac{\text{1}}{\text{4}}\text{of air remaining in the}\\ \text{cylinder each time.}\\ \text{Therefore, amounts of air in the cylinder after each}\\ \text{successive removal by the vacuum pump will be}\\ \text{V,}\frac{\text{3V}}{\text{4}}\text{,}\left(\frac{\text{3V}}{\text{4}}-\frac{\text{3V}}{4}×\frac{1}{4}\right)\text{,}\left\{\left(\frac{\text{3V}}{\text{4}}-\frac{\text{3V}}{4}×\frac{1}{4}\right)-\left(\frac{\text{3V}}{\text{4}}-\frac{\text{3V}}{4}×\frac{1}{4}\right)×\frac{1}{4}\right\}\text{,}...\\ \text{i.e., V,}\frac{\text{3V}}{\text{4}}\text{,}{\left(\frac{\text{3}}{\text{4}}\right)}^{\text{2}}\text{V,}{\left(\frac{\text{3}}{\text{4}}\right)}^{\text{3}}\text{V,}...\\ \text{Here, the differences between a term and its preceding term}\\ \text{are not equal. Therefore, list of numbers involved does}\\ \text{not make an A.P.}\\ \\ \text{(iii)}\\ \text{According to question, cost in rupees of digging first}\\ \text{metre and thereafter each subsequent metre are}\\ \text{respectively:}\\ \text{150, 150+50, 150+50+50, 150+50+50+50,}...\\ \text{150, 200, 250, 300,}...\\ \text{Differences between a term and its preceding term in the}\\ \text{above list of numbers are equal.}\\ \text{Therefore, 150, 200, 250, 300,}...\\ \text{forms an AP.}\\ \\ \text{(iv)}\\ \text{Amount at the end of first year=10000}{\left(\text{1+}\frac{\text{8}}{\text{100}}\right)}^{\text{1}}\text{=}₹\text{10800}\\ \text{Amount at the end of second year=10800}{\left(\text{1+}\frac{\text{8}}{\text{100}}\right)}^{\text{1}}\text{=}₹\text{11664}\\ \text{Amount at the end of third year=11664}{\left(\text{1+}\frac{\text{8}}{\text{100}}\right)}^{\text{1}}\text{=}₹\text{12597.12}\\ \text{Obviously, differences between amounts in any two successive}\\ \text{years are not equal. Therefore, list of numbers involved here}\\ \text{does not make an AP.}\end{array}$

## 1. What is the best way to find the Sum of an arithmetic progression?

To find the Sum of an arithmetic progression, you’ll need to know the first term’s value, the number of terms, and the common difference between each term. To arrive at the final solution, use the formula given below.

S = (n / 2)*(2a+(n−1)d)

How many types of progressions are there in Mathematics?

Ans: In Mathematics, there are three types of progressions. They are: s:

• Arithmetic progression (A.P.)
• Geometric progression (G.P.)
• Harmonic progression (H.P.)