Important Questions Class 10 Mathematics Chapter 5 – Arithmetic Progressions

Mathematics is the subject that tests the thinking ability of the students and helps them develop their cognitive skills. Hence, it is an important part of the academic curriculum and has been included in almost all the fields. This makes it the most demanding subject in higher studies. Everything about the chapter is covered in detail in the Important Questions Class 10 Mathematics Chapter 5.

Arithmetic progression is the systematic arrangement of the numerical and the patterns in the series. It helps in studying a particular group of data with ease. The important topics included in the Mathematics Class 10 Chapter 5 important questions include the definition of arithmetic progressions, nth terms of an A.P., the sum of n terms of an A.P. and lots of examples based on it.

Class 10 Chapter 5 is covered in detail in the Important Questions Class 10 Mathematics Chapter 5. It is crucial for the students to follow this chapter as it is one of the most important chapters of Mathematics. All the key points and main concepts are highlighted in it for students to easily grasp each concept. All the formulas are listed to quickly revise and recall before the examinations. This helps in building up the confidence of the students and thereby helps them score better in their examinations.

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Important Questions Class 10 Mathematics Chapter 5 – With Solutions

The following important questions and their solutions are included in the Class 10 Mathematics Chapter 

Question 1. For the AP,  d = –4, n = 7, an = 4, thus a is

(A) 6 

(B) 7 

(C) 20 

(D) 28

Answer 1:(D) 28

Explanation:We find the nth term for the AP is

an = a + (n – 1)d

here,

a = first term

an is the nth term

d is the common difference

 As per the question,

4 = a + (7 – 1)(- 4)

4 = a – 24

a = 24 + 4 = 28

Question 2. For the AP, a = 3.5, d = 0, n = 101, an will be

(A) 0 

(B) 3.5 

(C) 103.5

(D) 104.5

Answer 2:(B) 3.5

Explanation:We find the nth term of the AP. is

an= a + (n – 1)d

here,

a = first term

an is the nth term

d is the common difference

an = 3.5 + (101 – 1)0

= 3.5

(As, d = 0, it’s a constant of the AP.)

Question 3. The list for the numbers – 10, – 6, – 2, 2,… is

(A) an A.P. having d = – 16

(B) an A.P. having d = 4

(C) an A.P. having d = – 4

(D) not an A.P.

Answer 3: (B) an AP with d = 4

Explanation: As per the question,

a1 = – 10

a2 = – 6

a3 = – 2

a4 = 2

a2 – a1 = 4

a3 – a2 = 4

a4 – a3 = 4

a2 – a1 = a3– a2 = a4– a3= 4

Hence, it’s an A.P having d = 4

Question 4. The 11th term of the AP.: –5, (–5/2), 0, 5/2, …is

(A) –20 

(B) 20 

(C) –30 

(D) 30

Answer 4: (B) 20

Explanation:First term is a = – 5

The Common difference is

d = 5 – (-5/2) = 5/2

n = 11

Given that the nth term for the AP is

an = a + (n – 1)d

here,

a = first term

an is nth term

d is the common difference

a11 = – 5 + (11 – 1)(5/2)

a11 = – 5 + 25 = 20

Question 5. The first four terms of the AP., whose first term is –2 and their common difference is –2, are

(A) – 2, 0, 2, 4

(B) – 2, 4, – 8, 16

(C) – 2, – 4, – 6, – 8

(D) – 2, – 4, – 8, –16

Answer 5: (C) – 2, – 4, – 6, – 8

Explanation:For the First term, a = – 2

For the Second Term, d = – 2

a1 = a = – 2

Given that the nth term for the AP is

an = a + (n – 1)d

here,

a = first term

an is nth term

d is the common difference

So, we get,

a2 = a + d = – 2 + (- 2) = – 4

Same as

a3 = – 6

a4 = – 8

Thus, the A.P is

– 2, – 4, – 6, – 8

Question 6. The 21st term of the AP., whose first two terms are –3 and 4 is

(A) 17 

(B) 137 

(C) 143

 (D) –143

Answer 6:(B) 137

Explanation:First, two terms of the AP. are a1 = – 3 and a2 = 4.

We know, nth term for the AP is

an = a + (n – 1)d

here,

a = first term

an  is nth term

d is the common difference

a2 = a + d

4 = – 3 + d

d = 7

Common difference is d = 7

a21 = a + 20d

= – 3 + (20)(7)

= 137

Question 7. If the 2nd term of the AP. is 13 and the 5th term is 25, then what is the 7th term?

(A) 30 

(B) 33 

(C) 37 

(D) 38

Answer 7:(B) 33

Explanation:Given that the nth term for the AP is

an = a + (n – 1)d

here,

a = first term

an is nth term

d is the common difference

a2 = a + d = 13 …..(1)

a5 = a + 4d = 25 …… (2)

From the equation (1) we get,

a = 13 – d

By using it in the equation (2), we get,

13 – d + 4d = 25

13 + 3d = 25

3d = 12

d = 4

a = 13 – 4 = 9

a7= a + 6d

= 9 + 6(4)

= 9 + 24 = 33

Question 8. Which term of the AP .: 21, 42, 63, 84… is 210?

(A) 9th

(B) 10th

 (C) 11th 

(D) 12th

Answer 8: (B) 10th

Explanation: Let nth term for the given AP be 210.

As per the question,

first term is a = 21

common difference is d = 42 – 21 = 21 and an = 210

Given that the nth term for the AP is

an = a + (n – 1)d

here,

a = first term

an is the nth term

d is the common difference

210 = 21 + (n – 1)21

189 = (n – 1)21

n – 1 = 9

n = 10

Hence, 10th term for the AP is 210.

Question 9. If the common difference of the AP. is 5, what is a18 – a13?

(A) 5 

(B) 20 

(C) 25 

(D) 30

Answer 9: (C) 25

Explanation: Given that,

The common difference for the AP i.e., d = 5

Then,

We are aware that the nth term for the AP is

an = a + (n – 1)d

here a = first term

an is nth term

d is the common difference

a18 -a13 = a + 17d – (a + 12d)

= 5d

= 5(5)

= 25

Question 10: What is the common difference of the AP. for which a18 – a14 = 32?

(A) 8

(B) –8

(C) –4

(D) 4

Answer 10: (A) 8

Explanation:The nth term of the AP . is given by  an=a+(n−1)d.

Then,

 a18=a+(18−1)d, a18=a+17d  

Also,

 a14=a+(14−1)d, a14=a+13d  

We have,

a18 – a14 = 32

(a+17d)−(a+13d)=32

a+17d−a−13d=32

4d=32

d=8

 common difference for the AP is 8.

Hen

So, the correct answer is an option (A).

Question 11: Two AP’s have the same common difference. The first term for one of these is –1, and that for the other is –8. Then the difference for the 4th term is

(A) –1

(B) –8

(C) 7

(D) –9

Answer 11: (C) 7

Explanation: General terms of the AP. are a, a+d, a+2d, a+3d,……

1st A.P. with the first term −1 as well as the common difference d  is −1, −1+d, −1+2d,….

2nd A.P. with the first term −8 as well as the common difference d  is −8, −8+d, −8+2d,….

The nth term of the AP. is given by an=a+(n−1)d

The 4th term for the first A.P. is:

 a4 =−1+(3−1)d

a4 =−1+3d                       …..(1)

The 4th term for the second AP is: 

A4 =−8+(4−1)d

A4 =−8+3d               …..(2)

Subtracting the (2) from (1),

 a4−A4 =(−1+3d)−(−8+3d)           

              =−1+3d+8−3d           

              =7

The difference between their 4th term is 7.

Thus, the correct answer is an option (C).

Question 12: If 7 times the 7th term of the AP. is equal to 11 times the 11th term, the 18th term will be

(A) 7

(B) 11

(C) 18

(D) 0

Answer 12: (D) 0

Explanation: The nth term for the AP is given by 

an=a+(n−1)d.

We have,

7a7=11a11

7[a+(7−1)d]=11[a+(11−1)d]

7(a+6d)=11(a+10d)

7a+42d=11a+110d

4a+68d=0

a+17d=0

a+(18−1)d=0

a18=0

Thus, the 18th term of an AP is 0.

So, the correct answer is an option (D).

Question 13: 4th term of the end of the AP.: –11, –8, –5, …, 49 is

(A) 37

(B) 40

(C) 43

(D) 58

Answer 13: (B) 40

Explanation: The nth term from the end of the AP. =l−(n−1)d, 

here

l = last term

d = common difference

n = number of terms

Given that,

 A.P.: –11, –8, –5, …, 49, 

here

l=49

d =−8−(−11)  =−8+11=3 

The 4th term from the end =49−(4−1)×349−940

4th term from the end of the AP. is 40.

Thus, the correct answer is an option (B).

Question 14: The famous mathematician associated with finding the sum of the first 100 natural numbers is

(A) Pythagoras

(B) Newton

(C) Gauss

(D) Euclid

Answer 14: (C) Gauss

Explanation: Gauss is a famous mathematician associated for finding the sum of the first 100 natural numbers.

thus, the correct answer is an option (C).

Question 15: If the first term of the AP. is –5 and the common difference is 2, the sum of the first 6 terms is

(A) 0

(B) 5

(C) 6

(D) 15

Answer 15: (A) 0

Explanation: Given that, 

The first term of the AP. a=–5 and the common difference d=2.

Then, the sum of n terms of the AP. is 

Sn = n 2[2a+(n−1)d]

Thus, the sum of the first 6 terms is given by

S6 = 6 2[2×(−5)+(6−1)×2]S6

      = 3(−10+10)S6

      = 3×0 

      =0

The sum of the first 6 terms is 0.

So, the correct answer is an option (A).

Question 16: The sum of the first 16 terms of the A.P.: 10, 6, 2,… is

(A) –320

(B) 320

(C) –352

(D) –400

Answer 16: (A) –320

Explaination: The sum of the n terms of the AP is 

n 2[2a+(n−1)d].

Given that,

AP: 10, 6, 2,…

here

a=10,d=6−10=−4

 To find sum of the first 16 terms

S16 = 16 2[2×10+(16−1)×(−4)]

S16 = 8(20−60)S16

       = 8×(−40)S16

       = −320

 sum of the first 16 terms for the AP is −320.

Thus, the correct answer is option (A).

Question 17: For the AP a = 1, an = 20 and Sn = 399, n is

(A) 19

(B) 21

(C) 38

(D) 42

Answer 17:(C) 38

Explanation:Given that, For the AP a = 1, an = 20 and Sn = 399.

Then,

a+(n−1)d=20

1+(n−1)d=20

(n−1)d=19                     …..(1)

And

n 2[2a+(n−1)d]=399    

n[2×1+19]=399×2         [from (1)]

21n=798

   n=38

Thus, the correct answer is option (C).

Question 18: The sum of the first five multiples for the 3 is

(A) 45

(B) 55

(C) 65

(D) 75

Answer 18:

Explanation:If the first five multiples for the 3 are considered,

a = 3

n = 5

d = 3

Hence,

S5 =52[2×3+(5−1)3]               

[∵Sn=n2[2a+(n−1)d]]

52[6+12]52×1845]

Therefore, the correct answer is option (A).

Question 19. Which among the following forms an AP? State your answer.

(i) –1, –1, –1, –1,…

Answer 19 (i):

We now have,

 a1 = – 1 , a2 = – 1, a3 = – 1 and a4 = – 1

a2 – a1 = 0

a3 – a2 = 0

a4 – a3 = 0

Clearly, the difference for the successive terms is same, thus the given list for the numbers from an AP.

(ii) 0, 2, 0, 2,…

Answer (ii):

We now have 

a1 = 0, a2 = 2, a3 = 0 and a4 = 2

a2 – a1 = 2

a3 – a2 = – 2

a4 – a3 = 2

Clearly, the difference for the successive terms is not same, thus  the given list of numbers does not form an AP.

(iii) 1, 1, 2, 2, 3, 3…

Answer (iii):

We now have,

a1 = 1 , a2 = 1, a3 = 2, as well as a4 = 2

a2 – a1 = 0

a3 – a2 = 1

Clearly, the difference for the successive terms is not the same. Thus the given list of numbers does not form an AP.

(iv) 11, 22, 33…

Answer (iv):

We now have,

 a1 = 11, a2 = 22 and a3 = 33

a2 – a1 = 11

a3 – a2 = 11

Clearly, the difference for the successive terms is same. Thus the given list of numbers form an AP.

(v) 1/2,1/3,1/4, …

Answer (v):

We now have,

 a1 = ½ , a2 = 1/3 and a3 = ¼

a2 – a1 = -1/6

a3 – a2 = -1/12

Clearly, the difference for the successive terms is not the same. Thus the given list of numbers does not form an AP.

(vi) 2, 22, 23, 24, …

Answer (vi):

We now have,

a1 = 2 , a2 = 22, a3 = 23 and a4 = 24

a2 – a1 = 22 – 2 = 4 – 2 = 2

a3 – a2 = 23 – 22 = 8 – 4 = 4

Clearly, the difference for the successive terms is not the same. Thus the given list of numbers does not form an AP.

(vii) √3, √12, √27, √48, …

Answer (vii):

We now have,

a1 = √3, a2 = √12, a3 = √27 and a4 = √48

a2 – a1 = √12 – √3 = 2√3 – √3 = √3

a3 – a2 = √27 – √12 = 3√3 – 2√3 = √3

a4 – a3 = √48 – √27 = 4√3 – 3√3 = √3

Clearly, the difference for the successive terms is the same. Thus, the given list of numbers from an A.P.

Question 20. State whether it is true to show that –1, -3/2, –2, 5/2,… forms an AP as

a2 – a1 = a3 – a2.

Answer 20:

False

a1 = -1, a2 = -3/2, a3 = -2 and a4 = 5/2

a2 – a1 = -3/2 – (-1) = – ½

a3 – a2 = – 2 – (- 3/2) = – ½

a4 – a3 = 5/2 – (-2) = 9/2

Clearly, the difference for the successive terms is not the same, although a2 – a1 = a3 – a2, but for the a4 – a3 ≠ a3 – a2 thus,  it does not form an A.P.

Question  21. Of the AP.: –3, –7, –11, …, could we directly find a30 – a20 without actually finding a30 and a20? Justify your answer.

Answer 21:

Yes

Given that,

First term is  a = – 3

Common difference is d = a2 – a1 = – 7 – (- 3) = – 4

a30 – a20 = a + 29d – (a + 19d)

= 10d

= – 40

It is so because the difference between any two terms of the AP. is proportional to the common difference of the AP.

Question  22. Verify that each for the following is an AP, and write the next three terms.

(i) 0, 1/4, 1/2, 3/4,…

Answer 22 (i):

We have,

a1 = 0

a2 = ¼

a3 = ½

a4 = ¾

a2 – a1 = ¼ – 0 = ¼

a3 – a2 = ½ – ¼ = ¼

a4 – a3 = ¾ – ½ = ¼

As, difference for the successive terms are equal,

So, 0, 1/4, 1/2, 3/4… is an AP with a common difference ¼.

Thus, the next three term will be,

¾ + ¼ , ¾ + 2(¼), ¾ + 3(¼)

1, 5/4 , 3/2

(ii) 5, 14/3, 13/3, 4…

Answer 22 (ii):

We have,

a1 = 5

a2 = 14/3

a3 = 13/3

a4 = 4

a2 – a1 = 14/3 – 5 = -1/3

a3 – a2 = 13/3 – 14/3 = -1/3

a4  – a3 = 4 – 13/3 = -⅓

As the difference for the successive terms are equal,

So, 5, 14/3, 13/3, 4… is an AP with common difference -1/3.

Thus, the next three term will be,

4 + (-1/3), 4 + 2(-1/3), 4 + 3(-1/3)

11/3 , 10/3, 3

(iii) √3 , 2√3, 3√3,…

Answer 22 (iii):

We have,

a1 = √3

a2 = 2√3

a3 = 3√3

a4 = 4√3

a2 – a1 = 2√3 – √3 = √3

a3 – a2 = 3√3 – 2√3= √3

a4  – a3 = 4√3 – 3√3= √3

As, difference for the successive terms are equal,

So, √3 , 2√3, 3√3,… is an AP with common difference √3.

Thus, the next three term will be,

4√3 + √3, 4√3 + 2√3, 4√3 + 3√3

5√3, 6√3, 7√3

(iv) a + b, (a + 1) + b, (a + 1) + (b + 1), …

Answer 22 (iv):

We have,

a1 = a + b

a2 = (a + 1) + b

a3 = (a + 1) + (b + 1)

a2 – a1 = (a + 1) + b – (a + b) = 1

a3 – a2 = (a + 1) + (b + 1) – (a + 1) – b = 1

As the difference for the successive terms are equal,

So, a + b, (a + 1) + b, (a + 1) + (b + 1), … is an AP with common difference 1.

Thus, the next three term will be,

(a + 1) + (b + 1) + 1, (a + 1) + (b + 1) + 1(2), (a + 1) + (b + 1) + 1(3)

(a + 2) + (b + 1), (a + 2) + (b + 2), (a + 3) + (b + 2)

(v) a, 2a + 1, 3a + 2, 4a + 3,…

Answer 22 (v):

We have

 a1 = a

a2 = 2a + 1

 a3 = 3a + 2

a4 = 4a + 3

a2 – a1 = (2a + 1) – (a) = a + 1

a3 – a2 = (3a + 2) – (2a + 1) = a + 1

a4  – a3 = (4a + 3) – (3a+2) = a + 1

As the difference for the successive terms are equal,

So, a, 2a + 1, 3a + 2, 4a + 3,… is an AP with common difference a+1.

Thus, the next three term will be,

4a + 3 +(a + 1), 4a + 3 + 2(a + 1), 4a + 3 + 3(a + 1)

5a + 4, 6a + 5, 7a + 6

Question  23. Write the first four terms of the AP. If the first term a and the common difference d are given as:

(i) a = 10, d = 10

(ii) a = -2, d = 0

(iii) a = 4, d = – 3

(iv) a = -1 d = 1/2

(v) a = – 1.25, d = – 0.25

Answer 23:

(i) a = 10, d = 10

 Considering the Arithmetic Progression series be a1, a2, a3, a4, a5 …

a1 = a = 10

a2 = a1+d = 10+10 = 20

a3 = a2+d = 20+10 = 30

a4 = a3+d = 30+10 = 40

a5 = a4+d = 40+10 = 50

And so on…

Hence, the A.P. series will be 10, 20, 30, 40, 50 …

Also,

First four terms of the AP. will be 10, 20, 30, and 40.

(ii) a = – 2, d = 0

Considering, the Arithmetic Progression series a1, a2, a3, a4, a5 …

a1 = a = -2

a2 = a1+d = – 2+0 = – 2

a3 = a2+d = – 2+0 = – 2

a4 = a3+d = – 2+0 = – 2

Hence, the A.P. series will be – 2, – 2, – 2, – 2 …

Also,

 First four terms of this A.P. will be – 2, – 2, – 2 as well as – 2.

(iii) a = 4, d = – 3

Considering, the Arithmetic Progression series be a1, a2, a3, a4, a5 …

a1 = a = 4

a2 = a1+d = 4-3 = 1

a3 = a2+d = 1-3 = – 2

a4 = a3+d = -2-3 = – 5

Hence, the A.P. series will be 4, 1, – 2 – 5 …

Also, the first four terms of this A.P. will be 4, 1, – 2 and – 5.

(iv) a = – 1, d = 1/2

Considering, the Arithmetic Progression series a1, a2, a3, a4, a5 …

a2 = a1+d = -1+1/2 = -1/2

a3 = a2+d = -1/2+1/2 = 0

a4 = a3+d = 0+1/2 = 1/2

Hence, the A.P. series will be-1, -1/2, 0, 1/2

Also,

First four terms of this A.P. will be -1, -1/2, 0 and 1/2.

(v) a = – 1.25, d = – 0.25

Considering, the Arithmetic Progression series be a1, a2, a3, a4, a5 …

a1 = a = – 1.25

a2 = a1 + d = – 1.25-0.25 = – 1.50

a3 = a2 + d = – 1.50-0.25 = – 1.75

a4 = a3 + d = – 1.75-0.25 = – 2.00

Hence, the A.P series will be 1.25, – 1.50, – 1.75, – 2.00 ……..

Also, the first four terms of this A.P. will be – 1.25, – 1.50, – 1.75 and – 2.00.

Question  24. For the following AP.s, write the first term as well as the common difference.

(i) 3, 1, – 1, – 3 …

(ii) -5, – 1, 3, 7 …

(iii) 1/3, 5/3, 9/3, 13/3 ….

(iv) 0.6, 1.7, 2.8, 3.9 …

Answer 24:

(i) Given that the series,

3, 1, – 1, – 3 …

First term is a = 3

Common difference is d = Second term – First term

  1 – 3 = -2

 d = -2

(ii) Given that the series, – 5, – 1, 3, 7 …

First term is a = -5

Common difference is d = Second term – First term

( – 1)-( – 5) 

= – 1+5 = 4

(iii) Given that the series, 1/3, 5/3, 9/3, 13/3 ….

First term is a = 1/3

Common difference is d = Second term – First term

 5/3 – 1/3 

= 4/3

(iv) Given that the series, 0.6, 1.7, 2.8, 3.9 …

First term is a = 0.6

Common difference is d = Second term – First term

 1.7 – 0.6

1.1

Question  25. Which among the following are AP.s?  Find out the common difference d as well as write three more terms.

(i) 2, 4, 8, 16 …

(ii) 2, 5/2, 3, 7/2 ….

(iii) -1.2, -3.2, -5.2, -7.2 …

(iv) -10, – 6, – 2, 2 …

(v) 3, 3 + √2, 3 + 2√2, 3 + 3√2

(vi) 0.2, 0.22, 0.222, 0.2222 ….

(vii) 0, – 4, – 8, – 12 …

(viii) -1/2, -1/2, -1/2, -1/2 ….

(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 …

Answer 25:

(i) Given that,

2, 4, 8, 16 …

Where, the common difference is;

a2 – a1= 4 – 2 = 2

a3 – a2 = 8 – 4 = 4

a4 – a3 = 16 – 8 = 8

As an+1 – an or the common difference is not the same every time.

Hence, the given series are not forming an A.P.

(ii) Given that,

 2, 5/2, 3, 7/2 ….

Where,

a2 – a1 = 5/2-2 = 1/2

a3 – a2 = 3-5/2 = 1/2

a4 – a3 = 7/2-3 = ½

As an+1 – an or the common difference is the same every time.

Thus,, d = 1/2, and the given series are in A.P.

The next three terms are;

a5 = 7/2+1/2 = 4

a6 = 4 +1/2 = 9/2

a7  = 9/2 +1/2 = 5

(iii) Given that, -1.2, – 3.2, -5.2, -7.2 …

Where,

a2 – a1 = (-3.2)-(-1.2) = -2

a3 – a2 = (-5.2)-(-3.2) = -2

a4 – a3 = (-7.2)-(-5.2) = -2

As an+1 – an or common difference is the same every time.

Thus, d = -2 and the given series are in A.P.

So, next three terms are;

a5 = – 7.2-2 = -9.2

a6 = – 9.2-2 = – 11.2

a7 = – 11.2-2 = – 13.2

(iv) Given that,

 -10, – 6, – 2, 2 …

Where, the terms and their difference are;

a2 – a1 = (-6)-(-10) = 4

a3 – a2 = (-2)-(-6) = 4

a4 – a3 = (2 -(-2) = 4

As, an+1 – an or the common difference is the same every time.

Thus, d = 4 and the given numbers are in A.P.

So, next three terms are;

a5 = 2+4 = 6

a6 = 6+4 = 10

a7 = 10+4 = 14

(v) Given that, 3, 3+√2, 3+2√2, 3+3√2

Where,

a2 – a1 = 3+√2-3 = √2

a3 – a2 = (3+2√2)-(3+√2) = √2

a4 – a3 = (3+3√2) – (3+2√2) = √2

As an+1 – an or the common difference is the same every time.

Thus, d = √2, and the given series forms an A.P.

So, the next three terms are given;

a5 = (3+√2) +√2 = 3+4√2

a6 = (3+4√2)+√2 = 3+5√2

a7 = (3+5√2)+√2 = 3+6√2

(vi) 0.2, 0.22, 0.222, 0.2222 ….

Where,

a2 – a1 = 0.22-0.2 = 0.02

a3 – a2 = 0.222-0.22 = 0.002

a4 – a3 = 0.2222-0.222 = 0.0002

As an+1 – an or the common difference is not the same every time.

Thus, the given series doesn’t form an A.P.

(vii) 0, -4, -8, -12 …

Where,

a2 – a1 = (-4)-0 = -4

a3 – a2 = (-8)-(-4) = -4

a4 – a3 = (-12)-(-8) = -4

As an+1 – an or the common difference is the same every time.

Thus, d = -4, and the given series form an A.P.

So, the next three terms are given;

a5= -12-4 = -16

a6 = -16-4 = -20

a7 = -20-4 = -24

(viii) -1/2, -1/2, -1/2, -1/2 ….

Where,

a2 – a1 = (-1/2) – (-1/2) = 0

a3 – a2 = (-1/2) – (-1/2) = 0

a4 – a3 = (-1/2) – (-1/2) = 0

As an+1 – an or the common difference is the same every time.

Thus, d = 0, and the given series form an A.P.

So, next three terms are;

a5 = (-1/2)-0 = -1/2

a6 = (-1/2)-0 = -1/2

a7 = (-1/2)-0 = -1/2

(ix) 1, 3, 9, 27 …

Where,

a2 – a1 = 3-1 = 2

a3 – a2 = 9-3 = 6

a4 – a3 = 27-9 = 18

As an+1 – an or the common difference is not the same every time.

Thus, the given series doesn’t form an A.P.

(x) a, 2a, 3a, 4a …

Where,

a2 – a1 = 2a–a = a

a3 – a2 = 3a-2a = a

a4 – a3 = 4a-3a = a

As an+1 – an or the common difference is the same every time.

Thus, d = a, and the given series form an A.P.

So, the next three terms are;

a5 = 4a+a = 5a

a6 = 5a+a = 6a

a7  = 6a+a = 7a

(xi) a, a2, a3, a4 …

Where,

a2 – a1 = a2–a = a(a-1)

a3 – a2 = a3 – a2 = a2(a-1)

a4 – a3 = a4 – a3 = a3(a-1)

As the an+1 – an or the common difference is not the same every time.

Thus, the given series doesn’t form an A.P.

(xii) √2, √8, √18, √32 …

Where,

a2 – a1 = √8-√2  = 2√2-√2 = √2

a3 – a2 = √18-√8 = 3√2-2√2 = √2

a4 – a3 = 4√2-3√2 = √2

As an+1 – an or the common difference is same every time.

Thus, d = √2 and the given series forms a A.P.

So, next three terms are given;

a5 = √32+√2 = 4√2+√2 = 5√2 = √50

a6  = 5√2+√2 = 6√2 = √72

a7 = 6√2+√2 = 7√2 = √98

(xiii) √3, √6, √9, √12 …

Where,

a2 – a1 = √6-√3 = √3×√2-√3 = √3(√2-1)

a3 – a2 = √9-√6 = 3-√6 = √3(√3-√2)

a4 – a3 = √12 – √9 = 2√3 – √3×√3 = √3(2-√3)

As an+1 – an or the common difference is not the same every time.

Thus, the given series doesn’t form a A.P.

(xiv) 12, 32, 52, 72 …

Also, 1, 9, 25, 49 …..

Where,

a2 − a1 = 9−1 = 8

a3 − a2 = 25−9 = 16

a4 − a3 = 49−25 = 24

As an+1 – an or the common difference is not the same every time.

Thus,  the given series doesn’t form a A.P.

(xv) 12, 52, 72, 73 …

 1, 25, 49, 73 …

Where,

a2 − a1 = 25−1 = 24

a3 − a2 = 49−25 = 24

a4 − a3 = 73−49 = 24

As an+1 – an or the common difference is same every time.

Thus, d = 24 and the given series forms a A.P.

So, the next three terms are;

a5 = 73+24 = 97

a6 = 97+24 = 121

a7 = 121+24 = 145

Question  26. Write the first three terms of the APs if a and d are given below as:

1. a =1/2, d = -1/6
2. a = –5, d = –3
3. a = √2 , d = 1/√2

Answer 26:

(i) a =1/2, d = -1/6

We have,

First three terms for the AP are :

a, a + d, a + 2d

½, ½ + (-1/6), ½ + 2 (-1/6)

½, 1/3, 1/6

(ii) a = –5, d = –3

We have,

First three terms of the AP. are :

a, a + d, a + 2d

-5, – 5 + 1 (- 3), – 5 + 2 (- 3)

– 5, – 8, – 11

(iii) a = √2 , d = 1/√2

We have,

First three terms for the AP are :

a, a + d, a + 2d

√2, √2+1/√2, √2+2/√2

√2, 3/√2, 4/√2

Question  27. Find out a, b and c so that the following numbers are of the AP.: a, 7, b, 23, c.

Answer 27:

as For a, 7, b, 23, c… to be for AP

It has to satisfy the condition,

a5 – a4 = a4 – a3 = a3 – a2 = a2 – a1= d

here d is the common difference

7 – a = b – 7 = 23 – b = c – 23 …(1)

equating,

b – 7 = 23 – b

2b = 30

b = 15 (eqn 1)

And,

7 – a = b – 7

From the eqn 1

7 – a = 15 – 7

a = – 1

Also,

c – 23 = 23 – b

c – 23 = 23 – 15

c – 23 = 8

c = 31

Hence, a = – 1

b = 15

c = 31

Now, we say that, the sequence for the – 1, 7, 15, 23, 31 is an A.P.

Question  28. Determine the A.P. in which the fifth term is 19 as well as the difference of the eighth term from the thirteenth term is 20.

Answer 28:

We have,

The first term for the AP = a

Also, the common difference = d.

As per the question,

5th term for a5 = 19

By the nth term formula,

an = a + (n – 1)d

We find,

a + 4d = 19

a = 19 – 4d …(1)

And,

13th term – 8th term = 20

a + 12d – (a + 7d) = 20

5d = 20

d = 4

Putting d = 4 in the equation 1,

We have,

a = 19 – 4(4)

a = 3

Now, the AP becomes,

3, 3 + 4 , 3 + 2(4),…

3, 7, 11,…

Question  29. The eighth term of the AP. is half its second term and the eleventh term exceeds one third of the fourth term by 1. Find out the 15th term.

Answer 29:

We have,

First term for the AP = a

Common difference for the AP = d

nth  term for the AP, an = a + (n – 1)d

As per the question,

as = ½ a2

2a8 = a2

2(a + 7d) = a + d

2a + 14d = a + d

a = – 13d …(1)

And

a11 = 1/3 a4 + 1

3(a + 10d) = a + 3d + 3

3a + 30d = a + 3d + 3

2a + 27d = 3

Puting a = -13d in the equation,

2 (- 13d) + 27d = 3

d = 3

Now,

a = – 13(3)= – 39

Then,

a15  = a + 14d

= – 39 + 14(3)

= – 39 + 42

= 3

Thus, the 15th term is 3.

Question  30. An A.P. contains 37 terms. The sum of the three middle most terms is 225 and the sum of the last three is 429. Find out the A.P.

Answer 30:

We have,

First term for the AP = a

Common difference for the AP = d

nth term for the AP, an = a + (n – 1)d

As, n = 37 (odd),

Middle term will be (n+1)/2 = 19th term

So, the three middle most terms will be,

18th, 19th and 20th terms

As per the question,

a18 + a19 + a20 = 225

By using an = a + (n – 1)d

a + 17d + a + 18d + a + 19d = 225

3a + 54d = 225

3a = 225 – 54d

a = 75 – 18d … (1)

Then, we know that the last three terms will be 35th, 36th and 37th terms.

As per the question,

a35 + a36 + a37 = 429

a + 34d + a + 35d + a + 36d = 429

3a + 105d = 429

a + 35d = 143

Putting a = 75 – 18d from the equation 1,

75 – 18d + 35d = 143 [ using eqn1]

17d = 68

d = 4

So,

a = 75 – 18(4)

a = 3

Thus, the AP is a, a + d, a + 2d….

i.e. 3, 7, 11….

Question  31. Check if -150 is a term of the A.P. 11, 8, 5, 2, …

Answer 31:

For the given series, of the A.P. 11, 8, 5, 2..

First-term is, a = 11

Common difference is, d = a2−a1 = 8−11 = −3

Let −150 be the nth term of the AP.

As you know, for an A.P.,

an = a+(n−1)d

-150 = 11+(n -1)(-3)

-150 = 11-3n +3

-164 = -3n

n = 164/3

It is clear, n is not an integer but a fraction.

Thus, – 150 is not a term of this A.P.

Question  32. Find out the 31st term of the AP. whose 11th term is 38 and the 16th term is 73.

Answer 32:

It is given that,

11th term, a11 = 38

also 16th term, a16 = 73

We know,

an = a+(n−1)d

a11 = a+(11−1)d

38 = a+10d ………………………………. (i)

In the same way,

a16 = a +(16−1)d

73 = a+15d ………………………………………… (ii)

Subtracting the equation (i) from (ii), we find,

35 = 5d

d = 7

From the equation (i), we can write,

38 = a+10×(7)

38 − 70 = a

a = −32

a31 = a +(31−1) d

= − 32 + 30 (7)

= − 32 + 210

= 178

So, 31st term is 178.

Question  33. An A.P. consists of 50 terms of which the 3rd term is 12 and the last term is 106. Find the 29th term.

Answer 33: 

Given that,

3rd term is a3 = 12

50th term is  a50 = 106

We have,

an = a+(n−1)d

a3 = a+(3−1)d

12 = a+2d ……………………………. (i)

Similarly,,

a50 = a+(50−1)d

106 = a+49d …………………………. (ii)

On subtracting the equation (i) from (ii), we get

94 = 47d

d = 2 = common difference

From the equation (i), we can now write 

12 = a+2(2)

a = 12−4 = 8

a29 = a+(29−1) d

a29 = 8+(28)2

a29 = 8+56 = 64

Hence, the 29th term is 64.

Question  34. If the 3rd and the 9th terms of the AP. are 4 and − 8 respectively. Which term of the AP. is zero.

Answer 34:

Given that,

3rd term is a3 = 4

also 9th term is, a9 = −8

We have,

an = a+(n−1)d

Therefore,

a3= a+(3−1)d

4 = a+2d ……………………………………… (i)

a9 = a+(9−1)d

−8 = a+8d ………………………………………………… (ii)

Subtracting equation (i) from (ii), we get,

−12 = 6d

d = −2

From the equation (i), we can now write,

4 = a+2(−2)

4 = a−4

a = 8

Let the nth term of the AP. be zero.

an = a+(n−1)d

= 8+(n−1)(−2)

 = 8−2n+2

2n = 10

n = 5

So, the 5th term of the AP. is 0.

Question  35. If 17th term of the AP. exceeds its 10th term by 7. Find out the common difference.

Answer 35:

We have, for an A.P series;

an= a+(n−1)d

a17 = a+(17−1)d

a17 = a +16d

Similarly,

 a10 = a+9d

As per the question,

a17 − a10 = 7

Hence,

(a +16d)−(a+9d) = 7

7d = 7

d = 1

Thus, the common difference is 1.

Question  36. Which term of of the AP. 3, 15, 27, 39,.. will be 132 more than the 54th term?

Answer 36:

Given that,

 A.P. is 3, 15, 27, 39, …

first term is a = 3

common difference is d = a2 − a1 = 15 − 3 = 12

We have,

an = a+(n−1)d

Hence,

a54 = a+(54−1)d

⇒3+(53)(12)

⇒3+636 = 639

a54 = 639+132=771

We have to find the term of the AP. which is 132 more than a54, i.e.771.

Let the nth term be 771.

an = a+(n−1)d

771 = 3+(n −1)12

768 = (n−1)12

(n −1) = 64

n = 65

Thus, 65th term was 132 more than 54th term.

Or another method is;

Let the nth term be 132 more than the 54th term.

n = 54 + 132/2

= 54 + 11 

= 65th term

Question  37. The sum of the first five terms of the AP. and the sum of the first seven terms for the same A.P. is 167. If the sum of the first ten terms of the AP. is 235, find out the sum of the first twenty terms.

Answer 37:

We find, of the AP.,

First term = a

Common difference = d

Number of terms for the AP = n

As per the question,

We get,

S5+ S7 = 167

By the formula for sum of the n terms,

Sn = (n/2) [2a + (n-1)d]

Hence, we find,

(5/2) [2a + (5-1)d] + (7/2)[2a + (7-1)d] = 167

5(2a + 4d) + 7(2a + 6d) = 334

10a + 20d + 14a + 42d = 334

24a + 62d = 334

12a + 31d = 167

12a = 167 – 31d …(1)

We find,

S10 = 235

(10/2) [2a + (10-1)d] = 235

5[ 2a + 9d] = 235

2a + 9d = 47

Multiplying L.H.S and R.H.S with 6,

We find,

12a + 54d = 282

From the equation (1)

167 – 31d + 54d = 282

23d = 282 – 167

23d = 115

d = 5

Substituting the value for the d = 5 in the equation (1)

12a = 167 – 31(5)

12a = 167 – 155

12a = 12

a = 1

Given that,

S20 = (n/2) [2a + (20 – 1)d]

= 20/(2[2(1) + 19 (5)])

= 10[ 2 + 95]

= 970

Hence, the sum of the first 20 terms is 970.

Question  38. Find out the

(i) sum of the integers between 1 and 500 which are multiples of 2 as well as of 5.

(ii) sum of the integers from 1 to 500 which are multiples of 2 as well as of 5 .

(iii) sum of the integers from 1 to 500 which are multiples of 2 or 5.

[Hint (iii): These numbers will be: multiples for the 2 + multiples for the 5 – multiples for the 2 as well as of 5]

Answer 38:

(i) sum of the integers between 1 and 500 which are multiples for the 2 as well as of 5.

Given that,

Multiples for the 2 as well as of 5 = LCM of (2, 5) = 10

Multiples for the 2 as well as of 5 between 1 and 500 = 10, 20, 30…, 490.

Therefore,

We can conclude that 10, 20, 30…, 490 is an AP having the common difference, d = 10

First term is a = 10

Let the number for the terms in this AP = n

By the nth term formula,

an = a + (n – 1)d

490 = 10 + (n – 1)10

480 = (n – 1)10

n – 1 = 48

n = 49

sum of the AP,

Sn  = (n/2) [a + an], Where an is the last term, that is given]

= (49/2) × [10 + 490]

= (49/2) × [500]

= 49 × 250

= 12250

Hence, sum of the those integers between 1 and 500 which are multiples for the 2 as well as of 5 = 12250

(ii) sum of the integers from 1 to 500 which are multiples for the 2 as well as of 5.

Given that,

Multiples for the 2 as well as of 5 = LCM of (2, 5) = 10

Multiples for the 2 as well as of 5 from 1 and 500 = 10, 20, 30…, 500.

Therefore,

We conclude that 10, 20, 30…, 500 is an AP having the common difference, d = 10

First term is a = 10

Let the number of terms for the AP = n

By the nth term formula,

an = a + (n – 1)d

500 = 10 + (n – 1)10

490 = (n – 1)10

n – 1 = 49

n = 50

sum of the AP,

Sn = (n/2) [ a + an], Where an is the last term, that is given]

= (50/2) ×[10+500]

= 25× [10 + 500]

= 25(510)

= 12750

Hence, sum of the those integers from 1 to 500 which are multiples for the 2 as well as of 5 = 12750

(iii) Sum of those integers from 1 to 500 which are multiples for 2 or 5.

We are aware that,

Multiples for the 2 or 5 = Multiple for the 2 + Multiple of 5 – Multiple for the LCM (2, 5)

Multiples for the 2 or 5 = Multiple for the 2 + Multiple of 5 – Multiple for the LCM (10)

Multiples for the 2 or 5 from 1 to 500 = List for the multiple of 2 from 1 to 500 + List for the multiple

of 5 from 1 to 500 – List for the multiple of 10 ranging from 1 to 500

= (2, 4, 6… 500) + (5, 10, 15… 500) – (10, 20, 30… 500)

Required sum = sum of the (2, 4, 6,…, 500) + sum of the (5, 10, 15,…, 500) – sum of the (10, 20, 30,., 500)

Considering the first series,

2, 4, 6, …., 500

First term is a = 2

Common difference is d = 2

Let n be number of the terms

an = a + (n – 1)d

500 = 2 + (n – 1)2

498 = (n – 1)2

n – 1 = 249

n = 250

sum of the AP, Sn = (n/2) [ a + an]

Let the sum of the AP be S1,

S1 = S250 = (250/2) ×[2+500]

S1 = 125(502)

S1 = 62750 … (1)

Considering the second series,

5, 10, 15, …., 500

First term is a = 5

Common difference is d = 5

Let n be number of the terms

Using the nth term formula

an = a + (n – 1)d

500 = 5 + (n – 1)

495 = (n – 1)5

n – 1 = 99

n = 100

sum of the AP, Sn = (n/2) [ a + an]

Let the sum of the AP be S2,

S2 = S100 = (100/2) ×[5+500]

S2 = 50(505)

S2 = 25250 … (2)

Considering the third series,

10, 20, 30, …., 500

First term is a = 10

Common difference is d = 10

Let n be number of the terms

an = a + (n – 1)d

500 = 10 + (n – 1)10

490 = (n – 1)10

n – 1 = 49

n = 50

sum of the AP, Sn= (n/2) [ a + an]

Let the sum of the AP be S3,

S3 = S50 = (50/2) × [2+510]

S3 = 25(510)

S3 = 12750 … (3)

So, the required Sum is 

S = S1+ S2 – S3

S = 62750 + 25250 – 12750

= 75250

5. Find out the sum of the integers between 100 and 200 that are

 

(i) divisible by 9

(ii) not divisible by 9

[Hint (ii): These numbers obtained will be: Total numbers – Total numbers divisible by 9]

Answer:

(i) The number between 100 and 200 that is divisible with 9 = 108, 117, 126, …198

Let the number of the terms between 100 and 200 that is divisible with 9 = n

an = a + (n – 1)d

198 = 108 + (n – 1)9

90 = (n – 1)9

n – 1 = 10

n = 11

sum of the AP = Sn = (n/2) [ a + an]

Sn = (11/2) × [108 + 198]

= (11/2) × 306

= 11(153)

= 1683

(ii) sum of the integers between 100 and 200 which is not divisible by 9 = (sum of the total numbers between 100 and 200) – (sum of the total numbers between 100 and 200 which is divisible by 9)

Sum is S = S1 – S2

Where,

S1 = sum of the AP 101, 102, 103, – – – , 199

S2 = sum of the AP 108, 117, 126, – – – , 198

For an AP 101, 102, 103, – – – , 199

First-term is a = 101

Common difference is d = 199

Let number of the terms = n

Then,

an = a + (n – 1)d

199 = 101 + (n – 1)1

98 = (n – 1)

n = 99

sum of the AP = Sm = (n/2) [ a + an]

Sum of the AP.,

S1 = (99/2) × [199 + 101]

= (99/2) × 300

= 99(150)

= 14850

For an AP 108, 117, 126, – – – – , 198

First-term is a = 108

Common difference is d = 9

Last term is an = 198

Let number of the terms = n

Then,

an = a + (n – 1)d

198 = 108 + (n – 1)9

10 = (n – 1)

n = 11

sum of the AP = Sn = (n/2) [ a + an]

Sum of the AP.,

S2 = (11/2) × [108 + 198]

= (11/2) × (306)

= 11(153)

= 1683

Putting the value of S1 and S2 in this equation, we get,

S = S1– S2

S = S1+ S2

= 14850 – 1683

= 13167

Question  39. Two A.P.s have equal common difference. The difference for the 100th term is 100, what is the difference for the 1000th terms?

Answer 39:

Let, the first term for the two APs be a1 and a2 respectively

Their common difference for the APs be d.

For the first A.P.,we get,

an = a+(n−1)d

Therefore,

a100 = a1+(100−1)d

= a1 + 99d

a1000 = a1+(1000−1)d

a1000 = a1+999d

For the second A.P., we get,

an = a+(n−1)d

So,

a100 = a2+(100−1)d

= a2+99d

a1000= a2+(1000−1)d

= a2+999d

We know, the difference between 100th term of the two APs = 100

Therefore, (a1+99d) − (a2+99d) = 100

a1−a2 = 100……………………………………………………………….. (i)

Difference between 1000th terms for the two APs

(a1+999d) − (a2+999d) = a1−a2

From the equation (i),

The difference is a1−a2 = 100

So, the difference between 1000th terms for the two A.P. will be 100.

 

Question  40. How many three digit numbers are divisible by 7?

Answer 40:

First three-digit number which is divisible by 7 are;

For First number = 105

For Second number = 105+7 = 112

For Third number = 112+7 =119

Hence, 105, 112, 119, …

All are three-digit numbers are divisible by 7, and hence all these are terms of the AP. having first term as 105 as well as the common difference as 7.

Then, the largest possible three-digit number is 999.

When we divide 999 with 7, the remainder will be 5.

thus, 999-5 = 994 is the maximum possible three-digit number which is divisible by 7.

The series becomes,

105, 112, 119, …, 994

Let 994 be the nth term of the AP.

first term, a = 105

The common difference is d = 7

an = 994

n = ?

As 

Now,

an = a+(n−1)d

994 = 105+(n−1)7

889 = (n−1)7

(n−1) = 127

n = 128

Hence, 128 three-digit numbers are divisible by 7.

Question  41. How many multiples for the 4 lie between 10 and 250?

Answer 41:

The first multiple for the 4 which is greater than 10 is 12.

Next multiple for it will be 16.

Hence, the series formed as;

12, 16, 20, 24, …

All these are divisible with 4, and hence, all these are terms of the AP. by the first term as 12 and the common difference is 4.

When we divide 250 with 4, the remainder for it will be 2. Thus, 250 − 2 = 248 is divisible by 4.

The series becomes

12, 16, 20, 24, …, 248

Let 248 be the nth term of the AP.

The first term is a = 12

The common difference is d = 4

an = 248

Now,

an = a+(n−1)d

248 = 12+(n-1)×4

236/4 = n-1

59  = n-1

n = 60

Hence, there are 60 multiples for the 4 between 10 and 250.

Question  42. For what value of the n, are the nth terms for the two A.P.s 63, 65, 67, and 3, 10, 17, … same?

Answer 42:

Given the two APs are; 

63, 65, 67,… and 3, 10, 17,….

Taking the first AP,

63, 65, 67, …

The First term is a = 63

Common difference is d = a2−a1 = 65−63 = 2

We get, the nth term of the AP. = an = a+(n−1)d

an = 63+(n−1)2 = 63+2n−2

an = 61+2n ………………………………………. (i)

Taking the second AP,

3, 10, 17, …

First term is a = 3

Common difference is d = a2− a1 = 10 − 3 = 7

We have,

nth term of the AP. = 3+(n−1)7

an = 3+7n−7

an = 7n−4 ……………………………………………………….. (ii)

Given that the nth term of the A.P.s are equal to each other.

Equating both the equations, we find,

61+2n = 7n−4

61+4 = 5n

5n = 65

n = 13

Thus, 13th terms of both these A.P.s are equal to each other.

Question  43. Find out the A.P. whose third term is 16 and the 7th term exceeds the 5th term by 12.

Answer 43:

Given that,

The Third term is a3 = 16

Now,

a +(3−1)d = 16

a+2d = 16 ………………………………………. (i)

We have, 7th term exceeds the 5th term with 12.

a7− a5 = 12

[a+(7−1)d]−[a +(5−1)d]= 12

(a+6d)−(a+4d) = 12

2d = 12

d = 6

From the equation (i), we find,

a+2(6) = 16

a+12 = 16

a = 4

Hence, A.P. will be 4, 10, 16, 22, …

Question  44. Find out the 20th term from the last term of the AP. 3, 8, 13, …, 253.

Answer 44:

Given that,

 A.P. is3, 8, 13, …, 253

Common difference is d= 5.

So, we can write the AP in the reverse order as;

253, 248, 243, …, 13, 8, 5

In the case of the new AP,

first term is a = 253

and the common difference is d = 248 − 253 = −5

n = 20

So, using nth term formula, we find,

a20 = a+(20−1)d

a20 = 253+(19)(−5)

a20 = 253−95

a = 158

Hence, 20th term from the last term of the AP. 3, 8, 13, …, 253.is 158.

Question  45. The sum of  the 4th and 8th terms of of the AP. is 24 and the sum of the 6th and 10th terms is 44. Find out the first three terms of of the AP.

Answer 45:

We find, the nth term for the AP is;

an= a+(n−1)d

a4 = a+(4−1)d

a4 = a+3d

Similarly, we get,

a8 = a+7d

a6= a+5d

a10 = a+9d

Given that,

a4+a8= 24

a+3d+a+7d = 24

2a+10d = 24

a+5d = 12 …………………………………………………… (i)

a6+a10 = 44

a +5d+a+9d = 44

2a+14d = 44

a+7d = 22 …………………………………….. (ii)

On subtracting the equation (i) from (ii), we can find,

2d = 22 − 12

2d = 10

d = 5

From the equation (i), we can find,

a+5d = 12

a+5(5) = 12

a+25 = 12

a = −13

a2 = a+d = − 13+5 = −8

a3 = a2+d = − 8+5 = −3

Thus, the first three terms of the AP. are −13, −8, and −3.

Question  46. Subba Rao started her work in 1995 at an annual salary of Rs 5000 and received an increment for the Rs 200 each year. In which year did his income reach Rs 7000?

Answer 46:

From the given question, 

The incomes for the Subba Rao increases every year by Rs.200 and thus forms an A.P.

Hence, after 1995, the salaries for the each year are;

5000, 5200, 5400, …

Where, first term is a = 5000

and the common difference is d = 200

Assume after nth year, his salary will be Rs 7000.

Thus, by the nth term formula for the AP,

an = a+(n−1) d

7000 = 5000+(n−1)200

200(n−1)= 2000

(n−1) = 10

n = 11

So, for the 11th year, his salary will be Rs 7000.

Question  47. Ramkali saved Rs 5 in the first week of the year as well as increased her weekly saving by Rs 1.75. If in the nth week, her weekly savings become Rs 20.75, find out the n.

Answer 47:

Given,

 Ramkali saved Rs.5 in the first week as well as started saving each week by Rs.1.75.

So,

The first term is a = 5

and the common difference is d = 1.75

Also,

an = 20.75

To Find n 

Using the nth term formula,

an = a+(n−1)d

Hence,

20.75 = 5+(n -1)×1.75

15.75 = (n -1)×1.75

(n -1) = 15.75/1.75 = 1575/175

= 63/7 = 9

n -1 = 9

n = 10

Thus, n is 10.

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