NCERT Solutions for Class 10 Maths Chapter 5 – Arithmetic Progressions Exercise 5.3

NCERT Solutions for Class 10 Maths Chapter 5 – Arithmetic Progressions Exercise 5.3 are provided to help students strengthen their understanding of Arithmetic Progressions (AP) and prepare effectively for the Class 10 board exams. These solutions are prepared by subject experts and follow the latest CBSE syllabus and exam pattern.

Exercise 5.3 mainly focuses on finding the nth term of an Arithmetic Progression using the formula:

NCERT Solutions for Class 10 Maths Chapter 5 – Arithmetic Progressions Exercise 5.3

Arithmetic Progression

Medium

Q.

Fill in the blanks in the following table, given that a is the first term, d the common difference and a_n the nth term of the AP:

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Arithmetic Progression

Medium

Q.

$\begin{array}{l}\text{In an AP:}\\ \text{(i) given }\mathrm{a}=5,\mathrm{d}=3,{\mathrm{a}}_{\mathrm{n}}=50\text{, find n and }{\mathrm{S}}_{\mathrm{n}}\text{.}\\ \text{(ii) given }\mathrm{a}=7,{\mathrm{a}}_{13}=35\text{, find d and }{\mathrm{S}}_{13}\text{.}\\ \text{(iii) given }{\mathrm{a}}_{12}=37,\mathrm{d}=3\text{, find a and }{\mathrm{S}}_{12}\text{.}\\ \text{(iv) given }{\mathrm{a}}_3=15,{\mathrm{S}}_{10}=125,\text{ find }\mathrm{d}\text{ and }{\mathrm{a}}_{10}\text{.}\\ \text{(v) given }\mathrm{d}=5,{\mathrm{S}}_9=75,\text{ find }\mathrm{a}\text{ and }{\mathrm{a}}_9\text{.}\\ \text{(vi) given }\mathrm{a}=2,\mathrm{d}=8,{\mathrm{S}}_{\mathrm{n}}=90,\text{ find }\mathrm{n}\text{ and }{\mathrm{a}}_{\mathrm{n}}\text{.}\\ \text{(vii) given }\mathrm{a}=8,{\mathrm{a}}_{\mathrm{n}}=62,{\mathrm{S}}_{\mathrm{n}}=210,\text{ find }\mathrm{n}\text{ and }\mathrm{d}\text{.}\\ \text{(viii) given }{\mathrm{a}}_{\mathrm{n}}=4,\mathrm{d}=2,{\mathrm{S}}_{\mathrm{n}}=-14,\text{ find }\mathrm{n}\text{ and }\mathrm{a}\text{.}\\ \text{(ix) given }\mathrm{a}=3,\mathrm{n}=8,\mathrm{S}=192,\text{ find }\mathrm{d}\text{.}\\ \text{(x) given }\mathrm{l}=28,\mathrm{S}=144,\text{ and there are total 9 terms.}\\ \text{ Find\hspace{0.33em}}\mathrm{a}\text{.}\end{array}$

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Arithmetic Progression

Difficult

Q.

In an A.P. of 40 terms, the sum of first 9 terms is 153 and the sum of last 6 terms is 687. Determine the first term and common difference of A.P. Also, find the sum of all the terms of the A.P.

[CBSE - 2024]

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Arithmetic Progression

Medium

Q.

The sum of first and eighth terms of an A.P. is 32 and their product is 60 . Find the first term and common difference of the A.P. Hence, also find the sum of its first 20 terms.

[CBSE - 2024]

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Arithmetic Progression

Medium

Q.

If the sum of first four terms of an AP is 40 and that of first 14 terms is 280. Find the sum of its first n terms.

[CBSE - 2019]

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Arithmetic Progression

Medium

Q.

If $$\mathrm{S}_n,$$​ the sum of first n terms of an AP is given by $$\mathrm{S}_n=3 n^2-4 n,$$​ find the $$n^{\rm th}$$ term.

[CBSE - 2019]

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Arithmetic Progression

Medium

Q.

Which term of the AP 3, 15, 27, 39, … will be 120 more than its 21st term?

[CBSE - 2019]

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Arithmetic Progression

Easy

Q.

In an A.P., if the first term a = 7, $$n^{th}$$​ term $$a_n=84$$​ and the sum of first n terms $$s_n=\frac{2093}{2}$$​, then n is equal to:

[CBSE - 2024]

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Arithmetic Progression

Medium

Q.

If the sum of first m terms of an AP is $$\mathrm{AP}\text{ is }2\mathrm{~m}^2+3\mathrm{~m}{,}$$​ then its second term is:

[CBSE - 2025]

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Arithmetic Progression

Medium

Q.

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)]

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Arithmetic Progression

Medium

Q.

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

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Arithmetic Progression

Medium

Q.

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.

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Arithmetic Progression

Medium

Q.

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

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Arithmetic Progression

Medium

Q.

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

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Arithmetic Progression

Medium

Q.

$\begin{array}{l}{Choose the correct choice in the following and justify:}\\ {(i) 30th term of the AP: 10, 7, 4, . . . , is}\\ {     (A) 97     (B) 77      (C) }-{77      (D) }-{87}\\ {(ii) 11th term of the AP: }-{3, }\frac{-{1}}{2}{, 2, . . ., is}\\ {      (A) 28     (B) 22     (C) }-{38      (D) }-{48}\frac{1}{2}\end{array}$

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Arithmetic Progression

Medium

Q.

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

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Arithmetic Progression

Medium

Q.

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.

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Arithmetic Progression

Medium

Q.

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?

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Arithmetic Progression

Medium

Q.

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.

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Arithmetic Progression

Medium

Q.

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

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Arithmetic Progression

Medium

Q.

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

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Arithmetic Progression

Medium

Q.

How many multiples of 4 lie between 10 and 250?

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Arithmetic Progression

Medium

Q.

How many three-digit numbers are divisible by 7?

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Arithmetic Progression

Medium

Q.

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

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Arithmetic Progression

Medium

Q.

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

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Arithmetic Progression

Medium

Q.

$\begin{array}{l}\text{Find the number of terms in each of the following APs:}\\ \text{(i) 7, 13, 19, . . . , 205 (ii) 18, 15}\frac{1}{2},\text{ 13, . . . , }-\text{47}\end{array}$

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Arithmetic Progression

Medium

Q.

The sum of the first 'm' terms of an Arithmetic Progression (AP) is given by the expression $2m^2 + 3m$. Determine the second term of this AP.

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$$a_n = a + (n - 1)d$$

an​=a+(n−1)d

In this exercise, students learn how to:

• Find any particular term of an AP

• Determine whether a given number is a term of an AP

• Solve real-life problems using the concept of nth term

The solutions are explained in a clear, step-by-step manner, helping students understand the logic behind each problem.

Q1. Find the sum of the following APs.

(i) 2, 7, 12, … to 10 terms
a = 2, d = 5, n = 10

$$S_{10}=\frac{10}{2}[2a+(10-1)d]=5[4+45]=245$$

S10​=210​[2a+(10−1)d]=5[4+45]=245

✅ Sum = 245

(ii) −37, −33, −29, … to 12 terms
a = −37, d = 4, n = 12

$$S_{12}=\frac{12}{2}[2(-37)+(11)(4)]=6[-74+44]=6(-30)=-180$$

S12​=212​[2(−37)+(11)(4)]=6[−74+44]=6(−30)=−180

✅ Sum = −180

(iii) 0.6, 1.7, 2.8, … to 100 terms
a = 0.6, d = 1.1, n = 100

$$S_{100}=\frac{100}{2}[2(0.6)+99(1.1)]=50[1.2+108.9]=50(110.1)=5505$$

S100​=2100​[2(0.6)+99(1.1)]=50[1.2+108.9]=50(110.1)=5505

✅ Sum = 5505

(iv) 1/15, 1/12, 1/10, … to 11 terms
a = 1/15, d = 1/12 − 1/15 = 1/60, n = 11

$$S_{11}=\frac{11}{2}\left[2\cdot\frac{1}{15}+10\cdot\frac{1}{60}\right] =\frac{11}{2}\left[\frac{2}{15}+\frac{1}{6}\right] =\frac{11}{2}\left[\frac{3}{10}\right]=\frac{33}{20}$$

S11​=211​[2⋅151​+10⋅601​]=211​[152​+61​]=211​[103​]=2033​

✅ Sum = 33/20

Q2. Find the sums given below

(i) 7, 7

$\frac{1}{2}$

21​, 8, … , 84
a = 7, d = 1/2, l = 84

$$84=7+(n-1)\frac{1}{2}\Rightarrow 77=\frac{n-1}{2}\Rightarrow n=155$$

84=7+(n−1)21​⇒77=2n−1​⇒n=155

$$S=\frac{n}{2}(a+l)=\frac{155}{2}(7+84)=\frac{155}{2}\cdot 91=\frac{14105}{2}$$

S=2n​(a+l)=2155​(7+84)=2155​⋅91=214105​

✅ Sum = 14105/2

(ii) 34 + 32 + 30 + … + 10
a = 34, d = −2, l = 10

$$10=34+(n-1)(-2)\Rightarrow 10=34-2n+2\Rightarrow 2n=26\Rightarrow n=13$$

10=34+(n−1)(−2)⇒10=34−2n+2⇒2n=26⇒n=13

$$S=\frac{13}{2}(34+10)=\frac{13}{2}\cdot 44=286$$

S=213​(34+10)=213​⋅44=286

✅ Sum = 286

(iii) −5 + (−8) + (−11) + … + (−230)
a = −5, d = −3, l = −230

$$-230=-5+(n-1)(-3)\Rightarrow -225=-3(n-1)\Rightarrow n-1=75\Rightarrow n=76$$

−230=−5+(n−1)(−3)⇒−225=−3(n−1)⇒n−1=75⇒n=76

$$S=\frac{76}{2}(-5-230)=38(-235)=-8930$$

S=276​(−5−230)=38(−235)=−8930

✅ Sum = −8930

Q3. In an AP

(i) a = 5, d = 3,

$a_n$

an​=50; find n and

$S_n$

Sn​

$$50=5+(n-1)3\Rightarrow 45=3(n-1)\Rightarrow n=16$$

50=5+(n−1)3⇒45=3(n−1)⇒n=16

$$S_{16}=\frac{16}{2}(5+50)=8\cdot 55=440$$

S16​=216​(5+50)=8⋅55=440

✅ n = 16,

$S_{16}=440$

S16​=440

(ii)

$a_{13}=35$

a13​=35, a = 7; find d and

$S_{13}$

S13​

$$35=7+12d\Rightarrow d=\frac{28}{12}=\frac{7}{3}$$

35=7+12d⇒d=1228​=37​

$$S_{13}=\frac{13}{2}(7+35)=\frac{13}{2}\cdot 42=273$$

S13​=213​(7+35)=213​⋅42=273

✅ d = 7/3,

$S_{13}=273$

S13​=273

(iii)

$a_{12}=37$

a12​=37, d = 3; find a and

$S_{12}$

S12​

$$37=a+11\cdot 3 \Rightarrow a=4$$

37=a+11⋅3⇒a=4

$$S_{12}=\frac{12}{2}(4+37)=6\cdot 41=246$$

S12​=212​(4+37)=6⋅41=246

✅ a = 4,

$S_{12}=246$

S12​=246

(iv)

$a_3=15$

a3​=15,

$S_{10}=125$

S10​=125; find d and

$a_{10}$

a10​

$$a+2d=15 \quad ...(1)$$

a+2d=15...(1)

$$125=\frac{10}{2}[2a+9d]\Rightarrow 25=2a+9d \quad ...(2)$$

125=210​[2a+9d]⇒25=2a+9d...(2)

From (1): 2a+4d=30
Subtract from (2): (2a+9d)−(2a+4d)=25−30 ⇒ 5d=−5 ⇒ d=−1
Then a=15−2(−1)=17

$$a_{10}=a+9d=17-9=8$$

a10​=a+9d=17−9=8

✅ d = −1,

$a_{10}=8$

a10​=8

(v) d = 5,

$S_9=75$

S9​=75; find a and

$a_9$

a9​

$$75=\frac{9}{2}[2a+8\cdot 5]\Rightarrow 75=\frac{9}{2}(2a+40) \Rightarrow 150=9(2a+40)\Rightarrow 2a+40=\frac{50}{3} \Rightarrow a=-\frac{35}{3}$$

75=29​[2a+8⋅5]⇒75=29​(2a+40)⇒150=9(2a+40)⇒2a+40=350​⇒a=−335​

$$a_9=a+8d=-\frac{35}{3}+40=\frac{85}{3}$$

a9​=a+8d=−335​+40=385​

✅ a = −35/3,

$a_9=85/3$

a9​=85/3

(vi) a = 2, d = 8,

$S_n=90$

Sn​=90; find n and

$a_n$

an​

$$90=\frac{n}{2}[2\cdot2+(n-1)8] =\frac{n}{2}[4+8n-8] =\frac{n}{2}(8n-4)=n(4n-2)$$

90=2n​[2⋅2+(n−1)8]=2n​[4+8n−8]=2n​(8n−4)=n(4n−2)

$$4n^2-2n-90=0 \Rightarrow 2n^2-n-45=0 \Rightarrow (2n+9)(n-5)=0 \Rightarrow n=5$$

4n2−2n−90=0⇒2n2−n−45=0⇒(2n+9)(n−5)=0⇒n=5

$$a_5=2+4\cdot 8=34$$

a5​=2+4⋅8=34

✅ n = 5,

$a_n=34$

an​=34

(vii) a = 8,

$a_n=62$

an​=62,

$S_n=210$

Sn​=210; find n and d

$$210=\frac{n}{2}(8+62)=\frac{n}{2}\cdot 70 \Rightarrow n=6$$

210=2n​(8+62)=2n​⋅70⇒n=6

$$62=8+(6-1)d \Rightarrow 54=5d \Rightarrow d=\frac{54}{5}$$

62=8+(6−1)d⇒54=5d⇒d=554​

✅ n = 6, d = 54/5

(viii)

$a_n=4$

an​=4, d = 2,

$S_n=-14$

Sn​=−14; find n and a

$$4=a+(n-1)2 \Rightarrow a=6-2n$$

4=a+(n−1)2⇒a=6−2n

$$-14=\frac{n}{2}(a+4)=\frac{n}{2}(6-2n+4)=\frac{n}{2}(10-2n)=n(5-n)$$

−14=2n​(a+4)=2n​(6−2n+4)=2n​(10−2n)=n(5−n)

$$n(5-n)=-14 \Rightarrow n^2-5n-14=0 \Rightarrow (n-7)(n+2)=0 \Rightarrow n=7$$

n(5−n)=−14⇒n2−5n−14=0⇒(n−7)(n+2)=0⇒n=7

$$a=6-14=-8$$

a=6−14=−8

✅ n = 7, a = −8

(ix) a = 3, n = 8,

$S_n=192$

Sn​=192; find d

$$192=\frac{8}{2}[2\cdot 3+7d]=4(6+7d) \Rightarrow 48=6+7d \Rightarrow d=6$$

192=28​[2⋅3+7d]=4(6+7d)⇒48=6+7d⇒d=6

✅ d = 6

(x) l = 28,

$S_9=144$

S9​=144; find a

$$144=\frac{9}{2}(a+28)\Rightarrow 288=9(a+28)\Rightarrow a+28=32\Rightarrow a=4$$

144=29​(a+28)⇒288=9(a+28)⇒a+28=32⇒a=4

✅ a = 4

Q4. How many terms of the AP 9, 17, 25, … give sum 636?

a = 9, d = 8,

$S_n=636$

Sn​=636

$$636=\frac{n}{2}[18+(n-1)8] \Rightarrow 1272=n[18+8n-8]=n(8n+10)$$

636=2n​[18+(n−1)8]⇒1272=n[18+8n−8]=n(8n+10)

$$8n^2+10n-1272=0 \Rightarrow 4n^2+5n-636=0 \Rightarrow (n-12)(4n+53)=0 \Rightarrow n=12$$

8n2+10n−1272=0⇒4n2+5n−636=0⇒(n−12)(4n+53)=0⇒n=12

✅ n = 12

Q5. First term 5, last term 45, sum 400. Find n and d.

$$400=\frac{n}{2}(5+45)=\frac{n}{2}\cdot 50=25n \Rightarrow n=16$$

400=2n​(5+45)=2n​⋅50=25n⇒n=16

$$45=5+(16-1)d \Rightarrow 40=15d \Rightarrow d=\frac{8}{3}$$

45=5+(16−1)d⇒40=15d⇒d=38​

✅ n = 16, d = 8/3

Q6. First term 17, last term 350, d = 9. Find n and sum.

$$350=17+(n-1)9 \Rightarrow 333=9(n-1)\Rightarrow n-1=37 \Rightarrow n=38$$

350=17+(n−1)9⇒333=9(n−1)⇒n−1=37⇒n=38

$$S=\frac{38}{2}(17+350)=19\cdot 367=6973$$

S=238​(17+350)=19⋅367=6973

✅ n = 38, Sum = 6973

Q7. d = 7 and 22nd term = 149. Find sum of first 22 terms.

$$149=a+21\cdot 7 \Rightarrow a=149-147=2$$

149=a+21⋅7⇒a=149−147=2

$$S_{22}=\frac{22}{2}(2+149)=11\cdot 151=1661$$

S22​=222​(2+149)=11⋅151=1661

✅ Sum = 1661

Q8. Second term 14 and third term 18. Find sum of first 51 terms.

$$d=18-14=4,\quad a=14-4=10$$

d=18−14=4,a=14−4=10

$$S_{51}=\frac{51}{2}[2\cdot 10+50\cdot 4] =\frac{51}{2}(20+200)=\frac{51}{2}\cdot 220=5610$$

S51​=251​[2⋅10+50⋅4]=251​(20+200)=251​⋅220=5610

✅ Sum = 5610

Q9. $S_7=49$

S7​=49 and $S_{17}=289$

S17​=289. Find $S_n$

Sn​.

$$S_n=\frac{n}{2}[2a+(n-1)d]$$

Sn​=2n​[2a+(n−1)d]

From

$S_7=49$

S7​=49:

$$49=\frac{7}{2}(2a+6d)\Rightarrow 98=7(2a+6d)\Rightarrow 2a+6d=14 ...(1)$$

49=27​(2a+6d)⇒98=7(2a+6d)⇒2a+6d=14...(1)

From

$S_{17}=289$

S17​=289:

$$289=\frac{17}{2}(2a+16d)\Rightarrow 578=17(2a+16d)\Rightarrow 2a+16d=34 ...(2)$$

289=217​(2a+16d)⇒578=17(2a+16d)⇒2a+16d=34...(2)

Subtract (1) from (2): 10d=20 ⇒ d=2
Then 2a+12=14 ⇒ a=1
So,

$$S_n=\frac{n}{2}[2\cdot1+(n-1)2]=\frac{n}{2}[2+2n-2]=\frac{n}{2}(2n)=n^2$$

Sn​=2n​[2⋅1+(n−1)2]=2n​[2+2n−2]=2n​(2n)=n2

✅

$S_n=n^2$

Sn​=n2

Q10. Show sequences form an AP and find sum of first 15 terms.

(i)

$a_n=3n+4$

an​=3n+4
a₁ = 7, common difference d = 4 ⇒ AP

$$S_{15}=\frac{15}{2}[2\cdot 7+14\cdot 4] =\frac{15}{2}(14+56)=\frac{15}{2}\cdot 70=525$$

S15​=215​[2⋅7+14⋅4]=215​(14+56)=215​⋅70=525

✅ Sum = 525

(ii)

$a_n=9-5n$

an​=9−5n
a₁ = 4, common difference d = −5 ⇒ AP

$$S_{15}=\frac{15}{2}[2\cdot 4+14(-5)] =\frac{15}{2}(8-70)=\frac{15}{2}(-62)=-465$$

S15​=215​[2⋅4+14(−5)]=215​(8−70)=215​(−62)=−465

✅ Sum = −465

Q11. If $S_n=2n-4n^2$

Sn​=2n−4n2, find $S_1$

S1​, $S_2$

S2​, $a_2$

a2​, $a_3$

a3​, $a_{10}$

a10​, $a_n$

an​.

$$S_1=2(1)-4(1)^2=2-4=-2 \Rightarrow a_1=-2$$

S1​=2(1)−4(1)2=2−4=−2⇒a1​=−2

$$S_2=2(2)-4(4)=4-16=-12 \Rightarrow a_2=S_2-S_1=-12-(-2)=-10$$

S2​=2(2)−4(4)=4−16=−12⇒a2​=S2​−S1​=−12−(−2)=−10

$$S_3=2(3)-4(9)=6-36=-30 \Rightarrow a_3=S_3-S_2=-30-(-12)=-18$$

S3​=2(3)−4(9)=6−36=−30⇒a3​=S3​−S2​=−30−(−12)=−18

$$S_{10}=2(10)-4(100)=20-400=-380 \Rightarrow a_{10}=S_{10}-S_9$$

S10​=2(10)−4(100)=20−400=−380⇒a10​=S10​−S9​

$$S_9=2(9)-4(81)=18-324=-306 \Rightarrow a_{10}=-380-(-306)=-74$$

S9​=2(9)−4(81)=18−324=−306⇒a10​=−380−(−306)=−74

General term:

$$a_n=S_n-S_{n-1}=(2n-4n^2)-[2(n-1)-4(n-1)^2]=-8n+2$$

an​=Sn​−Sn−1​=(2n−4n2)−[2(n−1)−4(n−1)2]=−8n+2

✅

$S_1=-2,\; S_2=-12,\; a_2=-10,\; a_3=-18,\; a_{10}=-74,\; a_n=2-8n$

S1​=−2,S2​=−12,a2​=−10,a3​=−18,a10​=−74,an​=2−8n

Q12. Sum of first 40 positive integers divisible by 6

AP: 6, 12, 18, … (n=40)
a=6, d=6

$$S_{40}=\frac{40}{2}[2\cdot6+39\cdot6] =20(12+234)=20\cdot246=4920$$

S40​=240​[2⋅6+39⋅6]=20(12+234)=20⋅246=4920

✅ Sum = 4920

Q13. Sum of first 15 multiples of 8

AP: 8, 16, 24, … (n=15)
a=8, d=8

$$S_{15}=\frac{15}{2}[16+14\cdot 8] =\frac{15}{2}(16+112)=\frac{15}{2}\cdot128=960$$

S15​=215​[16+14⋅8]=215​(16+112)=215​⋅128=960

✅ Sum = 960

Q14. Sum of odd numbers between 0 and 50

Odd numbers: 1, 3, 5, …, 49
a=1, d=2, l=49

$$49=1+(n-1)2 \Rightarrow 48=2(n-1)\Rightarrow n=25$$

49=1+(n−1)2⇒48=2(n−1)⇒n=25

$$S=\frac{25}{2}(1+49)=\frac{25}{2}\cdot 50=625$$

S=225​(1+49)=225​⋅50=625

✅ Sum = 625

Q15. Penalty AP: 200, 250, 300, … (30 days). Find total penalty.

a=200, d=50, n=30

$$S_{30}=\frac{30}{2}[2\cdot200+29\cdot50] =15(400+1450)=15\cdot1850=27750$$

S30​=230​[2⋅200+29⋅50]=15(400+1450)=15⋅1850=27750

✅ Penalty = ₹27750

Q16. Total ₹700, 7 prizes, each ₹20 less than previous. Find prizes.

Let first prize = x
Prizes: x, x−20, x−40, … (7 terms), d=−20

$$700=\frac{7}{2}[2x+6(-20)] =\frac{7}{2}(2x-120) \Rightarrow 1400=7(2x-120) \Rightarrow 200=2x-120 \Rightarrow x=160$$

700=27​[2x+6(−20)]=27​(2x−120)⇒1400=7(2x−120)⇒200=2x−120⇒x=160

✅ Prizes: 160, 140, 120, 100, 80, 60, 40

Q17. Tree planting (3 sections of each class I to XII). Find total trees.

Totals per class: 3, 6, 9, …, 36 (n=12)
a=3, d=3

$$S_{12}=\frac{12}{2}[2\cdot 3+11\cdot 3] =6(6+33)=6\cdot 39=234$$

S12​=212​[2⋅3+11⋅3]=6(6+33)=6⋅39=234

✅ Total trees = 234

Q18. Spiral of 13 semicircles, radii: 0.5, 1.0, 1.5, … (take $\pi=\frac{22}{7}$

π=722​). Find total length.

Length of a semicircle =

$\pi r$

πr
So lengths form AP:

$0.5\pi, 1\pi, 1.5\pi, …$

0.5π,1π,1.5π,…
a = 0.5π, d = 0.5π, n = 13

$$S_{13}=\frac{13}{2}[2(0.5\pi)+12(0.5\pi)] =\frac{13}{2}[ \pi+6\pi ]=\frac{13}{2}\cdot 7\pi=\frac{91\pi}{2}$$

S13​=213​[2(0.5π)+12(0.5π)]=213​[π+6π]=213​⋅7π=291π​

With

$\pi=\frac{22}{7}$

π=722​:

$$S=\frac{91}{2}\cdot \frac{22}{7}=143$$

S=291​⋅722​=143

✅ Total length = 143 cm

Q19. 200 logs stacked: 20, 19, 18, … Find rows and logs in top row.

AP: a=20, d=−1,

$S_n=200$

Sn​=200

$$200=\frac{n}{2}[2\cdot 20+(n-1)(-1)] =\frac{n}{2}[40-(n-1)] =\frac{n}{2}(41-n)$$