Arithmetic Progressions – Exercise 5.3 | SJMaths

Arithmetic Progressions – Exercise 5.3

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This page provides comprehensive Arithmetic Progressions – Exercise 5.3. Free NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Exercise 5-3. Step-by-step explained answers for CBSE Board exams. Download PDF and practice now.

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Basics (Q1-3) Sums & Terms (Q4-10) Properties (Q11-14) Word Problems (Q15-20)

Basics: Finding Sums

1. Find the sum of the following APs:

(i) $2, 7, 12, \dots$ to 10 terms.

(ii) $-37, -33, -29, \dots$ to 12 terms.

(iii) $0.6, 1.7, 2.8, \dots$ to 100 terms.

(iv) $\frac{1}{15}, \frac{1}{12}, \frac{1}{10}, \dots$ to 11 terms.


2. Find the sums given below:

(i) $7 + 10\frac{1}{2} + 14 + \dots + 84$

(ii) $34 + 32 + 30 + \dots + 10$

(iii) $-5 + (-8) + (-11) + \dots + (-230)$


3. In an AP:

(i) Given $a=5, d=3, a_n=50$, find $n$ and $S_n$.

(ii) Given $a=7, a_{13}=35$, find $d$ and $S_{13}$.

(iii) Given $a_{12}=37, d=3$, find $a$ and $S_{12}$.

(iv) Given $a_3=15, S_{10}=125$, find $d$ and $a_{10}$.

(v) Given $d=5, S_9=75$, find $a$ and $a_9$.

(vi) Given $a=2, d=8, S_n=90$, find $n$ and $a_n$.

(vii) Given $a=8, a_n=62, S_n=210$, find $n$ and $d$.

(viii) Given $a_n=4, d=2, S_n=-14$, find $n$ and $a$.

(ix) Given $a=3, n=8, S=192$, find $d$.

(x) Given $l=28, S=144$, and there are total 9 terms. Find $a$.

Question 1
(i) $a=2, d=5, n=10$. $S_{10} = \frac{10}{2}[2(2) + 9(5)] = 5[4+45] = 5(49) = 245$.
(ii) $a=-37, d=4, n=12$. $S_{12} = \frac{12}{2}[2(-37) + 11(4)] = 6[-74+44] = 6(-30) = -180$.
(iii) $a=0.6, d=1.1, n=100$. $S_{100} = 50[1.2 + 99(1.1)] = 50[1.2 + 108.9] = 50(110.1) = 5505$.
(iv) $a=\frac{1}{15}, d=\frac{1}{60}, n=11$. $S_{11} = \frac{11}{2}[\frac{2}{15} + 10(\frac{1}{60})] = \frac{11}{2}[\frac{2}{15} + \frac{1}{6}] = \frac{11}{2}[\frac{4+5}{30}] = \frac{11}{2}(\frac{9}{30}) = \frac{33}{20}$.

Question 2
(i) $a=7, l=84, d=3.5$. $84 = 7 + (n-1)3.5 \Rightarrow 77 = 3.5(n-1) \Rightarrow n-1=22 \Rightarrow n=23$.
$S_{23} = \frac{23}{2}(7+84) = \frac{23}{2}(91) = 1046.5$.
(ii) $a=34, l=10, d=-2$. $10 = 34 + (n-1)(-2) \Rightarrow -24 = -2(n-1) \Rightarrow n=13$.
$S_{13} = \frac{13}{2}(34+10) = \frac{13}{2}(44) = 286$.
(iii) $a=-5, l=-230, d=-3$. $-230 = -5 + (n-1)(-3) \Rightarrow -225 = -3(n-1) \Rightarrow n=76$.
$S_{76} = \frac{76}{2}(-5 - 230) = 38(-235) = -8930$.

Question 3
(i) $50 = 5 + 3(n-1) \Rightarrow 45 = 3(n-1) \Rightarrow n=16$. $S_{16} = \frac{16}{2}(5+50) = 8(55) = 440$.
(ii) $35 = 7 + 12d \Rightarrow 28 = 12d \Rightarrow d = \frac{7}{3}$. $S_{13} = \frac{13}{2}(7+35) = \frac{13}{2}(42) = 273$.
(iii) $37 = a + 11(3) \Rightarrow a = 4$. $S_{12} = \frac{12}{2}(4+37) = 6(41) = 246$.
(iv) $a+2d=15$. $S_{10}=125 \Rightarrow 5[2a+9d]=125 \Rightarrow 2a+9d=25$. Solving: $d=-1, a=17$. $a_{10}=17+9(-1)=8$.
(v) $S_9=75 \Rightarrow \frac{9}{2}[2a+8(5)]=75 \Rightarrow 9(a+20)=75 \Rightarrow 3a+60=25 \Rightarrow 3a=-35 \Rightarrow a=-\frac{35}{3}$. $a_9 = -\frac{35}{3} + 40 = \frac{85}{3}$.
(vi) $90 = \frac{n}{2}[4 + (n-1)8] \Rightarrow 180 = n(8n-4) \Rightarrow 45 = n(2n-1) \Rightarrow 2n^2-n-45=0$. $n=5$. $a_5 = 2+32=34$.
(vii) $210 = \frac{n}{2}(8+62) \Rightarrow 420 = 70n \Rightarrow n=6$. $62 = 8 + 5d \Rightarrow 54=5d \Rightarrow d=\frac{54}{5}$.
(viii) $a_n=4 \Rightarrow a+(n-1)2=4 \Rightarrow a=6-2n$. $S_n=-14 \Rightarrow \frac{n}{2}(a+4)=-14$. Sub $a$: $\frac{n}{2}(10-2n)=-14 \Rightarrow n(5-n)=-14 \Rightarrow n^2-5n-14=0 \Rightarrow n=7$. $a=-8$.
(ix) $192 = \frac{8}{2}[6 + 7d] \Rightarrow 192 = 4(6+7d) \Rightarrow 48 = 6+7d \Rightarrow 42=7d \Rightarrow d=6$.
(x) $144 = \frac{9}{2}(a+28) \Rightarrow 32 = a+28 \Rightarrow a=4$.

Sums & Terms Conditions

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

5. 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.

6. 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?

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

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

9. 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.

10. Show that $a_1, a_2, \dots, a_n, \dots$ form an AP where $a_n$ is defined as below: (i) $a_n = 3 + 4n$ (ii) $a_n = 9 - 5n$. Also find the sum of the first 15 terms in each case.

4. $a=9, d=8, S_n=636$.
$636 = \frac{n}{2}[18 + (n-1)8] = n[9 + 4(n-1)] = n(4n+5)$.
$4n^2 + 5n - 636 = 0$. Using quadratic formula, $n = \frac{-5 \pm \sqrt{25 - 4(4)(-636)}}{8} = \frac{-5 \pm \sqrt{10201}}{8} = \frac{-5 \pm 101}{8}$.
$n = 12$ (since $n$ cannot be negative).

5. $a=5, l=45, S_n=400$.
$400 = \frac{n}{2}(5+45) \Rightarrow 800 = 50n \Rightarrow n=16$.
$l = a + (n-1)d \Rightarrow 45 = 5 + 15d \Rightarrow 40 = 15d \Rightarrow d = \frac{8}{3}$.

6. $a=17, l=350, d=9$.
$350 = 17 + (n-1)9 \Rightarrow 333 = 9(n-1) \Rightarrow 37 = n-1 \Rightarrow n=38$.
$S_{38} = \frac{38}{2}(17+350) = 19(367) = 6973$.

7. $d=7, a_{22}=149$.
$149 = a + 21(7) \Rightarrow 149 = a + 147 \Rightarrow a=2$.
$S_{22} = \frac{22}{2}(2+149) = 11(151) = 1661$.

8. $a_2=14, a_3=18 \Rightarrow d=4, a=10$.
$S_{51} = \frac{51}{2}[20 + 50(4)] = \frac{51}{2}(220) = 51(110) = 5610$.

9. $S_7=49 \Rightarrow 7^2$. $S_{17}=289 \Rightarrow 17^2$. This implies $S_n = n^2$.
Proof: $S_7 = \frac{7}{2}(2a+6d) = 7(a+3d) = 49 \Rightarrow a+3d=7$.
$S_{17} = \frac{17}{2}(2a+16d) = 17(a+8d) = 289 \Rightarrow a+8d=17$.
Subtracting: $5d=10 \Rightarrow d=2$. $a=1$.
$S_n = \frac{n}{2}[2(1) + (n-1)2] = \frac{n}{2}(2n) = n^2$.

10. (i) $a_n = 3+4n$. $a_1=7, a_2=11, a_3=15$. $d=4$.
$S_{15} = \frac{15}{2}[14 + 14(4)] = \frac{15}{2}(70) = 525$.
(ii) $a_n = 9-5n$. $a_1=4, a_2=-1, a_3=-6$. $d=-5$.
$S_{15} = \frac{15}{2}[8 + 14(-5)] = \frac{15}{2}(8-70) = \frac{15}{2}(-62) = -465$.

Properties & Multiples

11. If the sum of the first $n$ terms of an AP is $4n - n^2$, what is the first term (that is $S_1$)? What is the sum of first two terms? What is the second term? Similarly, find the 3rd, the 10th and the nth terms.

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

13. Find the sum of the first 15 multiples of 8.

14. Find the sum of the odd numbers between 0 and 50.

11. $S_n = 4n - n^2$.
$S_1 = 4(1) - 1 = 3$. First term $a_1 = 3$.
$S_2 = 4(2) - 4 = 4$. Sum of first two terms is 4.
$a_2 = S_2 - S_1 = 4 - 3 = 1$.
$S_3 = 4(3) - 9 = 3$. $a_3 = S_3 - S_2 = 3 - 4 = -1$.
$a_n = S_n - S_{n-1} = (4n - n^2) - [4(n-1) - (n-1)^2]$
$= 4n - n^2 - [4n - 4 - (n^2 - 2n + 1)] = 4n - n^2 - 4n + 4 + n^2 - 2n + 1 = 5 - 2n$.
$a_{10} = 5 - 20 = -15$.

12. $6, 12, 18, \dots$ (40 terms). $a=6, d=6$.
$S_{40} = \frac{40}{2}[12 + 39(6)] = 20(12 + 234) = 20(246) = 4920$.

13. $8, 16, 24, \dots$ (15 terms). $a=8, d=8$.
$S_{15} = \frac{15}{2}[16 + 14(8)] = \frac{15}{2}(16 + 112) = \frac{15}{2}(128) = 15(64) = 960$.

14. $1, 3, 5, \dots, 49$. $a=1, l=49$.
$49 = 1 + (n-1)2 \Rightarrow 48 = 2(n-1) \Rightarrow n=25$.
$S_{25} = \frac{25}{2}(1+49) = \frac{25}{2}(50) = 625$.

Word Problems

15. A contract on construction job specifies a penalty for delay: ₹ 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?

16. 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.

17. 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?

18. 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 Fig. 5.4. What is the total length of such a spiral made up of thirteen consecutive semicircles? (Take $\pi = \frac{22}{7}$)

19. 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. In how many rows are the 200 logs placed and how many logs are in the top row?

20. 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. 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?

15. $a=200, d=50, n=30$.
$S_{30} = \frac{30}{2}[400 + 29(50)] = 15(400 + 1450) = 15(1850) = 27750$.
Penalty: ₹ 27,750.

16. $n=7, S_7=700, d=-20$.
$700 = \frac{7}{2}[2a + 6(-20)] \Rightarrow 200 = 2a - 120 \Rightarrow 2a = 320 \Rightarrow a = 160$.
Prizes: ₹ 160, 140, 120, 100, 80, 60, 40.

17. Each class has 3 sections. Class $k$ plants $3 \times k$ trees.
AP: $3, 6, 9, \dots, 36$ (for Class XII). $n=12$.
$S_{12} = \frac{12}{2}(3+36) = 6(39) = 234$ trees.

18. Length of semicircle $l = \pi r$.
$l_1 = \pi(0.5), l_2 = \pi(1.0), \dots$
Total length $L = \pi(0.5) + \pi(1.0) + \dots$ (13 terms).
$L = \pi [0.5 + 1.0 + \dots]$. Inside is AP with $a=0.5, d=0.5, n=13$.
Sum of radii = $\frac{13}{2}[2(0.5) + 12(0.5)] = \frac{13}{2}(1+6) = \frac{91}{2}$.
Total Length = $\frac{22}{7} \times \frac{91}{2} = 11 \times 13 = 143$ cm.

19. $S_n=200, a=20, d=-1$.
$200 = \frac{n}{2}[40 + (n-1)(-1)] \Rightarrow 400 = n(41-n) \Rightarrow n^2 - 41n + 400 = 0$.
$(n-16)(n-25) = 0$.
If $n=25$, $a_{25} = 20 - 24 = -4$ (not possible).
So $n=16$. $a_{16} = 20 - 15 = 5$.
16 rows, 5 logs in top row.

20. Distances: $2 \times 5, 2 \times (5+3), 2 \times (5+6), \dots$
AP: $10, 16, 22, \dots$ (10 terms). $a=10, d=6$.
$S_{10} = \frac{10}{2}[20 + 9(6)] = 5(20+54) = 5(74) = 370$ m.
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