NCERT Solutions Class 10 Maths Chapter 5: Arithmetic Progressions
Detailed solutions for Exercise 5.3 (Sum of first $n$ terms of an AP).
#### **Q1. 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.**
#### **A1. Solution:**
**(i)** $a=2, d=7-2=5, n=10$.
$S_{10} = \frac{10}{2}[2(2) + (10-1)5]$
$= 5[4 + 45] = 5 \times 49 = 245$.
**Ans: 245**
**(ii)** $a=-37, d=-33 - (-37) = 4, n=12$.
$S_{12} = \frac{12}{2}[2(-37) + (11)4]$
$= 6[-74 + 44] = 6 \times (-30) = -180$.
**Ans: -180**
**(iii)** $a=0.6, d=1.7-0.6=1.1, n=100$.
$S_{100} = \frac{100}{2}[2(0.6) + 99(1.1)]$
$= 50[1.2 + 108.9] = 50 \times 110.1 = 5505$.
**Ans: 5505**
**(iv)** $a=\frac{1}{15}, d=\frac{1}{12} - \frac{1}{15} = \frac{5-4}{60} = \frac{1}{60}, n=11$.
$S_{11} = \frac{11}{2}[2(\frac{1}{15}) + 10(\frac{1}{60})]$
$= \frac{11}{2}[\frac{2}{15} + \frac{1}{6}] = \frac{11}{2}[\frac{4+5}{30}] = \frac{11}{2} \times \frac{9}{30} = \frac{33}{20}$.
**Ans: $\frac{33}{20}$**
#### **Q2. Find the sums given below:**
**(i) $7 + 10\frac{1}{2} + 14 + \dots + 84$**
**(ii) $34 + 32 + 30 + \dots + 10$**
**(iii) $-5 + (-8) + (-11) + \dots + (-230)$**
#### **A2. Solution:**
**(i)** $a=7, d=3.5 (\frac{7}{2}), l=84$.
Find $n$: $84 = 7 + (n-1)3.5 \Rightarrow 77 = (n-1)3.5 \Rightarrow 22 = n-1 \Rightarrow n = 23$.
$S_{23} = \frac{23}{2}(7 + 84) = \frac{23}{2}(91) = \frac{2093}{2} = 1046.5$.
**Ans: 1046.5**
**(ii)** $a=34, d=-2, l=10$.
Find $n$: $10 = 34 + (n-1)(-2) \Rightarrow -24 = (n-1)(-2) \Rightarrow 12 = n-1 \Rightarrow n = 13$.
$S_{13} = \frac{13}{2}(34 + 10) = \frac{13}{2}(44) = 13 \times 22 = 286$.
**Ans: 286**
**(iii)** $a=-5, d=-3, l=-230$.
Find $n$: $-230 = -5 + (n-1)(-3) \Rightarrow -225 = (n-1)(-3) \Rightarrow 75 = n-1 \Rightarrow n = 76$.
$S_{76} = \frac{76}{2}(-5 + (-230)) = 38 \times (-235) = -8930$.
**Ans: -8930**
#### **Q3. 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 $a$.**
#### **A3. Solution:**
**(i)** $50 = 5 + (n-1)3 \Rightarrow 45 = 3(n-1) \Rightarrow 15 = n-1 \Rightarrow n=16$.
$S_{16} = \frac{16}{2}(5+50) = 8 \times 55 = 440$.
**Ans: $n=16, S_{16}=440$**
**(ii)** $35 = 7 + 12d \Rightarrow 28 = 12d \Rightarrow d = \frac{28}{12} = \frac{7}{3}$.
$S_{13} = \frac{13}{2}(7+35) = \frac{13}{2}(42) = 13 \times 21 = 273$.
**Ans: $d=\frac{7}{3}, S_{13}=273$**
**(iii)** $37 = a + 11(3) \Rightarrow 37 = a + 33 \Rightarrow a = 4$.
$S_{12} = \frac{12}{2}(4+37) = 6 \times 41 = 246$.
**Ans: $a=4, S_{12}=246$**
**(iv)** $a+2d=15$ ...(1)
$S_{10} = \frac{10}{2}[2a+9d] = 125 \Rightarrow 5(2a+9d) = 125 \Rightarrow 2a+9d=25$ ...(2)
Multiply (1) by 2: $2a+4d=30$. Subtract from (2): $5d = -5 \Rightarrow d = -1$.
From (1): $a - 2 = 15 \Rightarrow a = 17$.
$a_{10} = 17 + 9(-1) = 8$.
**Ans: $d=-1, a_{10}=8$**
**(v)** $75 = \frac{9}{2}[2a + 8(5)] \Rightarrow \frac{150}{9} = 2a + 40 \Rightarrow \frac{50}{3} - 40 = 2a$.
$2a = \frac{50-120}{3} = -\frac{70}{3} \Rightarrow a = -\frac{35}{3}$.
$a_9 = -\frac{35}{3} + 8(5) = \frac{-35+120}{3} = \frac{85}{3}$.
**Ans: $a=-\frac{35}{3}, a_9=\frac{85}{3}$**
**(vi)** $90 = \frac{n}{2}[2(2) + (n-1)8] \Rightarrow 180 = n[4 + 8n - 8] \Rightarrow 180 = n[8n-4]$.
$180 = 4n(2n-1) \Rightarrow 45 = 2n^2 - n \Rightarrow 2n^2 - n - 45 = 0$.
$2n^2 - 10n + 9n - 45 = 0 \Rightarrow 2n(n-5) + 9(n-5) = 0$.
Since $n$ must be positive, $n=5$.
$a_5 = 2 + 4(8) = 34$.
**Ans: $n=5, a_5=34$**
**(vii)** $S_n = \frac{n}{2}(a+a_n) \Rightarrow 210 = \frac{n}{2}(8+62) \Rightarrow 210 = 35n \Rightarrow n=6$.
$a_6 = 8 + 5d = 62 \Rightarrow 5d = 54 \Rightarrow d = \frac{54}{5} = 10.8$.
**Ans: $n=6, d=\frac{54}{5}$**
**(viii)** $a_n = a + (n-1)2 = 4 \Rightarrow a = 4 - 2(n-1) \Rightarrow a = 6 - 2n$.
$S_n = \frac{n}{2}(a+4) = -14 \Rightarrow n(a+4) = -28$.
Substitute $a$: $n(6-2n+4) = -28 \Rightarrow n(10-2n) = -28 \Rightarrow 10n - 2n^2 = -28$.
$2n^2 - 10n - 28 = 0 \Rightarrow n^2 - 5n - 14 = 0$.
$(n-7)(n+2)=0$. Since $n>0, n=7$.
$a = 6 - 2(7) = -8$.
**Ans: $n=7, a=-8$**
**(ix)** $192 = \frac{8}{2}[2(3) + 7d] \Rightarrow 192 = 4[6+7d] \Rightarrow 48 = 6+7d$.
$42 = 7d \Rightarrow d = 6$.
**Ans: $d=6$**
**(x)** $S = \frac{n}{2}(a+l) \Rightarrow 144 = \frac{9}{2}(a+28)$.
$32 = a + 28 \Rightarrow a = 4$.
**Ans: $a=4$**
#### **Q4. How many terms of the AP: $9, 17, 25, \dots$ must be taken to give a sum of 636?**
#### **A4. Solution:**
$a=9, d=8, S_n=636$.
$636 = \frac{n}{2}[18 + (n-1)8] \Rightarrow 636 = n[9 + 4(n-1)] \Rightarrow 636 = n[4n+5]$.
$4n^2 + 5n - 636 = 0$.
$n = \frac{-5 \pm \sqrt{25 - 4(4)(-636)}}{8} = \frac{-5 \pm \sqrt{25 + 10176}}{8} = \frac{-5 \pm \sqrt{10201}}{8}$.
$n = \frac{-5 \pm 101}{8}$.
Taking positive: $n = \frac{96}{8} = 12$.
**Ans: 12 terms**
#### **Q5. 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.**
#### **A5. Solution:**
$a=5, l=45, S_n=400$.
$400 = \frac{n}{2}(5+45) \Rightarrow 400 = 25n \Rightarrow n = 16$.
$a_{16} = 5 + 15d = 45 \Rightarrow 15d = 40 \Rightarrow d = \frac{40}{15} = \frac{8}{3}$.
**Ans: $n=16, d=\frac{8}{3}$**
#### **Q6. 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?**
#### **A6. Solution:**
$a=17, l=350, d=9$.
$350 = 17 + (n-1)9 \Rightarrow 333 = (n-1)9 \Rightarrow 37 = n-1 \Rightarrow n = 38$.
$S_{38} = \frac{38}{2}(17+350) = 19 \times 367 = 6973$.
**Ans: $n=38, \text{Sum}=6973$**
#### **Q7. Find the sum of first 22 terms of an AP in which $d = 7$ and 22nd term is 149.**
#### **A7. Solution:**
$d=7, a_{22}=149$.
$a + 21(7) = 149 \Rightarrow a + 147 = 149 \Rightarrow a = 2$.
$S_{22} = \frac{22}{2}(2+149) = 11 \times 151 = 1661$.
**Ans: 1661**
#### **Q8. Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.**
#### **A8. Solution:**
$a_2 = 14, a_3 = 18 \Rightarrow d = 18 - 14 = 4$.
$a = 14 - 4 = 10$.
$S_{51} = \frac{51}{2}[2(10) + 50(4)] = \frac{51}{2}[20 + 200] = \frac{51}{2}(220) = 51 \times 110 = 5610$.
**Ans: 5610**
#### **Q9. 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.**
#### **A9. Solution:**
$S_7 = 49 \Rightarrow \frac{7}{2}(2a+6d) = 49 \Rightarrow 7(a+3d) = 49 \Rightarrow a+3d=7$.
$S_{17} = 289 \Rightarrow \frac{17}{2}(2a+16d) = 289 \Rightarrow 17(a+8d) = 289 \Rightarrow a+8d=17$.
Subtracting: $5d = 10 \Rightarrow d = 2$.
$a + 3(2) = 7 \Rightarrow a = 1$.
$S_n = \frac{n}{2}[2(1) + (n-1)2] = \frac{n}{2}[2 + 2n - 2] = \frac{n}{2}[2n] = n^2$.
**Ans: $n^2$**
#### **Q10. 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.**
#### **A10. Solution:**
**(i) $a_n = 3 + 4n$**
Terms: $a_1=7, a_2=11, a_3=15$.
$d = 11-7=4, 15-11=4$. Since difference is constant, it is an AP.
$S_{15} = \frac{15}{2}(7 + a_{15})$. $a_{15} = 3 + 60 = 63$.
$S_{15} = \frac{15}{2}(7+63) = \frac{15}{2}(70) = 15 \times 35 = 525$.
**Ans: 525**
**(ii) $a_n = 9 - 5n$**
Terms: $a_1=4, a_2=-1, a_3=-6$.
$d = -1-4=-5$. It is an AP.
$a_{15} = 9 - 5(15) = 9 - 75 = -66$.
$S_{15} = \frac{15}{2}(4 + (-66)) = \frac{15}{2}(-62) = 15 \times (-31) = -465$.
**Ans: -465**
#### **Q11. 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 $n^{th}$ terms.**
#### **A11. Solution:**
$S_n = 4n - n^2$.
$S_1 = 4(1) - 1^2 = 3$. First term $a_1 = 3$.
$S_2 = 4(2) - 2^2 = 8 - 4 = 4$.
$a_2 = S_2 - S_1 = 4 - 3 = 1$.
$S_3 = 4(3) - 3^2 = 12 - 9 = 3$.
$a_3 = S_3 - S_2 = 3 - 4 = -1$.
Common difference $d = a_2 - a_1 = 1 - 3 = -2$.
$a_{10} = 3 + 9(-2) = 3 - 18 = -15$.
$a_n = 3 + (n-1)(-2) = 3 - 2n + 2 = 5 - 2n$.
**Ans: $S_1=3, S_2=4, a_2=1, a_3=-1, a_{10}=-15, a_n=5-2n$**
#### **Q12. Find the sum of the first 40 positive integers divisible by 6.**
#### **A12. Solution:**
Series: $6, 12, 18, \dots$
$a=6, d=6, n=40$.
$S_{40} = \frac{40}{2}[2(6) + 39(6)] = 20[12 + 234] = 20(246) = 4920$.
**Ans: 4920**
#### **Q13. Find the sum of the first 15 multiples of 8.**
#### **A13. Solution:**
Series: $8, 16, 24, \dots$
$a=8, d=8, n=15$.
$S_{15} = \frac{15}{2}[2(8) + 14(8)] = \frac{15}{2}[16 + 112] = \frac{15}{2}(128) = 15 \times 64 = 960$.
**Ans: 960**
#### **Q14. Find the sum of the odd numbers between 0 and 50.**
#### **A14. Solution:**
Series: $1, 3, 5, \dots, 49$.
$a=1, d=2, l=49$.
$49 = 1 + (n-1)2 \Rightarrow 48 = 2(n-1) \Rightarrow 24 = n-1 \Rightarrow n = 25$.
$S_{25} = \frac{25}{2}(1 + 49) = \frac{25}{2}(50) = 25 \times 25 = 625$.
**Ans: 625**
#### **Q15. 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?**
#### **A15. Solution:**
$a=200, d=50, n=30$.
$S_{30} = \frac{30}{2}[2(200) + 29(50)] = 15[400 + 1450] = 15(1850) = 27750$.
**Ans: ₹ 27,750**
#### **Q16. 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.**
#### **A16. Solution:**
$n=7, S_7=700, d=-20$.
$700 = \frac{7}{2}[2a + 6(-20)]$
$200 = 2a - 120$ (Dividing by 3.5 or multiplying by 2/7)
$320 = 2a \Rightarrow a = 160$.
Prizes: $160, 140, 120, 100, 80, 60, 40$.
**Ans: ₹ 160, 140, 120, 100, 80, 60, 40**
#### **Q17. 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?**
#### **A17. Solution:**
Trees per class:
Class I: $1 \times 3 = 3$
Class II: $2 \times 3 = 6$
...
Class XII: $12 \times 3 = 36$
Series: $3, 6, 9, \dots, 36$.
$n=12$.
$S_{12} = \frac{12}{2}(3 + 36) = 6(39) = 234$.
**Ans: 234 trees**
#### **Q18. 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 figure. What is the total length of such a spiral made up of thirteen consecutive semicircles? (Take $\pi = \frac{22}{7}$)**
#### **A18. Solution:**
Length of semicircle = $\pi r$.
$L_1 = \pi(0.5)$, $L_2 = \pi(1.0)$, $L_3 = \pi(1.5)$.
$a = 0.5\pi, d = 0.5\pi, n = 13$.
Total Length $S_{13} = \frac{13}{2}[2(0.5\pi) + 12(0.5\pi)]$
$S_{13} = \frac{13}{2}[\pi + 6\pi] = \frac{13}{2}(7\pi) = \frac{13}{2} \times 7 \times \frac{22}{7}$
$S_{13} = 13 \times 11 = 143$.
**Ans: 143 cm**
#### **Q19. 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?**
#### **A19. Solution:**
$S_n = 200, a = 20, d = -1$.
$200 = \frac{n}{2}[2(20) + (n-1)(-1)]$
$400 = n[40 - n + 1] \Rightarrow 400 = n[41 - n] \Rightarrow 400 = 41n - n^2$.
$n^2 - 41n + 400 = 0$.
$n^2 - 16n - 25n + 400 = 0 \Rightarrow (n-16)(n-25) = 0$.
$n=16$ or $n=25$.
If $n=25$: $a_{25} = 20 + 24(-1) = -4$ (Logs cannot be negative).
Therefore, $n=16$.
$a_{16} = 20 + 15(-1) = 5$.
**Ans: 16 rows, 5 logs in top row**
#### **Q20. 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 10 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?**
#### **A20. Solution:**
Distance for 1st potato: $2 \times 5 = 10$.
Distance for 2nd potato: $2 \times (5+3) = 16$.
Distance for 3rd potato: $2 \times (5+3+3) = 22$.
Series: $10, 16, 22, \dots$ ($n=10, d=6$).
$S_{10} = \frac{10}{2}[2(10) + 9(6)] = 5[20 + 54] = 5(74) = 370$.
**Ans: 370 m**