Arithmetic Progressions - Exercise 5.3

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Overview

Master the summation of arithmetic series: $S_n = \dfrac{n}{2}[2a + (n - 1)d] = \dfrac{n}{2}(a + l)$. Solve problems on given parameter sets, quadratic formulations in $n$, multiples of integers, and real-life applied situations (construction penalties, tree planting, spiral lengths, and log stacking).

NCERT Exercise Practice questions with detailed step-by-step solutions

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Basics (Q1-3) Sums & Terms (Q4-10) Properties (Q11-14) Word Problems (Q15-20)
Questions 1, 2 & 3 Fundamental Sum Formulas & Parameter Drilling

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) $\dfrac{1}{15}, \dfrac{1}{12}, \dfrac{1}{10}, \dots$ to $11$ terms.

2. Find the sums given below:
(i) $7 + 10\dfrac{1}{2} + 14 + \dots + 84$
(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 Solutions
(i) $a = 2, d = 5, n = 10$:
$S_{10} = \dfrac{10}{2}[2(2) + 9(5)] = 5[4 + 45] = 5(49) = \mathbf{245}$.

(ii) $a = -37, d = 4, n = 12$:
$S_{12} = \dfrac{12}{2}[2(-37) + 11(4)] = 6[-74 + 44] = 6(-30) = \mathbf{-180}$.

(iii) $a = 0.6, d = 1.1, n = 100$:
$S_{100} = \dfrac{100}{2}[2(0.6) + 99(1.1)] = 50[1.2 + 108.9] = 50(110.1) = \mathbf{5505}$.

(iv) $a = \dfrac{1}{15}, d = \dfrac{1}{12} - \dfrac{1}{15} = \dfrac{5 - 4}{60} = \dfrac{1}{60}, n = 11$:
$S_{11} = \dfrac{11}{2}\left[\dfrac{2}{15} + 10\left(\dfrac{1}{60}\right)\right] = \dfrac{11}{2}\left[\dfrac{2}{15} + \dfrac{1}{6}\right] = \dfrac{11}{2}\left[\dfrac{4 + 5}{30}\right] = \dfrac{11}{2} \times \dfrac{9}{30} = \mathbf{\dfrac{33}{20}}$.
Question 2 Solutions
(i) $a = 7, l = 84, d = 3.5 = \dfrac{7}{2}$.
$84 = 7 + (n - 1)\left(\dfrac{7}{2}\right) \implies 77 = \dfrac{7}{2}(n - 1) \implies n - 1 = 22 \implies n = 23$.
$S_{23} = \dfrac{23}{2}(7 + 84) = \dfrac{23 \times 91}{2} = \mathbf{1046\dfrac{1}{2}} = \mathbf{1046.5}$.

(ii) $a = 34, l = 10, d = -2$.
$10 = 34 + (n - 1)(-2) \implies -24 = -2(n - 1) \implies n - 1 = 12 \implies n = 13$.
$S_{13} = \dfrac{13}{2}(34 + 10) = \dfrac{13}{2}(44) = 13 \times 22 = \mathbf{286}$.

(iii) $a = -5, l = -230, d = -3$.
$-230 = -5 + (n - 1)(-3) \implies -225 = -3(n - 1) \implies n - 1 = 75 \implies n = 76$.
$S_{76} = \dfrac{76}{2}[-5 + (-230)] = 38(-235) = \mathbf{-8930}$.
Question 3 Solutions
(i) $50 = 5 + (n - 1)3 \implies 45 = 3(n - 1) \implies n = 16$. $S_{16} = \dfrac{16}{2}(5 + 50) = 8(55) = \mathbf{440}$.
(ii) $35 = 7 + 12d \implies 12d = 28 \implies d = \mathbf{\dfrac{7}{3}}$. $S_{13} = \dfrac{13}{2}(7 + 35) = 13 \times 21 = \mathbf{273}$.
(iii) $37 = a + 11(3) \implies a = 37 - 33 = \mathbf{4}$. $S_{12} = \dfrac{12}{2}(4 + 37) = 6(41) = \mathbf{246}$.
(iv) $a + 2d = 15$  (1);  $S_{10} = 5[2a + 9d] = 125 \implies 2a + 9d = 25$  (2). Solving: $d = \mathbf{-1}, a = 17 \implies a_{10} = 17 + 9(-1) = \mathbf{8}$.
(v) $S_9 = \dfrac{9}{2}[2a + 8(5)] = 75 \implies 9(a + 20) = 75 \implies 3a + 60 = 25 \implies a = \mathbf{-\dfrac{35}{3}}$. $a_9 = -\dfrac{35}{3} + 40 = \mathbf{\dfrac{85}{3}}$.
(vi) $90 = \dfrac{n}{2}[4 + (n - 1)8] \implies 90 = n(4n - 2) \implies 2n^2 - n - 45 = 0 \implies (2n + 9)(n - 5) = 0 \implies n = \mathbf{5}$. $a_5 = 2 + 4(8) = \mathbf{34}$.
(vii) $210 = \dfrac{n}{2}(8 + 62) = 35n \implies n = \mathbf{6}$. $62 = 8 + 5d \implies 5d = 54 \implies d = \mathbf{\dfrac{54}{5}}$.
(viii) $a_n = 4 \implies a + 2(n - 1) = 4 \implies a = 6 - 2n$. $S_n = \dfrac{n}{2}(a + 4) = -14 \implies n(10 - 2n) = -28 \implies n^2 - 5n - 14 = 0 \implies n = \mathbf{7}$. $a = 6 - 14 = \mathbf{-8}$.
(ix) $192 = \dfrac{8}{2}[6 + 7d] = 4(6 + 7d) \implies 48 = 6 + 7d \implies 7d = 42 \implies d = \mathbf{6}$.
(x) $144 = \dfrac{9}{2}(a + 28) \implies 16 \times 2 = a + 28 \implies a = 32 - 28 = \mathbf{4}$.
Questions 4 to 10 Sum Conditions & Algebraic Formulations

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 $22^{\text{nd}}$ 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 $a_1, a_2, \dots, a_n$ 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.

Question 4 Solution
Here $a = 9, d = 8, S_n = 636$.
$636 = \dfrac{n}{2}[2(9) + (n - 1)8] = \dfrac{n}{2}[18 + 8n - 8] = n(4n + 5)$
$4n^2 + 5n - 636 = 0$.
Using the quadratic formula: $n = \dfrac{-5 \pm \sqrt{25 - 4(4)(-636)}}{8} = \dfrac{-5 \pm \sqrt{10201}}{8} = \dfrac{-5 \pm 101}{8}$.
Taking the positive integer: $n = \dfrac{96}{8} = 12$.
$\therefore 12$ terms must be taken.
Question 5 Solution
$a = 5, l = 45, S_n = 400$.
$S_n = \dfrac{n}{2}(a + l) \implies 400 = \dfrac{n}{2}(5 + 45) = 25n \implies n = 16$.
Now, $l = a + (n - 1)d \implies 45 = 5 + 15d \implies 15d = 40 \implies d = \dfrac{40}{15} = \dfrac{8}{3}$.
$\therefore$ Number of terms $n = 16$, common difference $d = \dfrac{8}{3}$.
Question 6 Solution
$a = 17, l = 350, d = 9$.
$l = a + (n - 1)d \implies 350 = 17 + (n - 1)9 \implies 333 = 9(n - 1) \implies n - 1 = 37 \implies n = 38$.
$S_{38} = \dfrac{38}{2}(17 + 350) = 19 \times 367 = 6973$.
$\therefore$ There are $38$ terms and their sum is $6973$.
Question 7 Solution
$d = 7, a_{22} = 149 \implies a + 21(7) = 149 \implies a + 147 = 149 \implies a = 2$.
$S_{22} = \dfrac{22}{2}(a + a_{22}) = 11(2 + 149) = 11 \times 151 = 1661$.
$\therefore S_{22} = 1661$.
Question 8 Solution
$a_2 = 14$ and $a_3 = 18 \implies d = 18 - 14 = 4$.
$a = a_2 - d = 14 - 4 = 10$.
$S_{51} = \dfrac{51}{2}[2(10) + 50(4)] = \dfrac{51}{2}[20 + 200] = \dfrac{51}{2} \times 220 = 51 \times 110 = 5610$.
$\therefore S_{51} = 5610$.
Question 9 Solution
$S_7 = \dfrac{7}{2}[2a + 6d] = 7(a + 3d) = 49 \implies a + 3d = 7$    (1)
$S_{17} = \dfrac{17}{2}[2a + 16d] = 17(a + 8d) = 289 \implies a + 8d = 17$    (2)
Subtracting (1) from (2): $5d = 10 \implies d = 2$.
From (1): $a + 6 = 7 \implies a = 1$.
$S_n = \dfrac{n}{2}[2(1) + (n - 1)2] = \dfrac{n}{2}[2n] = n^2$.
$\therefore S_n = n^2$.
Question 10 Solution
(i) $a_n = 3 + 4n$:
$a_1 = 7, a_2 = 11, a_3 = 15$. Difference $a_{k+1} - a_k = 4$ is constant, so it forms an AP ($a = 7, d = 4$).
$S_{15} = \dfrac{15}{2}[2(7) + 14(4)] = \dfrac{15}{2}[14 + 56] = \dfrac{15}{2}(70) = 525$.

(ii) $a_n = 9 - 5n$:
$a_1 = 4, a_2 = -1, a_3 = -6$. Difference $a_{k+1} - a_k = -5$ is constant, so it forms an AP ($a = 4, d = -5$).
$S_{15} = \dfrac{15}{2}[2(4) + 14(-5)] = \dfrac{15}{2}[8 - 70] = \dfrac{15}{2}(-62) = -465$.
Questions 11 to 14 Sum Functions & Natural 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 $3^{\text{rd}}$, the $10^{\text{th}}$ and the $n^{\text{th}}$ 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 $50$.

Question 11 Solution
Given $S_n = 4n - n^2$.
First term $a_1 = S_1 = 4(1) - 1^2 = \mathbf{3}$.
Sum of first two terms $S_2 = 4(2) - 2^2 = 8 - 4 = \mathbf{4}$.
Second term $a_2 = S_2 - S_1 = 4 - 3 = \mathbf{1}$.
Common difference $d = a_2 - a_1 = 1 - 3 = -2$.
$3^{\text{rd}}$ term $a_3 = a + 2d = 3 + 2(-2) = \mathbf{-1}$.
$10^{\text{th}}$ term $a_{10} = a + 9d = 3 + 9(-2) = 3 - 18 = \mathbf{-15}$.
$n^{\text{th}}$ term $a_n = a + (n - 1)d = 3 + (n - 1)(-2) = \mathbf{5 - 2n}$.
Question 12 Solution
Positive integers divisible by $6$: $6, 12, 18, \dots$ to $40$ terms.
$a = 6, d = 6, n = 40$.
$S_{40} = \dfrac{40}{2}[2(6) + 39(6)] = 20[12 + 234] = 20(246) = \mathbf{4920}$.
Question 13 Solution
Multiples of $8$: $8, 16, 24, \dots$ to $15$ terms.
$a = 8, d = 8, n = 15$.
$S_{15} = \dfrac{15}{2}[2(8) + 14(8)] = \dfrac{15}{2}[16 + 112] = \dfrac{15}{2}(128) = 15 \times 64 = \mathbf{960}$.
Question 14 Solution
Odd numbers between $0$ and $50$: $1, 3, 5, \dots, 49$.
$a = 1, d = 2, l = 49$.
$49 = 1 + (n - 1)2 \implies 48 = 2(n - 1) \implies n = 25$.
$S_{25} = \dfrac{25}{2}(1 + 49) = \dfrac{25}{2}(50) = 25 \times 25 = \mathbf{625}$.
Questions 15 to 20 Real-World Applied AP 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. Class I plants 1, Class II plants 2... Class XII plants 12). 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 $0.5\text{ cm}, 1.0\text{ cm}, 1.5\text{ cm}, 2.0\text{ cm}, \dots$. What is the total length of such a spiral made up of $13$ consecutive semicircles? (Take $\pi = \dfrac{22}{7}$)

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 $5\text{ m}$ from the first potato, and the other potatoes are placed $3\text{ m}$ 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, and repeats. What is the total distance the competitor has to run?

Question 15 Solution
Penalties form an AP: $200, 250, 300, \dots$ for $n = 30$ days.
$a = 200, d = 50, n = 30$.
$S_{30} = \dfrac{30}{2}[2(200) + 29(50)] = 15[400 + 1450] = 15(1850) = \mathbf{₹27,750}$.
Question 16 Solution
Total sum $S_7 = 700, n = 7, d = -20$.
$S_7 = \dfrac{7}{2}[2a + 6(-20)] = 700 \implies \dfrac{1}{2}[2a - 120] = 100 \implies a - 60 = 100 \implies a = 160$.
The values of the seven prizes are:
₹160, ₹140, ₹120, ₹100, ₹80, ₹60, and ₹40.
Question 17 Solution
Since there are $3$ sections per class, Class $k$ plants $3 \times k$ trees.
Sequence: $3(1), 3(2), 3(3), \dots, 3(12) \implies 3, 6, 9, \dots, 36$.
This is an AP with $a = 3, d = 3, n = 12, l = 36$.
$S_{12} = \dfrac{12}{2}(3 + 36) = 6(39) = \mathbf{234}\text{ trees}$.
Question 18 Solution
Length of a semicircle of radius $r$ is $l = \pi r$.
Total length $L = \pi(0.5) + \pi(1.0) + \pi(1.5) + \dots + \pi(r_{13}) = \pi(0.5 + 1.0 + 1.5 + \dots \text{ to } 13 \text{ terms})$.
Sum of radii: $S_{13} = \dfrac{13}{2}[2(0.5) + 12(0.5)] = \dfrac{13}{2}[1 + 6] = \dfrac{91}{2}$.
$L = \dfrac{22}{7} \times \dfrac{91}{2} = 11 \times 13 = \mathbf{143}\text{ cm}$.
Question 19 Solution
Number of logs per row from bottom up: $20, 19, 18, \dots$
$a = 20, d = -1, S_n = 200$.
$200 = \dfrac{n}{2}[2(20) + (n - 1)(-1)] = \dfrac{n}{2}[41 - n] \implies 400 = 41n - n^2$
$n^2 - 41n + 400 = 0 \implies (n - 16)(n - 25) = 0 \implies n = 16$ or $n = 25$.
If $n = 25$, $a_{25} = 20 + 24(-1) = -4$ logs (impossible).
Therefore, $n = 16$.
Logs in the top row: $a_{16} = 20 + 15(-1) = 5$.
$\therefore$ The logs are placed in $16$ rows, and there are $5$ logs in the top row.
Question 20 Solution
Distance run for each potato involves a round trip (to potato and back):
$1^{\text{st}}$ potato: $2 \times 5 = 10\text{ m}$
$2^{\text{nd}}$ potato: $2 \times (5 + 3) = 16\text{ m}$
$3^{\text{rd}}$ potato: $2 \times (5 + 6) = 22\text{ m}$
This forms an AP: $10, 16, 22, \dots$ for $n = 10$ potatoes.
$a = 10, d = 6, n = 10$.
$S_{10} = \dfrac{10}{2}[2(10) + 9(6)] = 5[20 + 54] = 5(74) = \mathbf{370}\text{ metres}$.
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