CBSE Class 10 Mathematics Arithmetic Progressions VBQs Set J

SUM OF FIRST n TERMS OF AN AP

• If first term of an AP be \( a \) and its common difference is \( d \), then the sum \( S_n \) of the first \( n \) terms of an AP is given by
  \( S = \frac{n}{2}[2a + (n - 1)d] \) or, \( S = \frac{n}{2}(a + a_n) \) where \( a_n = n\text{th term of the AP.} \)
• If \( l \) is the last term of an AP of \( n \) terms, then the sum of all ‘\( n \)’ terms can also be given by
  \( S_n = \frac{n}{2}(a + l) \). Sometimes \( S_n \) is also denoted by \( S \).
• The sum of first \( n \) positive integers is given by
  \( S_n = \frac{n(n + 1)}{2} \).
• If \( S_n \) is the sum of the first \( n \) terms of an AP, then its \( n\text{th} \) term is given by \( a_n = S_n - S_{n-1} \), i.e., the \( n\text{th} \) term of an AP is the difference of the sum to first \( n \) terms and the sum to first \( (n - 1) \) terms of it.

Question. Find the sum of the AP: \( 7 + 10\frac{1}{2} + 14 + \dots + 84 \)
Answer: Let \( a \) be the first term, \( d \) the common difference and \( a_n \) the last term of given AP.
We have, \( a = 7, d = 10\frac{1}{2} - 7 = \frac{21}{2} - 7 = \frac{21 - 14}{2} = \frac{7}{2} \) and \( a_n = 84 \)
Now, \( a_n = a + (n - 1)d \Rightarrow 84 = 7 + (n - 1) \times \frac{7}{2} \)
\( \Rightarrow 77 = (n - 1) \times \frac{7}{2} \Rightarrow 11 \times 2 = (n - 1) \Rightarrow 22 = n - 1 \)
\( \therefore n = 22 + 1 = 23 \)
Now, \( S_n = \frac{n}{2}[2a + (n - 1)d] \)
\( \Rightarrow S = \frac{23}{2} \left[ 2 \times 7 + (23 - 1) \times \frac{7}{2} \right] \)
\( \Rightarrow S = \frac{23}{2} [14 + 22 \times \frac{7}{2}] = \frac{23}{2} [14 + 77] = \frac{23}{2} \times 91 = \frac{2093}{2} = 1046\frac{1}{2} \)

Question. How many terms of the AP: \( 9, 17, 25, \dots \) must be taken to give a sum of \( 636 \)?
Answer: Let sum of \( n \) terms be \( 636 \).
Then, \( S_n = 636, a = 9, d = 17 - 9 = 8 \)
\( \Rightarrow \frac{n}{2}[2a + (n - 1)d] = 636 \Rightarrow \frac{n}{2}[2 \times 9 + (n - 1) \times 8] = 636 \)
\( \Rightarrow \frac{n}{2} \times 2[9 + (n - 1) \times 4] = 636 \Rightarrow n(9 + 4n - 4) = 636 \)
\( \Rightarrow n[5 + 4n] = 636 \Rightarrow 5n + 4n^2 = 636 \Rightarrow 4n^2 + 5n - 636 = 0 \)
\( \therefore n = \frac{-5 \pm \sqrt{(5)^2 - 4 \times 4 \times (-636)}}{2 \times 4} = \frac{-5 \pm \sqrt{25 + 10176}}{8} = \frac{-5 \pm \sqrt{10201}}{8} = \frac{-5 \pm 101}{8} \)
\( n = \frac{96}{8}, \frac{-106}{8} = 12, -\frac{53}{4} \)
But \( n \neq -\frac{53}{4} \), So, \( n = 12 \)
Thus, the sum of \( 12 \) terms of the given AP is \( 636 \).

Question. 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.
Answer: We have, \( S_7 = 49 \)
\( \Rightarrow 49 = \frac{7}{2}[2a + (7 - 1) \times d] \Rightarrow 7 \times 2 = [2a + 6d] \)
\( \Rightarrow 14 = 2a + 6d \Rightarrow a + 3d = 7 \dots(i) \)
and \( S_{17} = 289 \)
\( \Rightarrow 289 = \frac{17}{2}[2a + (17 - 1)d] \Rightarrow 2a + 16d = \frac{289 \times 2}{17} = 34 \)
\( \Rightarrow a + 8d = 17 \dots(ii) \)
Now subtracting equation (i) from (ii), we get
\( 5d = 10 \Rightarrow d = 2 \)
Putting the value of \( d \) in equation (i), we get
\( a + 3 \times 2 = 7 \Rightarrow a = 7 - 6 = 1 \)
Here \( a = 1 \) and \( d = 2 \)
Now, \( S_n = \frac{n}{2}[2a + (n - 1)d] \)
\( = \frac{n}{2}[2 \times 1 + (n - 1) \times 2] = \frac{n}{2}[2 + 2n - 2] = \frac{n}{2} \times 2n = n^2 \)

Question. Find the sum of the integers between \( 100 \) and \( 200 \) that are:
(i) divisible by \( 9 \)
(ii) not divisible by \( 9 \)
Answer: (i) Numbers divisible by \( 9 \) between \( 100 \) and \( 200 \) are \( 108, 117, 126, \dots, 198 \).
Here, \( a = 108, d = 9, a_n = 198 \)
\( a_n = a + (n - 1)d \Rightarrow 198 = 108 + (n - 1)9 \)
\( \Rightarrow 198 = 108 + 9n - 9 \Rightarrow 198 = 99 + 9n \)
\( \Rightarrow 198 - 99 = 9n \Rightarrow \frac{99}{9} = n \)
\( \Rightarrow n = 11 \)
Thus, \( S_n = \frac{n}{2}[2a + (n - 1)d] \)
\( S_{11} = \frac{11}{2}[2 \times 108 + 10 \times 9] = \frac{11}{2}[216 + 90] = 1683 \)
(ii) Numbers between \( 100 \) and \( 200 \) are \( 101, 102, \dots, 199 \).
Here, \( a = 101, d = 1, a_n = 199 \)
Now, \( a_n = a + (n - 1)d \)
\( \Rightarrow 199 = 101 + (n - 1)1 \Rightarrow 199 = 100 + n \Rightarrow n = 99 \)
So, \( S_n = \frac{n}{2}(a + l) \), where \( l \) is the last term
\( = \frac{99}{2}(101 + 199) = \frac{99}{2} \times 300 = 14850 \)
Sum of the numbers which are not divisible by \( 9 \)
= Sum of total numbers – sum of numbers which are divisible by \( 9 \)
\( = S_{99} - S_{11} = 14850 - 1683 = 13167 \)

Question. The sum of the first \( n \) terms of an AP is \( 3n^2 + 6n \). Find the \( n\text{th} \) term of this AP.
Answer: We have, \( S_n = 3n^2 + 6n \)
\( S_{n-1} = 3(n - 1)^2 + 6(n - 1) \)
\( = 3(n^2 + 1 - 2n) + 6n - 6 \)
\( = 3n^2 + 3 - 6n + 6n - 6 = 3n^2 - 3 \)
The \( n\text{th} \) term will be \( a_n \)
\( a_n = S_n - S_{n-1} \)
\( = 3n^2 + 6n - (3n^2 - 3) = 6n + 3 \)

Question. In an AP, the sum of first ten terms is \( -150 \) and the sum of next ten terms is \( -550 \). Find the AP.
Answer: Let \( a \) be the first term and \( d \) the common difference of the AP.
We have, \( S_{10} = -150 \)
\( \Rightarrow \frac{10}{2}[2a + 9d] = -150 \Rightarrow 2a + 9d = -30 \dots(i) \)
and \( S_{20} - S_{10} = -550 \)
\( \Rightarrow S_{20} = -550 - 150 = -700 \)
\( \Rightarrow \frac{20}{2}[2a + 19d] = -700 \)
\( \Rightarrow 2a + 19d = -70 \dots(ii) \)
From (i) and (ii), \( d = -4 \) and \( a = 3 \)
So, the AP is: \( 3, -1, -5, \dots \)

Question. The first and the last terms of an AP are \( 5 \) and \( 45 \) respectively. If the sum of all its terms is \( 400 \), find its common difference ‘\( d \)’.
Answer: We have, \( a = 5, T_n = 45, S_n = 400 \)
\( \therefore T_n = a + (n - 1)d \)
\( \Rightarrow 45 = 5 + (n - 1)d \Rightarrow (n - 1)d = 40 \dots(i) \)
\( S_n = \frac{n}{2}(a + T_n) \)
\( \Rightarrow 400 = \frac{n}{2}(5 + 45) \Rightarrow n = 2 \times 8 = 16 \)
Substituting the value of \( n \) in (i), we get
\( (16 - 1)d = 40 \Rightarrow 15d = 40 \)
\( d = \frac{40}{15} = \frac{8}{3} \)

Question. The sum of the first \( n \) terms of an AP is given by \( S_n = 3n^2 - 4n \). Determine the AP and the \( 12\text{th} \) term.
Answer: We have, \( S_n = 3n^2 - 4n \dots(i) \)
Replacing \( n \) by \( n - 1 \), we get
\( S_{n-1} = 3(n - 1)^2 - 4(n - 1) \dots(ii) \)
Since, \( a_n = S_n - S_{n-1} = \{3n^2 - 4n\} - \{3(n - 1)^2 - 4(n - 1)\} \)
\( = \{3n^2 - 4n\} - \{3n^2 + 3 - 6n - 4n + 4\} \)
\( = 3n^2 - 4n - 3n^2 - 3 + 6n + 4n - 4 = 6n - 7 \)
So, \( n\text{th} \) term, \( a_n = 6n - 7 \dots(iii) \)
Substituting \( n = 1, 2, 3, \dots \) respectively in (iii), we get
\( a_1 = 6 \times 1 - 7 = -1, a_2 = 6 \times 2 - 7 = 5 \)
\( a_3 = 6 \times 3 - 7 = 11 \)
Hence, AP is \( -1, 5, 11, \dots \)
\( 12\text{th} \) term, \( a_{12} = 6 \times 12 - 7 = 72 - 7 = 65 \)

Question. The sum of \( n, 2n, 3n \) terms of an AP are \( S_1, S_2 \) and \( S_3 \) respectively. Prove that \( S_3 = 3(S_2 - S_1) \).
Answer: Let \( a \) be the first term and \( d \) the common difference of the AP
\( \therefore S_1 = \frac{n}{2}[2a + (n - 1)d] \dots(i) \)
\( S_2 = \frac{2n}{2}[2a + (2n - 1)d] \dots(ii) \)
\( S_3 = \frac{3n}{2}[2a + (3n - 1)d] \dots(iii) \)
Now, \( S_2 - S_1 = \frac{2n}{2}[2a + (2n - 1)d] - \frac{n}{2}[2a + (n - 1)d] \)
\( = \frac{n}{2}[2\{2a + (2n - 1)d\} - \{2a + (n - 1)d\}] \)
\( = \frac{n}{2}[4a + 4nd - 2d - 2a - nd + d] \)
\( = \frac{n}{2}[2a + 3nd - d] = \frac{n}{2}[2a + (3n - 1)d] \)
\( \therefore 3(S_2 - S_1) = \frac{3n}{2}[2a + (3n - 1)d] = S_3 \)
\( \Rightarrow 3(S_2 - S_1) = S_3 \)

Question. Find the common difference of an AP whose first term is \( 5 \) and the sum of its first four terms is half the sum of the next four terms.
Answer: Given the first term of the AP, \( a = 5 \). Let \( d \) be the common difference.
Then, as per the question
\( \sum_{n=1}^{4} a_n = \frac{1}{2} \sum_{n=5}^{8} a_n \)
\( \Rightarrow a_1 + a_2 + a_3 + a_4 = \frac{1}{2} [a_5 + a_6 + a_7 + a_8] \)
\( \Rightarrow [a + (a + d) + (a + 2d) + (a + 3d)] = \frac{1}{2} [(a + 4d) + (a + 5d) + (a + 6d) + (a + 7d)] \)
\( \Rightarrow 4a + 6d = \frac{1}{2} (4a + 22d) \Rightarrow 2(4a + 6d) = 4a + 22d \)
\( \Rightarrow 2(20 + 6d) = 20 + 22d \) [\( a = 5 \) (given)]
\( \Rightarrow 40 + 12d = 20 + 22d \Rightarrow 20 = 22d - 12d \)
\( \Rightarrow 20 = 10d \Rightarrow d = 2 \)

Exercise 

I. Very Short Answer Type Questions 

Question. The sum of first five terms of the AP: \( 3, 7, 11, 15, \dots \) is:
(a) 44
(b) 55
(c) 22
(d) 11
Answer: (b) 55

Question. If the first term of an AP is \( 1 \) and the common difference is \( 2 \), then the sum of first \( 26 \) terms is
(a) 484
(b) 576
(c) 676
(d) 625
Answer: (c) 676

Question. If the sum to \( n \) terms of an AP is \( 3n^2 + 4n \), then the common difference of the AP is
(a) 7
(b) 5
(c) 8
(d) 6
Answer: (d) 6

Question. If \( a, b, c \) are in AP then \( ab + bc = \)
(a) \( b \)
(b) \( b^2 \)
(c) \( 2b^2 \)
(d) \( \frac{1}{b} \)
Answer: (c) \( 2b^2 \)

Question. The sum of all natural numbers which are less than \( 100 \) and divisible by \( 6 \) is
(a) 412
(b) 510
(c) 672
(d) 816
Answer: (d) 816

Question. Assertion (A): Sum of the first \( 10 \) terms of the arithmetic progression \( -0.5, -1.0, -1.5, \dots \) is \( 27.5 \).
Reason (R): Sum of first \( n \) terms of an AP is given as \( S_n = \frac{n}{2} [2a + (n - 1)d] \) where \( a = \text{first term}, d = \text{common difference.} \)
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).
(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).
(c) Assertion (A) is true but reason (R) is false.
(d) Assertion (A) is false but reason (R) is true.
Answer: (a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).

Question. Assertion (A): The sum of the first \( n \) terms of an AP is given by \( S_n = 3n^2 - 4n \). Then its \( n\text{th} \) term, \( a_n = 6n - 7 \).
Reason (R): \( n\text{th} \) term of an AP, whose sum of \( n \) terms is \( S_n \), is given by \( a_n = S_n - S_{n-1} \).
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).
(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).
(c) Assertion (A) is true but reason (R) is false.
(d) Assertion (A) is false but reason (R) is true.
Answer: (a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).

Question. Assertion (A): Sum of first hundred even natural numbers divisible by \( 5 \) is \( 500 \).
Reason (R): Sum of the first \( n \) terms of an AP is given by \( S_n = \frac{n}{2} [a + l] \) where \( l = \text{last term.} \)
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).
(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).
(c) Assertion (A) is true but reason (R) is false.
(d) Assertion (A) is false but reason (R) is true.
Answer: (d) Assertion (A) is false but reason (R) is true.

Question. Find the sum of first \( 10 \) terms of the AP: \( 2, 7, 12, \dots \)
Answer: \( a = 2, d = 7 - 2 = 5, n = 10 \).
\( S_{10} = \frac{10}{2} [2(2) + (10 - 1)5] = 5 [4 + 45] = 5 \times 49 = 245 \).

Question. If the sum of first \( m \) terms of an AP is \( 2m^2 + 3m \), then what is its second term?
Answer: \( S_m = 2m^2 + 3m \).
\( S_1 = a_1 = 2(1)^2 + 3(1) = 5 \).
\( S_2 = a_1 + a_2 = 2(2)^2 + 3(2) = 8 + 6 = 14 \).
\( a_2 = S_2 - S_1 = 14 - 5 = 9 \).

Case Study Based Questions

Pollution—A Major Problem: One of the major serious problems that the world is facing today is the environmental pollution. Common types of pollution include light, noise, water and air pollution. In a school, students thoughts of planting trees in and around the school to reduce noise pollution and air pollution.
Condition I: It was decided that the number of trees that each section of each class will plant 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 a section of class XII will plant 12 trees.
Condition II: It was decided that the number of trees that each section of each class will plant be the double of the class in which they are studying, e.g. a section of class I will plant 2 trees, a section of class II will plant 4 trees and so on a section of class XII will plant 24 trees.

Question. Refer to Condition I: The AP formed by sequence i.e. number of plants by students is
(a) 0, 1, 2, 3, ..., 12
(b) 1, 2, 3, 4, ..., 12
(c) 0, 1, 2, 3, ..., 15
(d) 1, 2, 3, 4, ..., 15
Answer: (b) 1, 2, 3, 4, ..., 12

Question. Refer to Condition I: If there are two sections of each class, how many trees will be planted by the students?
(a) 126
(b) 152
(c) 156
(d) 184
Answer: (c) 156

Question. Refer to Condition I: If there are three sections of each class, how many trees will be planted by the students?
(a) 234
(b) 260
(c) 310
(d) 326
Answer: (a) 234

Question. Refer to Condition II: If there are two sections of each class, how many trees will be planted by the students?
(a) 422
(b) 312
(c) 360
(d) 540
Answer: (b) 312

Question. Refer to Condition II: If there are three sections of each class, how many trees will be planted by the students?
(a) 468
(b) 590
(c) 710
(d) 620
Answer: (a) 468

Your elder brother wantsto buy a car and plansto take loan from a bank for his car. He repays histotal loan of ` 1,18,000 by paying every month starting with the first instalment of ` 1000. If he increases the instalment by ` 100 every month , answer the following:

Question. The amount paid by him in 30th installment is
(a) ` 3900
(b) ` 3500
(c) ` 3700
(d) ` 3600
Answer: (a) ` 3900

Question. The total amount paid by him upto 30 installments is
(a) ` 37000
(b) ` 73500
(c) ` 75300
(d) ` 75000
Answer: (b) ` 73500

Question. What amount does he still have to pay after 30th installment?
(a) ` 45500
(b) ` 49000
(c) ` 44500
(d) ` 54000
Answer: (c) ` 44500

Question. If total installments are 40, then amount paid in the last installment is
(a) ` 4900
(b) ` 3900
(c) ` 5900
(d) ` 9400
Answer: (a) ` 4900

Question. The ratio of the 1st installment to the last installment is
(a) 1 : 49
(b) 10 : 49
(c) 10 : 39
(d) 39 : 10
Answer: (b) 10 : 49

IMPORTANT FORMULAE

• The \( n\text{th} \) term of an AP, \( a_n = a + (n - 1)d \)
• The \( n\text{th} \) term of an AP from end, \( a_n = l - (n - 1)d \)
• Sum of finite terms of an AP
  \( S_n = \frac{n}{2}[2a + (n - 1)d] \) or \( S_n = \frac{n}{2}(a + a_n) \)
• If there are only \( n \) terms in an AP, then
  \( a_n = l \), the last term
  \( S_n = \frac{n}{2}(a + l) \)

Note: \( a_n = S_n - S_{n-1} \) where, \( a = \text{first term, } n = \text{number of terms, } d = \text{common difference, and } a_n = n\text{th term, } l = \text{last term.} \)