CBSE Class 10 Mathematics Arithmetic Progressions Worksheet Set 15

Multiple Choice Questions

Choose and write the correct option in the following questions.

Question. Which of the following two numbers do you think should be the succeeding terms of the pattern given below?
-1/3, -1/12, 1/6, 5/12, __, __
(a) 2/3 and 12/11
(b) 4/6 and 11/4
(c) 2/3 and 11/12
(d) 1/2 and 24/22
Answer: (c) 2/3 and 11/12

Question. The first term of an AP is \( p \) and the common difference is \( q \), then its 10th term is 
(a) \( q + 9p \)
(b) \( p - 9q \)
(c) \( p + 9q \)
(d) \( 2p + 9q \)
Answer: (c) \( p + 9q \)

Question. The common difference of an AP, whose \( n^{th} \) term is \( a_n = (3n + 7) \), is 
(a) 3
(b) 7
(c) 10
(d) 6
Answer: (a) 3

Question. Find the nth term of the AP shown below.
__, __, -36, __, -44, ......... up to n
(a) \( 4n - 24 \)
(b) \( -4n - 24 \)
(c) \( 24 - 4n \)
(d) \( -40 + 4n \)
Answer: (b) \( -4n - 24 \)

Question. Which term of the AP 21, 42, 63, 84, ... is 210? 
(a) 9th
(b) 10th
(c) 11th
(d) 12th
Answer: (b) 10th

Question. The value of \( p \) for which \( (2p + 1) \), 10 and \( (5p + 5) \) are three consecutive terms of an AP is 
(a) -1
(b) -2
(c) 1
(d) 2
Answer: (d) 2

Question. The 4th term from the end of the AP -11, -8, -5, ..., 49 is 
(a) 37
(b) 40
(c) 43
(d) 58
Answer: (b) 40

Question. The next term of the AP \( \sqrt{8}, \sqrt{18}, \sqrt{32}, .... \) is
(a) \( 5\sqrt{2} \)
(b) \( 5\sqrt{3} \)
(c) \( 3\sqrt{3} \)
(d) \( 5\sqrt{3} \)
Answer: (a) \( 5\sqrt{2} \)

Question. The first four terms of an AP, whose first term is -2 and the common difference is -2 are 
(a) -2, 0, 2, 4
(b) -2, 4, -8, 16
(c) -2, -4, -6, -8
(d) -2, -4, -8, -16
Answer: (c) -2, -4, -6, -8

Question. The \( n^{th} \) term of the AP \( a, 3a, 5a \) is
(a) \( na \)
(b) \( (2n - 1)a \)
(c) \( (2n + 1)a \)
(d) \( 2na \)
Answer: (b) \( (2n - 1)a \)

Question. Two APs have same common difference. The first term of one of these is -1 and that of the other is - 8. Then the difference between their 4th term is 
(a) -1
(b) -8
(c) 7
(d) -9
Answer: (c) 7

Question. The 21st term of the AP whose first two terms are -3 and 4, is 
(a) 17
(b) 137
(c) 143
(d) -143
Answer: (b) 137

Question. What would be the last term of an arithmetic progression with 10 terms whose second term is -23 and the third term is -35?
(a) -119
(b) 119
(c) -650
(d) 350
Answer: (a) -119

Question. The first term of an AP is 5 and the last term is 45. If the sum of all the terms is 400, the number of terms is
(a) 20
(b) 8
(c) 10
(d) 16
Answer: (d) 16

Question. Rakesh loves to travel and he travels every year. He has seen 8 different cities in his first year. Thereafter every year he has seen 2 cities. If he had followed this pattern, how many cities did he see by the end of 10 years of travel? 
(a) \( [5 \{ 8 + 9(2) \}] \) cities
(b) \( [5 \{ 8 + 10(2) \}] \) cities
(c) \( [5 \{ 16 + 9(2) \}] \) cities
(d) \( [5 \{ 16 + 10(2) \}] \) cities
Answer: (c) \( [5 \{ 16 + 9(2) \}] \) cities

Question. The sum of first 16 terms of the AP 10, 6, 2, ... is 
(a) -320
(b) 320
(c) -352
(d) -400
Answer: (a) -320

Question. What is the sum of all the three-digit numbers divisible by 12? 
(a) \( \left[ \frac{73}{2} (108 + 996) \right] \)
(b) \( \left[ \frac{75}{2} (108 + 996) \right] \)
(c) \( \left[ \frac{73}{2} (216 + 996) \right] \)
(d) \( \left[ \frac{75}{2} (216 + 996) \right] \)
Answer: (b) \( \left[ \frac{75}{2} (108 + 996) \right] \)

Question. If the sum of \( p \) terms of an AP is \( q \) and the sum of \( q \) terms is \( p \), then the sum of \( p + q \) terms will be
(a) 0
(b) \( p - q \)
(c) \( p + q \)
(d) \( - (p + q) \)
Answer: (d) \( - (p + q) \)

Question. The first term of an AP is -76 and the sum of first 45 terms is -9360. Which of the following is the last term of this AP?
(a) 416 + 76
(b) 416 - 76
(c) - 416 + 76
(d) - 416 - 76
Answer: (d) - 416 - 76

Question. Mr. Sharma buys a property every year for 12 years. Every year he buys \( x \) acres more than the previous year. If in the 8th year he bought 45 acres of land and in the 5th year he bought 30 acres of land, how many acres did he buy in the last year? 
(a) \( [5 + 11(5)] \) acres
(b) \( [10 + 11(5)] \) acres
(c) \( [5 + 11(10)] \) acres
(d) \( [5 + 12(10)] \) acres
Answer: (b) \( [10 + 11(5)] \) acres

Short Answer Questions-

Question. Find \( a, b \) and \( c \) if it is given that the numbers \( a, 7, b, 23, c \) are in AP. 
Answer: Sol. Given that \( a, 7, b, 23, c \) are in AP.
\( \therefore \) Second term \( (a_2) = 7 \)
\( \implies \) \( a + (2 - 1) d = 7 \)
\( \implies \) \( a + d = 7 \) ...(i)
Also, 4th term \( (a_4) = 23 \)
\( \implies \) \( a + 3d = 23 \) ...(ii)
From equation (i) and (ii), we have
\( a + d = 7 \)
\( \underline{a + 3d = 23} \)
- 2d = - 16 \( \implies \) \( d = 8 \)
Putting \( d = 8 \) in equation (i), we get
\( a + 8 = 7 \)
\( \implies \) \( a = -1 \)
Now, \( b = 7 + d = 7 + 8 = 15 \)
\( c = 23 + d = 23 + 8 = 31 \)

Question. If \( m \) times the \( m^{th} \) term of an AP is equal to \( n \) times its \( n^{th} \) term, show that the \( (m + n)^{th} \) term of the AP is zero. 
Answer: Sol. Given \( m[a + (m - 1)d] = n[a + (n - 1)d] \)
\( \implies \) \( a(m - n) + d(m^2 - m - n^2 + n) = 0 \)
\( \implies \) \( (m - n) [a + (m + n - 1)d] = 0 \)
\( \because \) \( m \neq n \implies a + (m + n - 1)d = 0 \)
\( \implies \) \( a_{m+n} = 0 \)

Question. The digits of a positive number of three digits are in AP and their sum is 15. The number obtained by reversing the digits is 594 less than the original number. Find the number. 
Answer: Sol. Let three digits of 3 digit no be - \( a-d, a, a+d \).
Their sum = 15
\( a-d + a + a+d = 15 \) \( \implies \) \( 3a = 15 \) \( \implies \) \( a = 5 \)
Required 3 digit no = \( 100(a-d) + 10a + a+d \)
\( = 111a - 99d \)
No obtained by reversing digit = \( 100(a+d) + 10a + a-d \)
\( = 111a + 99d \)
ATQ-
\( 111a + 99d = 111a - 99d - 594 \)
\( \implies \) \( 594 = 111a - 99d - 111a - 99d \)
\( 594 = -198d \)
\( \frac{594}{-198} = d \)
\( -3 = d \)
The no = \( 111a - 99d \)
\( = 111 \times 5 - 99 \times (-3) \)
\( = 555 + 297 = 852 \).
No \( \implies \) 852

Question. If the \( m \)th term of an AP is \( \frac{1}{n} \) and \( n \)th term is \( \frac{1}{m} \), then show that its \( (mn) \)th term is 1.
Answer: Let \( a \) and \( d \) be the first term and common difference respectively of the given AP. Then
\( a_m = a + (m - 1)d \)
\( \implies \) \( a + (m - 1)d = \frac{1}{n} \) ...(i)
and \( a_n = a + (n - 1)d \)
\( \implies \) \( a + (n - 1)d = \frac{1}{m} \) ...(ii)
Subtracting (ii) from (i), we have
\( (m - n)d = \frac{1}{n} - \frac{1}{m} \)
\( \implies \) \( (m - n)d = \frac{m - n}{mn} \)
\( \therefore d = \frac{1}{mn} \)
Putting \( d = \frac{1}{mn} \) in (i), we get
\( a + (m - 1) \frac{1}{mn} = \frac{1}{n} \)
\( \implies \) \( a + \frac{m}{mn} - \frac{1}{mn} = \frac{1}{n} \)
\( \implies \) \( a - \frac{1}{mn} = 0 \) or \( a = \frac{1}{mn} \)
\( \therefore mn \)th term \( = a + (mn - 1)d = \frac{1}{mn} + (mn - 1) \frac{1}{mn} = \frac{1 + mn - 1}{mn} \)
\( \implies \) \( (mn) \)th term \( = 1 \)

Question. Which term of the arithmetic progression 3, 15, 27, 39 .... will be 120 more than its 21st term? 
Answer: We have, \( a = 3 \) and \( d = 12 \)
\( \therefore a_{21} = a + 20d = 3 + 20 \times 12 = 243 \)
Let \( n \)th term of the given AP be 120 more than its 21st term. Then,
\( a_n = 120 + a_{21} \)
\( \therefore 3 + (n - 1)d = 120 + 243 \)
\( \implies 3 + 12(n - 1) = 363 \)
\( \implies 12(n - 1) = 360 \)
\( \implies n - 1 = 30 \)
\( \implies n = 31 \)
Hence, 31st term of the given AP is 120 more than its 21st term.

Question. Show that the sum of all terms of an AP whose first term is \( a \), the second term is \( b \) and the last term is \( c \) is equal to \( \frac{(a + c)(b + c - 2a)}{2(b - a)} \).
Answer: In an AP, we have
First term \( = a \), Second term \( = b \)
\( \therefore \) Common difference, \( d = b - a \)
and last term, \( a_n = c \)
\( \therefore a_n = a + (n - 1)d \)
\( \implies c = a + (n - 1)(b - a) \)
\( \implies c - a = (n - 1)(b - a) \)
\( \implies n - 1 = \frac{c - a}{b - a} \)
\( \implies n = \frac{c - a}{b - a} + 1 \)
\( \implies n = \frac{c - a + b - a}{b - a} = \frac{b + c - 2a}{b - a} \)
\( \therefore n = \frac{b + c - 2a}{b - a} \)
\( \therefore \) Sum of all terms of an AP be
\( S_n = \frac{n}{2}(a + l) \)
\( = \frac{\left( \frac{b + c - 2a}{b - a} \right)}{2}(a + c) \)
\( S_n = \frac{(a + c)(b + c - 2a)}{2(b - a)} \)

Question. The sum of the first 30 terms of an AP is 1920. If the fourth term is 18, find its 11th term. 
Answer: \( \frac{30}{2} [2a + 29d] = 1920 \)
\( \implies 2a + 29d = 128 \) ...(i)
Also, \( a_4 = 18 \implies a + 3d = 18 \) ...(ii)
From equation (i) & (ii)
\( a = 6, d = 4 \)
\( \therefore a_{11} = a + 10d = 46 \) 

Question. Find the sum of all two digit natural numbers which are divisible by 4. 
Answer: Here \( a = 12, d = 4, a_n = 96 \)
The formula is \( a_n = a + (n - 1)d \)
Therefore \( 96 = 12 + (n - 1) \times 4 \)
\( \implies 96 = 8 + 4n \)
\( \implies n = \frac{88}{4} \)
\( \implies n = 22 \)
Apply the formula for sum,
\( S_n = \frac{n}{2} [2a + (n - 1)d] \)
Hence, \( S_{22} = 11[24 + 21 \times 4] = 11[24 + 84] \)
\( = 11 \times 108 = 1188 \).

Question. If the ratio of the sum of first \( n \) terms of two AP’s is \( (7n + 1) : (4n + 27) \), find the ratio of their \( m \)th terms. 
Answer: \( \frac{S_n}{S'_n} = \frac{\frac{n}{2}(2a + (n - 1)d)}{\frac{n}{2}(2a' + (n - 1)d')} = \frac{7n + 1}{4n + 27} \)
\( \implies \frac{a + \frac{n - 1}{2}d}{a' + \frac{n - 1}{2}d'} = \frac{7n + 1}{4n + 27} \) ...(i)
Since \( \frac{t_m}{t'_m} = \frac{a + (m - 1)d}{a' + (m - 1)d'} \), So replacing \( \frac{n - 1}{2} \) by \( m - 1 \implies n \) by \( 2m - 1 \) in (i)
\( = \frac{a + (m - 1)d}{a' + (m - 1)d'} = \frac{7(2m - 1) + 1}{4(2m - 1) + 27} \)
\( \implies \frac{t_m}{t'_m} = \frac{14m - 6}{8m + 23} \)

Question. For an AP, it is given that the first term \( (a) = 5 \), common difference \( (d) = 3 \), and the \( n \)th term \( (a_n) = 50 \). Find \( n \) and sum of first \( n \) terms \( (S_n) \) of the AP. 
Answer: We have, \( a = 5, d = 3 \) and \( a_n = 50 \)
\( \therefore a_n = 50 \)
\( \implies a + (n - 1)d = 50 \)
\( \implies 5 + (n - 1) \times 3 = 50 \)
\( \implies (n - 1) \times 3 = 45 \)
\( \implies n - 1 = 15 \)
\( \implies n = 16 \)
\( S_n = \frac{n}{2}(a + l) = \frac{16}{2}(5 + 50) = 8 \times 55 = 440 \)

Question. Find the sum of all two digit natural numbers which when divided by 3 yield 1 as remainder.
Answer: Two digit natural numbers which when divided by 3 yield 1 as remainder are:
10, 13, 16, 19, ....., 97, which forms an AP.
with \( a = 10, d = 3, a_n = 97 \)
\( a_n = 97 \implies a + (n - 1)d = 97 \)
or \( 10 + (n - 1)3 = 97 \)
\( \implies (n - 1) = \frac{87}{3} = 29 \)
\( \implies n = 30 \)
Now, \( S_{30} = \frac{30}{2} [2 \times 10 + 29 \times 3] = 15(20 + 87) = 15 \times 107 = 1605 \)

Long Answer Questions

Each of the following questions are of 5 marks.

Question. The sum of four consecutive numbers in AP is 32 and the ratio of the product of the first and last terms to the product of two middle terms is 7 : 15. Find the numbers. 
Answer: Let four consecutive numbers in AP are \( a - 3d, a - d, a + d, a + 3d \).
\( \therefore \) Sum \( = (a - 3d) + (a - d) + (a + d) + (a + 3d) \)
\( 32 = 4a \)
\( \implies a = \frac{32}{4} = 8 \)
ATQ, \( \frac{(a - 3d)(a + 3d)}{(a - d)(a + d)} = \frac{7}{15} \)
\( \implies \frac{a^2 - 9d^2}{a^2 - d^2} = \frac{7}{15} \)
\( \implies \frac{8^2 - 9d^2}{8^2 - d^2} = \frac{7}{15} \)
\( \implies 1 - \frac{8d^2}{8^2 - d^2} = 1 - \frac{7}{15} \) (Subtract from 1 both sides)
\( \implies \frac{8^2 - d^2 - 8^2 + 9d^2}{8^2 - d^2} = \frac{15 - 7}{15} \)
\( \implies \frac{8d^2}{64 - d^2} = \frac{8}{15} \)
\( \implies \frac{d^2}{64 - d^2} = \frac{1}{15} \)
\( \implies 15d^2 = 64 - d^2 \)
\( \implies 16d^2 = 64 \)
\( \implies d^2 = \frac{64}{16} = 4 \)
\( \therefore d = \pm 2 \)
When \( a = 8, d = 2 \) then four numbers are
\( 8 - 6, 8 - 2, 8 + 2, 8 + 6 \)
2, 6, 10, 14
When \( a = 8, d = -2 \), the four numbers are
14, 10, 6, 2

Assertion-Reason Questions

The following questions consist of two statements—Assertion(A) and Reason(R). Answer these questions selecting the appropriate option given below:
(a) Both A and R are true and R is the correct explanation for A.
(b) Both A and R are true but R is not the correct explanation for A.
(c) A is true but R is false.
(d) A is false but R is true.

Question. Assertion (A) : Common difference of the AP \( -5, -1, 3, 7, ... \) is 4.
Reason (R) : Common difference of the AP \( a, a+d, a+2d, .... \) is given by \( d = \text{2nd term } - \text{1st term} \).
Answer: Solution : Common difference, \( d = -1 - (-5) = 4 \)
So, both A and R are true and R is the correct explanation for A.
Hence, option (a) is correct.

Question. Assertion (A) : If \( a_n - a_{n-1} \) is not independent of \( n \) then the given sequence is an AP.
Reason (R) : Common difference \( d = a_n - a_{n-1} \) is constant or independent of \( n \).
Answer: Solution : We have, common difference of an AP \( d = a_n - a_{n-1} \) is independent of \( n \) or constant. So, A is false but R is true.
Hence, option (d) is correct.

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 to \( n \) terms is \( S_n \), is given by \( a_n = S_n - S_{n-1} \).
Answer: Solution : \( n\text{th} \) term of an AP be \( a_n = S_n - S_{n-1} \)

\( \implies a_n = 3n^2 - 4n - 3(n - 1)^2 + 4(n - 1) \)

\( \implies a_n = 6n - 7 \)
So, both A and R are true and R is the correct explanation for A.
Hence, option (a) is correct.

Question. Assertion (A) : Common difference of an AP in which \( a_{21} - a_7 = 84 \) is 14.
Reason (R) : \( n\text{th} \) term of an AP is given by \( a_n = a + (n - 1)d \).
Answer: Solution : We have, \( a_n = a + (n - 1)d \)
\( \therefore a_{21} - a_7 = \{a + (21 - 1)d\} - \{a + (7 - 1)d\} = 84 \)

\( \implies a + 20d - a - 6d = 84 \)

\( \implies 14d = 84 \)

\( \implies d = \frac{84}{14} = 6 \)

\( \implies d = 6 \)
So, A is false but R is true.
Hence, option (d) is correct.

Question. Assertion (A) : If three consecutive terms \( 2k+1, 3k + 3 \) and \( 5k - 1 \) form an AP then \( k \) is equal to 6.
Reason (R) : In an AP \( a, a+d, a+2d, ... \), the sum of \( n \) terms of the AP be \( S_n = \frac{n}{2}(2a + (n-1)d) \).
Answer: Solution : \( 2k+1, 3k + 3 \) and \( 5k - 1 \) form an AP, so
\( (3k + 3) - (2k + 1) = (5k - 1) - (3k + 3) \)

\( \implies k + 2 = 2k - 4 \)

\( \implies 2 + 4 = 2k - k = k \)

\( \implies k = 6 \)
So, both A and R are true but R is not the correct explanation for A.
Hence, option (b) is correct.