CBSE Class 10 Mathematics Arithmetic Progressions VBQs Set L

Question. What is the sum of first \( n \) odd natural numbers?
(a) \( n^2 – 1 \)
(b) \( n^2 \)
(c) \( n^2 – 2 \)
(d) None of the options
Answer: (b) \( n^2 \)
Explanation :
Solution: Series of odd natural numbers
\( = 1 + 3 + 5 + 7 + \dots + n \).
Here, \( a = 1, d = 2 \)
So, \( S = \frac{n}{2}\{2a+(n-1)d\} \)
\( = \frac{n}{2}\{2+(n-1)2\} \)
\( = \frac{n}{2}\{2+2n-2\} = n^2 \)

Question. If \( t_n = n(n + 3) \), then difference of its \( 5^{th} \) term and \( 2^{nd} \) term :
(a) 20
(b) 30
(c) – 30
(d) 10
Answer: (b) 30
Explanation :
Solution: \( t_n = n(n + 3) \)
\( \therefore t_5 = 5(5 + 3) = 40 \)
\( \therefore t_2 = 2(2 + 3) = 10 \)
\( \therefore t_5 – t_2 = 40 – 10 = 30 \)

Question. The \( n^{th} \) term of an A.P. \( \frac{1}{m}, \frac{m+1}{m}, \frac{2m+1}{m}, \dots \) is :
(a) \( \frac{m+1-mn}{m} \)
(b) \( \frac{mn-m+1}{m} \)
(c) \( \frac{mn-m-n}{m} \)
(d) \( \frac{mn+m-n}{m} \)
Answer: (b) \( \frac{mn-m+1}{m} \)
Explanation :
Solution: \( a = \frac{1}{m} \)
\( d = \frac{m+1}{m} - \frac{1}{m} = \frac{m+1-1}{m} = \frac{m}{m} = 1 \)
\( \therefore t_n = a + (n – 1)d \)
\( = \frac{1}{m} + (n-1) \times 1 = \frac{1+m(n-1)}{m} = \frac{1+mn-m}{m} = \frac{mn-m+1}{m} \)

Question. If the numbers \( 3k + 4, 7k + 1 \) and \( 12k – 5 \) are in A.P., then the value of \( k \) is :
(a) 2
(b) 3
(c) 4
(d) 5
Answer: (b) 3
Explanation :
Solution: \( 3k + 4, 7k + 1 \) and \( 12k – 5 \) are in A.P.
So,
\( (7k + 1) – (3k + 4) = (12k – 5) – (7k + 1) \)
\( 4k – 3 = 5k – 6 \)
\( k = 3 \)

Question. The \( 4^{th} \) term of an A.P. is equal to 3 times the first term and \( 7^{th} \) term excess which the \( 3^{rd} \) term by 1. Find its \( n^{th} \) term :
(a) \( n + 2 \)
(b) \( 3n + 1 \)
(c) \( 2n + 1 \)
(d) \( 3n + 2 \)
Answer: (c) \( 2n + 1 \)
Explanation :
Solution: \( t_4 = 3t_1 \)
\( \Rightarrow a + 3d = 3a \)
\( \text{or } d = \frac{2}{3}a \)
\( t_7 = 2t_3 + 1 \)
\( \Rightarrow a + 6d = 2(a + 2d) + 1 \)
\( \Rightarrow a + 2d - 1 = 0 \)
\( 2 \times \frac{2}{3}a - 1 = a \text{ or } a = 3 \)
\( d = \frac{2}{3} \times 3 = 2 \)
\( \therefore t_n = a + (n – 1)d \)
\( = 3 + (n – 1) \times 2 = (2n + 1) \)

Question. If \( t_n = 6n + 5 \), then \( t_{n + 1} = \)
(a) \( 6n – 1 \)
(b) \( 6n + 11 \)
(c) \( 6n + 6 \)
(d) \( 6n – 5 \)
Answer: (b) \( 6n + 11 \)
Explanation :
Solution: Here, \( t_n = 6n + 5 \)
\( t_{n + 1} = 6(n + 1) + 5 = 6n + 11 \)

Question. If the \( k^{th} \) term of the arithmetic progression 25, 50, 75, 100, .......... is 1000, then \( k \) is .......... .
(a) 20
(b) 30
(c) 40
(d) 50
Answer: (c) 40
Explanation :
Solution: Here, \( t_k = a + (k – 1)d \)
\( 1000 = 25 + (k – 1)25 \)
\( 1000 = 25 + 25k – 25 \)
\( 1000 = 25k \)
\( k = \frac{1000}{25} = 40 \)

Question. A man saves ₹ 320 during the first month, ₹ 360 in the second month, ₹ 400 in the third month. If he continues his savings in this sequence, in how many months will he save ₹ 20,000?
(a) 28
(b) 25
(c) 22
(d) 20
Answer: (b) 25
Explanation :
Solution: \( a = 320, d = 40 \)
\( S_n = 20,000 \)
It is required to find \( n \).
\( S_n = \frac{n}{2}[2 \times 320 + (n-1) \times 40] = 20,000 \)
\( \Rightarrow 500 = \frac{n}{2}[16 + (n-1)] \)
\( \Rightarrow 1000 = 16n + n^2 – n \)
\( \Rightarrow n^2 + 15n – 1000 = 0 \)
\( \Rightarrow n = 25 \)

Question. If \( t_n \) is the \( n^{th} \) term of an A.P., then \( t_{2n} – t_n \) is :
(a) \( (n + 1)d \)
(b) \( nd \)
(c) \( (n – 1)d \)
(d) None of the options
Answer: (b) \( nd \)
Explanation :
Solution:
\( t_{2n} – t_n = \{a + (2n – 1)d\} – \{a + (n – 1)d\} \)
\( t_{2n} – t_n = a + (2n – 1)d – a – (n – 1)d \)
\( t_{2n} – t_n = (2n – 1)d – (n – 1)d \)
\( t_{2n} – t_n = d\{(2n – 1) – (n – 1)\} \)
\( t_{2n} – t_n = d\{2n – 1 – n + 1\} \)
\( t_{2n} – t_n = nd \)

Question. The common difference of a constant A.P. is :
(a) 1
(b) 2
(c) 0
(d) None of the options
Answer: (c) 0
Explanation :
Solution: Constant AP means no increasing and no decreasing.
So common difference is 0.

Question. If \( a \) and \( l \), are first and last terms of an A.P., then number of terms :
(a) \( \left( \frac{l-a}{d} \right) + 1 \)
(b) \( \left( \frac{l-a}{d} \right) - 1 \)
(c) \( \left( \frac{l+a}{d} \right) + 1 \)
(d) None of the options
Answer: (a) \( \left( \frac{l-a}{d} \right) + 1 \)
Explanation :
We know,
\( l = a + (n – 1)d \)
\( l – a = (n – 1)d \)
\( n – 1 = \frac{l-a}{d} \)
\( n = \frac{l-a}{d} + 1 \)

Question. A mother divides ₹ 207 into three parts such that the amount are in A.P. and gives it to her three children. The product of the two least amounts that the children had ₹ 4623. Find the amount received by each child.
(a) ₹ 66, ₹ 68 and ₹ 70
(b) ₹ 67, ₹ 69 and ₹ 71
(c) ₹ 60, ₹ 64 and ₹ 68
(d) ₹ 57, ₹ 59 and ₹ 61
Answer: (b) ₹ 67, ₹ 69 and ₹ 71
Explanation :
Let the amount received by the three children be in the form of A.P.
is given by \( a – d, a, a + d \). Since, sum of the amount is ₹ 207, we have
\( (a – d) + a + (a + d) = 207 \)
\( 3a = 207 \)
\( \Rightarrow a = 69 \)
It is given that product of the two least amounts is 4623.
\( (a – d)a = 4623 \)
\( (69 – d)69 = 4623 \)
\( d = 2 \)
Therefore, amount given by the mother of her three children are ₹ (69 – 2), ₹ 69, ₹ (69 + 2). That is ₹ 67, ₹ 69 and ₹ 71.

Question. If every term of an A.P. is multiplied by 3, then the common difference of the new A.P., is :
(a) same
(b) increase by three
(c) 3 times of the previous A.P.
(d) None of the options
Answer: (c) 3 times of the previous A.P.

Question. If \( x, 10, y, 24, z \) are in A.P. then value of \( x, y \) and \( z \) are :
(a) 3, 17, 31,
(b) 31, 17, 3
(c) 3, 17, 30
(d) None of the options
Answer: (a) 3, 17, 31
Explanation :
\( 10 = x + d \dots (i) \)
\( 24 = x + 3d \dots (ii) \)
\( 2d = 14 \)
\( d = 7 \) and \( x = 3 \) so \( y = 17 \) and \( z = 31 \)

Question. The ratio of \( 6^{th} \) and \( 8^{th} \) term of an A.P. is 7 : 9. Find the ratio of \( 9^{th} \) term to \( 13^{th} \) term.
(a) 5 : 6
(b) 5 : 7
(c) 7 : 5
(d) None of the options
Answer: (b) 5 : 7
Explanation :
Solution: \( \frac{a+5d}{a+7d} = \frac{7}{9} \)
\( \Rightarrow 9a + 45d = 7a + 49d \)
\( \Rightarrow 2a = 4d \)
\( \Rightarrow a = 2d \)
So, ratio of \( 9^{th} \) term and \( 13^{th} \) term be
\( \frac{a+8d}{a+12d} = \frac{2d+8d}{2d+12d} = \frac{10d}{14d} = \frac{5}{7} \)

Question. The sum of the series \( 0.40 + 0.43 + 0.46 + \dots + 1 \) :
(a) 14.7
(b) 1.47
(c) 7.41
(d) None of the options
Answer: (a) 14.7
Explanation :
Solution: Here the value of \( n \) is not given.
But the last term is given. From this, we can find the value of \( n \).
Given, \( a = 0.40 \) and \( l = 1 \),
we find \( d = 0.43 – 0.40 = 0.03 \)
Therefore, \( n = \left( \frac{l-a}{d} \right) + 1 \)
\( = \left( \frac{1-0.40}{0.03} \right) + 1 = 21 \)
Sum of first \( n \) terms of an A.P.
\( S_n = \frac{n}{2}[a + l] \)
\( n = 21 \)
Therefore, \( S_n = \frac{21}{2}[0.40 + 1] = 14.7 \)
So, the sum of 21 terms of the given series is 14.7.

Very Short Answer Type Questions

Question. The first three terms of A.P. are \( (3y – 1), (3y + 5) \) and \( (5y + 1) \). Then find \( y \).
Answer: Sol. The terms of an A.P. are \( (3y – 1), (3y + 5) \) and \( (5y + 1) \)
Thus, \( d = (3y + 5) – (3y – 1) = (5y + 1) – (3y + 5) \)
\( \Rightarrow 6 = 2y – 4 \)
\( \Rightarrow 2y = 10 \)
\( \Rightarrow y = 5 \).

Question. If \( k, 2k – 1 \) and \( 2k + 1 \) are three consecutive terms of an A.P., then find the value of \( k \).
Answer: Sol. The terms of an A.P. are \( k, (2k – 1) \) and \( (2k + 1) \)
Thus, \( d = (2k – 1) – k = 2k + 1 – (2k – 1) \)
\( \Rightarrow 2k – 1 – k = 2k + 1 – 2k + 1 \)
\( \Rightarrow k – 1 = 2 \)
\( \Rightarrow k = 3 \).

Question. If the \( n^{th} \) term of an A.P. is \( (2n + 1) \), then find the sum of its first three terms.
Answer: Sol. Here, \( a_n = 2n + 1 \)
\( a_1 = 2(1) + 1 = 3 \)
\( a_2 = 2(2) + 1 = 5 \)
\( a_3 = 2(3) + 1 = 7 \)
\( \therefore a_1 + a_2 + a_3 = 3 + 5 + 7 = 15 \)

Question. In an A.P., if \( d = – 2, n = 5 \) and \( t_n = 0 \), then find the value of \( a \).
Answer: Sol. Given, \( d = – 2, n = 5 \) and \( t_n = 0 \)
We know that \( t_n = a + (n – 1) d \)
Thus \( t_5 = a + (5 – 1) (– 2) = 0 \)
\( \Rightarrow a + (4) (– 2) = 0 \)
\( \Rightarrow a – 8 = 0 \)
\( \Rightarrow a = 8 \).

Question. Find the \( 9^{th} \) term from the end (towards the first term) of the A.P. 5, 9, 13, \dots, 185.
Answer: Sol. Given, A.P. is 5, 9, 13, ...., 185
Here, \( l = 185 \) and \( d = 5 – 9 = 9 – 13 = – 4 \)
Then, \( a_9 = l + (n – 1)d \)
\( = 185 + (9 – 1) (– 4) \)
\( = 185 + 8 (– 4) \)
\( \therefore a_9 = 153 \)

Question. For what value of \( p \) are \( 2p + 1, 13, 5p – 3 \) are the three consecutive terms of an A.P. ?
Answer: Sol. For \( 2p + 1, 13, 5p – 3 \) to be consecutive, the common difference should be the same.
Thus \( 13 – (2p + 1) = 5p – 3 – 13 \)
\( \Rightarrow 12 – 2p = 5p – 16 \)
\( \Rightarrow 7p = 28 \)
\( \Rightarrow p = 4 \).

Question. If the sum of first \( p \) terms of A.P. is \( ap^2 + bp \), find its common difference.
Answer: Sol. Given, \( S_p = ap^2 + bp \)
\( \therefore S_1 = a(1)^2 + b(1) = a + b \)
and \( S_2 = a(2)^2 + b(2) = 4a + 2b \)
Now, we know that \( a_1 = S_1 = a + b \)
\( a_2 = S_2 – S_1 = 4a + 2b – a – b = 3a + b \)
Now, \( d = a_2 – a_1 = 3a + b – a – b = 2a \)
Hence, the common difference is 2a.

Question. Find the common difference of the Arithmetic Progression (A.P.)* \( \frac{1}{a}, \frac{3-a}{3a}, \frac{3-2a}{3a}, \dots \) (\( a \neq 0 \)).
Answer: Sol. Given, A.P. is \( \frac{1}{a}, \frac{3-a}{3a}, \frac{3-2a}{3a}, \dots \)
\( d = \frac{3-a}{3a} - \frac{1}{a} = \frac{3-a-3}{3a} = \frac{-a}{3a} = \frac{-1}{3} \).

Question. What is the common difference of an A.P. in which \( a_{21} – a_7 = 84 \) ?
Answer: Sol. Given, \( a_{21} – a_7 = 84 \)
\( \Rightarrow (a + 20d) – (a + 6d) = 84 \)
\( \Rightarrow a + 20d – a – 6d = 84 \)
\( \Rightarrow 20d – 6d = 84 \)
\( \Rightarrow 14d = 84 \)
\( \Rightarrow d = \frac{84}{14} = 6 \)
Hence, common difference = 6.

Question. The first and the last terms of an A.P. are 5 and 45 respectively. If the sum of all the terms is 400, find the common difference.
Answer: Sol. Given, \( S_n = 400, a = 5 \) and \( l = 45 \)
We know \( S_n = \frac{n}{2} [a + l] \)
\( \Rightarrow 400 = \frac{n}{2} [5 + 45] \)
\( \Rightarrow n = \frac{400 \times 2}{50} = 16 \)
Thus, applying \( t_n = a + (n – 1)d \), we have
\( 45 = 5 + (16 – 1)d \)
\( \Rightarrow 15d = 40 \)
\( \Rightarrow 3d = 8 \).