Question

If the sum of first \( n \) terms of an A.P. is \( 4n^2 + 2n \), then the common difference of A.P. is:

• A. \( 6 \)
• B. \( 14 \)
• C. \( 8 \)
• D. \( 4 \)

Hint:

To find the common difference from the sum formula: \[ a_n = S_n - S_{n-1} \] Then, compute \( d = a_2 - a_1 \).

Answer 1

The Correct Option is C

The first term of an arithmetic progression is obtained by finding \( S_1 \): \[ S_n = 4n^2 + 2n \] \[ S_1 = 4(1)^2 + 2(1) = 4 + 2 = 6 \] Thus, \( a = 6 \). The \( n \)th term of an A.P. is given by: \[ a_n = S_n - S_{n-1} \] Finding \( S_{n-1} \): \[ S_{n-1} = 4(n-1)^2 + 2(n-1) \] So, \[ a_n = [4n^2 + 2n] - [4(n-1)^2 + 2(n-1)] \] Expanding: \[ 4n^2 + 2n - (4(n^2 - 2n + 1) + 2n - 2) \] \[ = 4n^2 + 2n - (4n^2 - 8n + 4 + 2n - 2) \] \[ = 4n^2 + 2n - 4n^2 + 8n - 4 - 2n + 2 \] \[ = 8n - 2. \] Now, the common difference is: \[ d = a_2 - a_1. \] Finding \( a_2 \): \[ a_2 = 8(2) - 2 = 16 - 2 = \] Finding \( a_1 \): \[ a_1 = 8(1) - 2 = 8 - 2 = 6. \] \[ d = 14 - 6 = \]