If the sum of 2n terms of the A.P. 2, 5, 8, 11,... is equal to the sum

1.	If the sum of 2n terms of the A.P. 2, 5, 8, 11,... is equal to the sum of n terms of A.P. 57, 59, 61, 63, ..., then n is equal to(a) 10 (b) 11 (c) 12 (d) 13
Answer» Answer : (b) 11 For the 1st A.P., 2, 5, 8, 11, ...... , First term (a₁) = 2, common difference (d₁) = 3 ∴ The sum of this A.P. to 2n terms = \( \frac{2n}{2}[2a_1+(2n-1)d_1]\) = n[4+(2n - 1)3] = 4n + 6n² – 3n = 6n² + n = n (6n + 1) For the second A.P., 57, 59, 61, 63, ....... , First term (b₁) = 57, common difference (d₂) = 2 ∴ The sum of this A.P. to n term is \( Sum_n = \frac{n}{2}(2b_1+(n-1)d_2]\) = \(\frac{n}{2}\)[114 + (n - 1) 2] = 57n + n² – n = n² + 56n = n (n + 56) Given, S_2n = Sum_n ⇒ n(6n + 1) = n(n + 56) ⇒ 6n + 1 = n + 56 ⇒ 5n = 55 ⇒ n = 11.

If the sum of 2n terms of the A.P. 2, 5, 8, 11,... is equal to the sum of n terms of A.P. 57, 59, 61, 63, ..., then n is equal to(a) 10 (b) 11 (c) 12 (d) 13

Answer»

Answer : (b) 11

For the 1st A.P., 2, 5, 8, 11, ...... ,

First term (a₁) = 2, common difference (d₁) = 3

∴ The sum of this A.P. to 2n terms

= \( \frac{2n}{2}[2a_1+(2n-1)d_1]\)

= n[4+(2n - 1)3]

= 4n + 6n² – 3n = 6n² + n = n (6n + 1)

For the second A.P., 57, 59, 61, 63, ....... ,

First term (b₁) = 57, common difference (d₂) = 2

∴ The sum of this A.P. to n term is

\( Sum_n = \frac{n}{2}(2b_1+(n-1)d_2]\)

= \(\frac{n}{2}\)[114 + (n - 1) 2]

= 57n + n² – n = n² + 56n = n (n + 56)

Given, S_2n = Sum_n

⇒ n(6n + 1) = n(n + 56)

⇒ 6n + 1 = n + 56

⇒ 5n = 55

⇒ n = 11.