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, S2n = Sumn
⇒ n(6n + 1) = n(n + 56)
⇒ 6n + 1 = n + 56
⇒ 5n = 55
⇒ n = 11.