If the sum of first 2n terms of the A.P. 2, 5, 8, …, is equal to the sum of first n terms of the A.P. 57, 59, 61,…., then n equals
10
12
11
13
Sum to 2n terms of 2, 5, 8,,,… is
2n2[2(2)+(2n−1)(3)]=n(6n+1)
and sum to n terms of 57, 59, 61,…., is
n2[2(57)+(n−1)(2)]=n(56+n)
According to given condition
n(6n+1)=n(56+n) ⇒ 6n+1=56+n⇒ n=11