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

Sum to 2n terms of 2, 5, 8,,,… is2n2[2(2)+(2n−1)(3)]=n(6n+1)and sum to n terms of 57, 59, 61,…., isn2[2(57)+(n−1)(2)]=n(56+n)According to given conditionn(6n+1)=n(56+n) ⇒ 6n+1=56+n⇒ n=11