The sum of first 2n terms of A.P. 2, 5, 8, . . . is equal to the sum of first n terms of A.P. 57, 59, 61, . .. ., then n is equal to

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Solution

The correct option is B 11
$G i v e n \frac{2 n}{2} [4 + (2 n - 1) 3] = \frac{n}{2} [114 + (n - 1) 2] \Rightarrow 2 [6 n + 1] = 2 n + 112 \Rightarrow 10 n = 110 n = 11$

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