If the sum of first $2 n$ 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

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Solution

The correct option is C 11
Sum to n terms of the an A.P is
$S_{n} = \frac{n}{2} (2 a + (n - 1) d)$
Now equating sum of 2n terms of first AP to sum of n terms of second a.p we get;
$\frac{2 n}{2} (2 a_{1} + (2 n - 1) d_{1}) = \frac{n}{2} (2 a_{2} + (n - 1) d_{2})$
$\frac{2 n}{2} (4 + (2 n - 1) 3) = \frac{n}{2} (114 + (n - 1) 2)$
On solving, we get $n = 11$

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