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Given sum of the first n terms of an A.P is $2 n + 3 n^{2}$ Another A.P is formed with the same first term and double of the common difference , the sum of n terms of the new A.P is

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$6 n^{2} - n^{}$

$n + 4 n^{2}$

$3 n^{} + 2 {n^{2}}^{}$

$n^{2} + 4 n^{}$

answer is B.

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Detailed Solution

Given ; $S_{n} = 2 n + 3 n^{2}$
Now ,first term $S_{1} = 2 + 3 = 5; S_{2} = 2 (2) + 3 (4) = 16$
$a_{2} = 16 - 5 = 11; d = 11 - 5 = 6$
New A.p
$a_{2} = 5 d_{1} = 12 \Rightarrow s_{n} = \frac{n}{2} [10 + (n - 1) 12] = 6 n^{2} - n$

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