Given the sum of the first n terms of an A.P. | vedclass

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(B) For the first $A.P.$,$S_n = 3n^2 + 2n$.The first term $a = S_1 = 3(1)^2 + 2(1) = 5$.The sum of the first two terms $S_2 = 3(2)^2 + 2(2) = 12 + 4 =

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$6n^2 - n$

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(B) For the first $A.P.$,$S_n = 3n^2 + 2n$.The first term $a = S_1 = 3(1)^2 + 2(1) = 5$.The sum of the first two terms $S_2 = 3(2)^2 + 2(2) = 12 + 4 = 16$.The second term $a_2 = S_2 - S_1 = 16 - 5 = 11$.The common difference $d = a_2 - a = 11 - 5 = 6$.For the new $A.P.$,the first term $a' = a = 5$ and the common difference $d' = 2d = 2(6) = 12$.The sum of $n$ terms of the new $A.P.$ is $S_n' = \frac{n}{2} [2a' + (n - 1)d']$.$S_n' = \frac{n}{2} [2(5) + (n - 1)12] = \frac{n}{2} [10 + 12n - 12] = \frac{n}{2} [12n - 2] = 6n^2 - n$.

Given the sum of the first $n$ terms of an $A.P.$ is $S_n = 2n + 3n^2$. Another $A.P.$ is formed with the same first term and double the common difference. The sum of $n$ terms of the new $A.P.$ is:

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