The sum of the first n terms of an A.P. is gi | vedclass

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(A) The $n^{th}$ term of an $A.P.$ can be calculated using the formula $T_n = S_n - S_{n-1}$,where $S_n$ is the sum of the first $n$ terms.Given $S_n

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$6n + 2$

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(A) The $n^{th}$ term of an $A.P.$ can be calculated using the formula $T_n = S_n - S_{n-1}$,where $S_n$ is the sum of the first $n$ terms.Given $S_n = 3n^2 + 5n$.Then,$S_{n-1} = 3(n-1)^2 + 5(n-1)$.$S_{n-1} = 3(n^2 - 2n + 1) + 5n - 5$.$S_{n-1} = 3n^2 - 6n + 3 + 5n - 5$.$S_{n-1} = 3n^2 - n - 2$.Now,$T_n = S_n - S_{n-1} = (3n^2 + 5n) - (3n^2 - n - 2)$.$T_n = 3n^2 + 5n - 3n^2 + n + 2$.$T_n = 6n + 2$.

The sum of the first $n$ terms of an $A.P.$ is given by $S_n = 3n^2 + 5n$. Then,the $n^{th}$ term of the $A.P.$ $T_n = \ldots$

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