In an AP,if S_{n} = 3n^{2} + 5n and a_{k} = 1 | vedclass

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(D) The $n^{th}$ term of an $AP$ is given by $a_{n} = S_{n} - S_{n-1}$.Given $S_{n} = 3n^{2} + 5n$.Then $S_{n-1} = 3(n-1)^{2} + 5(n-1) = 3(n^{2} - 2n

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$27$

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(D) The $n^{th}$ term of an $AP$ is given by $a_{n} = S_{n} - S_{n-1}$.Given $S_{n} = 3n^{2} + 5n$.Then $S_{n-1} = 3(n-1)^{2} + 5(n-1) = 3(n^{2} - 2n + 1) + 5n - 5 = 3n^{2} - 6n + 3 + 5n - 5 = 3n^{2} - n - 2$.Now,$a_{n} = (3n^{2} + 5n) - (3n^{2} - n - 2) = 3n^{2} + 5n - 3n^{2} + n + 2 = 6n + 2$.Since $a_{k} = 164$,we substitute $n = k$ in the expression for $a_{n}$:$6k + 2 = 164$.$6k = 164 - 2 = 162$.$k = 162 / 6 = 27$.

In an $AP$,if $S_{n} = 3n^{2} + 5n$ and $a_{k} = 164$,find the value of $k$.

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