$\begin{array}{1 1}25 \\ 26 \\ 27 \\ 28 \end{array} $

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- $S_n=\large\frac{n}{2}$$[2a+(n-1)d]$
- In any series $S_n-S_{n-1}=t_n$

Given: sum of $n$ terms $S_n=3n^2+5n$

$\Rightarrow\:S_{n-1}=3(n-1)^2+5(n-1)$

We know that in any series $S_n-S_{n-1}=t_n$

$\Rightarrow\:t_n=\big[3n^2+5n\big]-\big[3(n-1)^2+5(n-1)\big]$

$\Rightarrow\:t_n=\big[3n^2+5n\big]-\big[3n^2-6n+3+5n-5\big]$

$\Rightarrow\:t_n=6n+2$

Also given that $m^{th}\:term=164$

$\Rightarrow\:t_m=164=6m+2$

$\Rightarrow\:m=27$

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