Arithmetic Progression

Let the first term of the arithmetic progression be $a$ and its common difference be $d$ . Then the sum of the first $n$ terms is given by $S_n = \dfrac {n(2a+(n-1)d)}2$ . Therefore,

$\begin{aligned} \frac {n(2a+(n-1)d)}2 & = 3n^2 + 5n & \small \color{#3D99F6} \text{For }n \ne 0 \\ \frac {2a+(n-1)d}2 & = 3n + 5 \\ a+{\color{#3D99F6}\frac d2} n - \frac d2 & = {\color{#3D99F6}3}n + 5 & \small \color{#3D99F6} \text{Equating the coefficient of }n \\ \implies d & = 6 & \small \color{#3D99F6} \text{Equating the constant terms} \\ \implies a & = 8 \end{aligned}$

Now, the $m$ th term is given by:

$\begin{aligned} a_m & = a + (m-1)d \\ 8 + 6(m-1) & = 164 \\ \implies m & = \frac {164-8}6 + 1 = \boxed{27} \end{aligned}$