###
**The sum of n terms of an A.P. is 3$\mathbf{n}^{\mathbf{2}}$ + 5n, then 164 is its.**

A. 24th term
B. 27th term
C. 26th term
D. 25th term
**Answer: Option B**

## Show Answer

Solution(By Apex Team)

$\begin{array}{l}
\text { Sum of } n \text { terms }\left(S_{n}\right)=3 n^{2}+5 n\\
\therefore \text { Sum of }(n-1) \text { terms }\left(S_{n-1}\right)\\
=3(n-1)^{2}+5(n-1)\\
=3\left(n^{2}-2 n+1\right)+5 n-5\\
=3 n^{2}-6 n+3+5 n-5\\
=3 n^{2}-n-2\\
\therefore n^{t h} \text { term }=S_{n}-S_{n-1}\\
\Rightarrow a_{n}=3 n^{2}+5 n-3 n^{2}+n+2\\
a_{n}=6 n+2, \text { But } a_{n}=164\\
\Rightarrow 6 n+2=164\\
\Rightarrow 6 n=164-2\\
\Rightarrow 6 n=162
\end{array}$
So, n = $\Large\frac{162}{6}$
= 27th term

## Related Questions On Progressions

### How many terms are there in 20, 25, 30 . . . . . . 140?

A. 22B. 25

C. 23

D. 24

### Find the first term of an AP whose 8th and 12th terms are respectively 39 and 59.

A. 5B. 6

C. 4

D. 3

### Find the 15th term of the sequence 20, 15, 10 . . .

A. -45B. -55

C. -50

D. 0

### The sum of the first 16 terms of an AP whose first term and third term are 5 and 15 respectively is

A. 600B. 765

C. 640

D. 680