**Question:**

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

(a) 24th term

(b) 27th term

(c) 26th term

(d) 25th term

**Solution:**

Here, the sum of first *n* terms is given by the expression,

We need to find which term of the A.P. is 164.

Let us take 164 as the *n**th* term.

So we know that the *n**th*term of an A.P. is given by,

So,

$164=S_{n}-S_{n-1}$

$164=3 n^{2}+5 n-\left[3(n-1)^{2}+5(n-1)\right]$

Using the property,

$(a-b)^{2}=a^{2}+b^{2}-2 a b$

We get,

$164=3 n^{2}+5 n-\left[3\left(n^{2}+1-2 n\right)+5(n-1)\right]$

$164=3 n^{2}+5 n-\left[3 n^{2}+3-6 n+5 n-5\right]$

$164=3 n^{2}+5 n-\left(3 n^{2}-n-2\right)$

$164=3 n^{2}+5 n-3 n^{2}+n+2$

$164=6 n+2$

Further solving for *n*, we get

$6 n=164-2$

$n=\frac{162}{6}$

$n=27$

Therefore, 164 is the $27^{\text {th }}$ term of the given A.P.

Hence the correct option is (b).

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