**Question:**

Find the $r^{\text {th }}$ term of the AP, the sum of whose first $n$ terms is $\left(3 n^{2}+2 n\right)$.

**Solution:**

Given: The sum of first n terms.

To Find: The $\mathrm{r}^{\text {th }}$ term.

Let the first term be a and common difference be d

Put n = 1 to get the first term

$a=S_{1}=3+2=5$

Put $n=2$ to get $a+(a+d) 2 a+d=12+4=1610+d=16 d=6 t_{r}=a+(r-1) d$

$t \gg r=5+(r-1) 6=5+6 r-6=6 r-1$

The $\mathrm{r}^{\text {th }}$ term is given by $6 \mathrm{r}-1$.