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 $6 \mathrm{r}-1$.
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