Question

In an $A.P.$, the sum of first $n$ terms is $\frac{3n^2}{2}+\frac{5n}{2}$. Find its ${25}^{th}$ term.

Answer 1

Let $S_n$ denote the sum of $n$ terms of an A.P. whose $n^{th}$ term is $a_n$.

We have,
$\Rightarrow S_n=\frac{3n^2}{2}+\frac{5n}{2}$
$⟹S_{n-1}=\frac{3}{2}(n-1{)}^2+\frac{5}{2}(n-1)$ [Replasing $n$ by $(n-1)$]
$\therefore S_n-S_{n-1}=\{\frac{3n^2}{2}+\frac{5n}{2}\}-\left\{\frac{3}{2}\right(n-1{)}^2+\frac{5}{2}(n-1)\}=3n+1$
$⟹a_n=3n+1$ $[\because a_n=S_n-S_{n-1}]$
$⟹a_{25}=3\times 25+1=76$ [Replacing $n$ by $25$]