The 5th and 13th terms of an AP are 5 and –3 respectively

To Find: AP and its $30^{\text {th }}$ term (i.e. $a_{30}=?$ )

Given: $a_{5}=5$ and $a_{13}=-3$

Formula Used: $a_{n}=a+(n-1) d$

(Where $a=a_{1}$ is first term, $a_{2}$ is second term, $a_{n}$ is nth term and $d$ is common difference of given $\mathrm{AP}$ )

By using the above formula, we have

$a_{5}=5=a+(5-1) d$, and $a_{13}=-3=a+(13-1) d$

$a+4 d=5$ and $a+12 d=-3$

On solving above 2 equation, we and $a+12 d=-3$ get

$a=9$ and $d=(-1)$

So $a_{30}=9+29(-1)=-20$

AP is $(9,8,7,6,5,4 \ldots \ldots)$ and $30^{\text {th }}$ term $=-20$