**Question:**

Find the $16^{\text {th }}$ term from the end of the AP $7,2,-3,-8,-13, \ldots,-113$

**Solution:**

To Find : 28th term from the end of the AP.

Given: The AP is $7,2,-3,-8,-13, \ldots,-113$

$a_{1}=7, a_{2}=2, d=2-7=-5$ and $I=-113$

Formula Used: nth term from the end $=1-(n-1) d$

(Where $/$ is last term and $d$ is common difference of given $A P$ )

By using $n$th term from the end $=1-(n-1) d$ formula

16th term from the end $=(-113)-15 d \rightarrow(-113)-15 \times(-5)=-38$

So $16^{\text {th }}$ term from the end is equal to $-38$.