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 $-38$.
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