Question:
Find the $28^{\text {th }}$ term from the end of the AP $6,9,12,15,18, \ldots ., 102 .$
Solution:
To Find $28^{\text {th }}$ term from the end of the AP.
Given: The AP is $6,9,12,15,18, \ldots ., 102$
$a_{1}=6, a_{2}=9, d=9-6=3$ and $I=102$
Formula Used: $n$th term from the end $=1-(n-1) d$
(Where lis last term and $d$ is common difference of given AP)
By using $n$th term from the end $=1-(n-1) d$ formula
28th term from the end $=102-27 \mathrm{~d} \rightarrow 102-27^{\times} 3=21$
So $28^{\text {th }}$ term from the end is equal to 21 .