Find the 28th term from the end of the AP

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 .

 

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