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Which term of the AP: 3, 15, 27, 39, … will be 132 more than its 54th term?
Given:
Given A.P. is $3, 15, 27, 39, ….$
To do:
We have to find which term of the given A.P. will be 132 more than its 54th term.
Solution:
Here,
$a_1=3, a_2=15, a_3=27$
Common difference $d=a_2-a_1=15-3=12$
We know that,
nth term $a_n=a+(n-1)d$
Therefore,
$a_{54}=3+(54-1)(12)$
$=3+53(12)$
$=3+636$
$=639$
132 more than the 54th term $=132+639=771$
This implies,
$a_{n}=3+(n-1)12$
$771=3+12n-12$
$12n=771+9$
$n=\frac{780}{12}$
$n=65$
Hence, 65th term is 132 more than the 54th term. 
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