Question

Which term of the AP : 3, 15, 27, 39, ……… will be 132 more than its 54th term?

Answer 1

Given A.P. is $3, 15, 27, 39, ……..$
$a = 3$ and $d = a_2 − a_1 = 15 − 3 = 12$
$a_{54} = a + (54 − 1) d$
$a_{54} = 3 + (53) (12)$
$a_{54}= 3 + 636 = 639$
Now $a_{54} +132 = 639 + 132 = 771$
We have to find the term of this A.P. which is $771$.
Let nth term be $771$.
$a_n = a + (n − 1) d$
$771 = 3 + (n − 1) 12$
$768 = (n − 1) 12$
$n − 1 = \frac {768}{12}$
$n − 1 = 64$
$n = 65$

Therefore, 65th term was $132$ more than 54th term.

Alternatively,
Let nth term be $132$ more than 54th term.
$n = 54+\frac {132}{12}$
$n = 54+11$
$n = 65 ^{th}$term.