Question

Which term of an $A.P$ $3,15,27,39,....willbe132$ more than its 54th term?

• A. $1st$
• B. $63rd$
• C. $65th$
• D. None of these

Answer 1

Answer: (C)

$65th$

Here, $a=3,d=15–3=12.$

$54th$ term can be given as follows: $a54=a+53d=3+53x12=3+636=639$. So, the required term $=639+132=771$,

$771=a+(n–1)d,$

$771=3+(n-1)12,$

$(n – 1)12 = 771 – 3 = 768,$

$n–1=64$ Or, $n=65.$

Thus, the required term is $65th$ term.