Question

The sum of first $q$ terms of an A.P is $162$. The ratio of its $6$th term to is $13$th term is $1:2$. Find the $15$th term of the A.P.

Answer 1

Given,

$S_9=\frac{9}{2}[2a+(9-1\left)d\right]=162$

$162=\frac{9}{2}[2a+8d]$........(1)

$a_1+a_9=36$

$a_n=a+(n-1)d$

$a_6=a+(6-1)d=a+5d$

$a_{13}=a+(13-1)d=a+12d$

From given, we have,

$\frac{a_6}{a_{13}}=\frac{a+5d}{a+12d}$

$\frac{1}{2}=\frac{a+5d}{a+12d}$

$a+12d=2a+10d$

solving the above equation, we get,

$a=2d$

using this value in (1), we get,

$162=\frac{9}{2}\left[2\right(2d)+8d]$

solving the above equation, we get,

$d=3$

$a=2\left(3\right)=6$

$a_{15}=6+(15-1)3=48$