Question

The eighth terms of an AP is half of its second term and 11th term exceeds one third of fourth term by 1. Find the 15th term.

Answer 1

$a_8=\frac{1}{2}{a}_2$

$a_{11}+1=\frac{1}{4}{a}_4$

$a+7d=\frac{1}{2}(a+d)$

$2a+14d=a+d$

$a+13d=0$

$a+10d+1=\frac{1}{3}(a+3d)$

$30a+30d+3=a+3$

$2a+27d+3=0$

$2a+26d+d+3=0$

$2(a+13d)+d+3=0$

$0+d+3=0$

$d=-3$

$a=39$

$a_{15}=a+14d=39+14(-3)=-3$