Question

Determine an $AP$ whose third term is $9$ and when fifth term is subtracted from $8th$ term, we get $6$.

Answer 1

Given: $t_3=a+2d=9$

$t_8-t_5=a+7d-a-4d=6$

$\Rightarrow 3d=6$

$\Rightarrow d=\frac{6}{3}=2$

$\therefore d=2$

Put $d=2$ in $t_3=a+2d=9$

$=a+2\times 2=9$

$\Rightarrow a=9-4=5$

$\therefore a=5,d=2$

The terms are $5,5+2,5+2\times 2,5+3\times 2,...$

or $5,7,9,11,...$