Question

Let $a_{1}, a_{2}, a_{3}, \ldots$ be in A.P. and $g_{1}, g_{2}, g_{3}, \ldots$ be in G.P.
If $a_{1}=g_{1}=3$ and $a_{8}=g_{4}=24$, then find $\frac{g_{7}}{a_{4}}$.

Answer 1

For A.P.,
$a_{1}=a=3 $
$a_{8}=24$
$\Rightarrow a+7 d=24\,\,\, ...(i)$
Substituting $a=3$ in (i),
we get $d=3$
$a_{4}=a+3 d=3+3(3)=12$
For G.P.,
$g_{1}=a=3$
$g_{4}=24$
$\Rightarrow a r^{3}=24$
$\Rightarrow 3 r^{3}=24$
$\Rightarrow r^{3}=8$
$\Rightarrow r=2$
$g_{7}=a r^{6}$
$=3(2)^{6} $
$=192$
$\Rightarrow \frac{g_{7}}{a_{4}}=\frac{192}{12}=16$