Question

Let $a_1,a_2,a_3,\dots$ be in A.P. and $g_1,g_2,g_3,\dots$ 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

Answer: 16

For A.P.,
$a_1=a=3$
$a_8=24$
$\Rightarrow a+7d=24...(i)$
Substituting $a=3$ in (i),
we get $d=3$
$a_4=a+3d=3+3(3)=12$
For G.P.,
$g_1=a=3$
$g_4=24$
$\Rightarrow a{r}^3=24$
$\Rightarrow 3r^3=24$
$\Rightarrow r^3=8$
$\Rightarrow r=2$
$g_7=a{r}^6$
$=3(2{)}^6$
$=192$
$\Rightarrow \frac{g_7}{a_4}=\frac{192}{12}=16$