Question

In a certain G.P. if $S_6=126$ and $S_3=14$, then find $a$ and $r$.

Answer 1

$S_6=126$ and $S_3=14$

We know that, for a G.P.,
$S_n=\frac{a(r^n-1)}{r-1}$
$\therefore S_6=\frac{a(r^6-1)}{r-1}$
$\therefore 126=\frac{a(r^6-1)}{r-1}$ ......$\left(1\right)$
and $S_3=\frac{a(r^3-1)}{r-1}$
$\therefore 14=\frac{a(r^3-1)}{r-1}$ .......$\left(2\right)$
Dividing $\left(1\right)$ by $\left(2\right)$, we get
$\frac{126}{14}=\frac{a(r^6-1)}{r-1}÷\frac{a(r^3-1)}{r-1}$
$\therefore \frac{126}{14}=\frac{a(r^6-1)}{r-1}\times \frac{r-1}{a(r^3-1)}$
$9=\frac{r^6-1}{r^3-1}$
$9=\frac{(r^3-1)(r^3+1)}{(r^3-1)}$
$\therefore 9=r^3+1$
$r^3+1=9$
$r^3=9-1$
$r^3=8$
$\therefore r=2$ ............(taking cube roots on both side)
$\therefore 14=\frac{a(2^3-1)}{2-1}$ ...........[Putting $r=2$]
$14=\frac{a(8-1)}{1}$
$7a=14$
$\therefore a=\frac{14}{7}$
$\therefore a=2$
$\therefore a=2,r=2$.