**Question:**

The sum of an infinite GP is 57, and the sum of their cubes is 9747. Find the GP.

**Solution:**

Let the first term Of G.P. be a, and common ratio be r.

$\frac{\mathrm{a}}{\therefore 1-\mathrm{r}}=57 \ldots$ (1)

On cubing each term will become,

$a^{3,} a^{3} r^{3}, \ldots$

$\therefore$ This sum $=\frac{\mathrm{a}^{3}}{1-\mathrm{r}^{3}}=9747 \ldots$ (2)

a=57(1-r) put this in equation 2 we get

$\frac{(57 \times(1-r))^{3}}{1-r^{3}}=9747$

$\Rightarrow \frac{57^{3} \times(1-r)^{3}}{1-r^{3}}=9747$

$\Rightarrow \frac{(1-r) \times(1-r)^{2}}{(1-r)\left(1+r+r^{2}\right)}=\frac{9747}{57 \times 57 \times 57}=\frac{1}{19}$

$\Rightarrow 19\left(1-2 r+r^{2}\right)=1+r+r^{2}$

$\Rightarrow 19 r^{2}-r^{2}-38 r-r+19-1=0$

$\Rightarrow 18 r^{2}-39 r+18=0$

$\Rightarrow 6 r^{2}-13 r+6=0$

$\Rightarrow(2 r-3)(3 r-2)=0$

$\Rightarrow r=2 / 3,3 / 2$

But $-1

⇒ r=2/3

Substitute this value of r in equation 1 we get

$a=57 \times\left(1-\frac{2}{3}\right)=19$

Thus the first term of G.P. is 19, and the common ratio is 2/3

$\therefore G . P=19, \frac{38}{3}, \frac{76}{9}, \ldots$

$19, \frac{38}{3}, \frac{76}{9}, \ldots$