Question

The sum of an infinite number of terms in G.P. is $15$, and the sum of their squares is $45$, find the series.

Answer 1

GIven:

$\frac{a}{1-r}=15$ ...... (1)

$\frac{a^2}{1-r^2}=45$ ...... (2)

From equation (1),

$a=15(1-r)$ ...... (3)

Substitute this value in the equation (2).

$\frac{\left[15\right(1-r){]}^2}{1-r^2}=45$

$\left[15\right(1-r){]}^2=45(1-r^2)$

$5(1-2r+r^2)=1-r^2$

$6r^2-10r+4=0$

$r=1or\frac{2}{3}$

Here, $r$ cannot be $1$. Therefore,

$r=\frac{2}{3}$

From equation (3),

$a=15(1-\frac{2}{3})$

$a=5$

Therefore, the required series is,

$\Rightarrow 5,\frac{10}{3},\frac{20}{9},\frac{40}{27},\frac{80}{81}........$