If the sum of the series 3 + 3x + 3x

Sum of series $3+3 x+3 x^{2}+\ldots \quad=\frac{45}{8} \quad$ (given)

Since a = 3

r = x

and sum of infinite g.p is $\frac{a}{1-r}$

$\therefore \frac{a}{1-r}=\frac{45}{8}$

$\Rightarrow \frac{3}{1-x}=\frac{45}{8}$

$8 \times 3=45(1-x)$

i.e $8=15(1-x)$

$8=15-15 x$

i. e $15 x=7$

i. e $x=\frac{7}{15}$