Question:
If the sum of the series $3+3 x+3 x^{2}+$ ____________ to $\infty$ is $\frac{45}{8}$, than $x=$ _______________
Solution:
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 $x=\frac{7}{15}$