Question

Solve:

$(0\cdot 16{)}^{lo{g}_{0.16}2\cdot 5}(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}.....\infty )$=?

Answer 1

We have,

$(0.16{)}^{{\log }_{0.16}2.5}(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+......\infty )$

We know that the sum of infinite terms of G.P

$=\frac{a}{1-r}$

Therefore,

$=2.5\times \frac{\frac{1}{3}}{1-\frac{1}{3}}$ $(a^{{\log }_{a}b}=b)$

$=2.5\times \frac{\frac{1}{3}}{\frac{2}{3}}$

$=2.5\times \frac{1}{2}$

$=1.25$

Hence, this is the answer.