Question

If y = $(0.64{)}^{lo{g}_{0.25}(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}.....\infty )}$ then y

• A. 0.25
• B. 0.6
• C. 0.8
• D. 0.9

Answer 1

Answer: (C)

0.8

We have,

$y=(0.64{)}^{{\log }_{0.25}}(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+......\infty )$

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

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

Therefore,

$y=(0.64{)}^{{\log }_{0.25}}$$\left(\frac{\frac{1}{3}}{1-\frac{1}{3}}\right)$

$y=(0.64{)}^{{\log }_{0.25}}$$\frac{\frac{1}{3}}{\frac{2}{3}}$

$y=(0.64{)}^{{\log }_{0.25}}$$\frac{1}{2}$

$y=(0.8{)}^{2{\log }_{0.25}}$$\frac{1}{2}$

$y=(0.8{)}^{{\log }_{0.25}}$$\frac{1}{4}$

$y=(0.8{)}^{{\log }_{0.25}0.25}$

$y=0.8$

Hence, this is the answer.