The value of (0.16^(log 2.5 (1/3 + 1/3^2 + 1/3^3 + ...... + ∞)))^(1/2) is:

Question

The value of \(\bigg[0.16^{\text{log}\,2.5\big(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+.........+∞\big)}\bigg]^{\frac{1}{2}}\) is :

(a) –1 

(b) 0 

(c) 1 

(d) None of these

Answer 1

(d) None of these

\(\bigg[(0.16)^{\text{log}\,2.5\big(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+.........+∞\big)}\bigg]^{\frac{1}{2}}\)

= \(\bigg[(0.4)^{2\,\text{log}\,2.5\bigg[\frac{\frac{1}{3}}{1-\frac{1}{3}}\bigg]}\bigg]^{\frac{1}{2}}\)

\(\bigg[\because \text{Sum of infinite GP} = \frac{\text{First term}}{\text{1-common ratio}}\bigg]\)