Question

JEE Mains Maths question is in the image

Answer 1

we have to find the value of $0.16^{log_{2.5}\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+......+\infty\right)}$

solution : first find the value of 1/3 + 1/3² + 1/3³ + ... ∞

this is a geometric progression of infinite terms.

we know, S_∞ = a/(1 - r)

here first term, a = 1/3 and r = 1/3

so, S_∞ = (1/3)/(1 - 1/3) = (1/3)/(2/3) = 1/2

now $0.16^{log_{2.5}\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+......+\infty\right)}=0.16^{log_{2.5}\frac{1}{2}}$

= $\left(\frac{2}{5}\right)^{2log_{2.5}(0.5)}$

= $\left(\frac{2}{5}\right)^{log_{2.5}(0.5)^2}$

= $\left(\frac{2}{5}\right)^{log_{5/2}(0.25)}$

= $\left(\frac{5}{2}\right)^{-log_{5/2}(0.25)}$

= $\left(\frac{5}{2}\right)^{log_{5/2}(0.25)^{-1}}$

= (0.25)¯¹

= 4

Therefore the value of given algebraic expression is 4.