JEE MATHEMATICS MCQ: SEQUENCE OF SERIES

1 Easy

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Application +4 | -1

Value of

y=(0.36)^{\log _{0.25}\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\ldots \text { upto } \infty\right)}

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Correct option: C

0.6

Option C (0.6) is correct because after simplifying the geometric series and applying logarithmic properties, the value of y is found to be 0.6.

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