Mathematics · 2020
JEE Main · 3 September 2020, Shift 1 · Q73
The value of (0.16)^log_2.5(1/3+1/3^2+1/3^3+… to ∞) is equal to \ \ \ \.
The value of $\displaystyle (0.16)^{\log _{2.5}\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\ldots \text { to } \infty\right)}$ is equal to
\_\_\_\_.
Official answer
4
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JEE Main 2020 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.