JEE Mains · Maths · STD 11 - basic of algoritham
The value of \((0.16)^{\log _{2.5}\left(\frac{1}{3}+\frac{1}{3^{2}}+\frac{1}{3^{3}}+\ldots . to \infty\right)}\) is equal to
- A \(-4\)
- B \(2\)
- C \(-2\)
- D \(4\)
Answer & Solution
Correct Answer
(D) \(4\)
Step-by-step Solution
Detailed explanation
\((0.16)^{\log _{2.5}\left(\frac{1}{3}+\frac{1}{3^{2}}+\ldots \ldots \ldots .+0 \infty\right)}\) \(=\left(\frac{4}{25}\right)^{\log _{\left(\frac{5}{2}\right)}\left(\frac{1}{2}\right)}\)…
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