JEE Mains · Maths · STD 11 - 8. sequence and series
Let \(\alpha=\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\ldots\infty\) and \(\beta=\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\ldots\infty\). Then the value of \((0.2)^{\log_{\sqrt{5}}(\alpha)}+(0.04)^{\log_5(\beta)}\) is equal to:
- A \(4\)
- B \(5\)
- C \(8\)
- D \(25\)
Answer & Solution
Correct Answer
(C) \(8\)
Step-by-step Solution
Detailed explanation
The sum of an infinite geometric progression is given by \(S_{\infty} = \dfrac{a}{1-r}\). For \(\alpha\), \(a = \dfrac{1}{4}\) and \(r = \dfrac{1}{2}\): \(\alpha = \dfrac{\dfrac{1}{4}}{1 - \dfrac{1}{2}} = \dfrac{1}{2}\) For \(\beta\), \(a = \dfrac{1}{3}\) and…
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