JEE Mains · Maths · STD 11 - 8. sequence and series
Let \(a _1, \frac{ a _2}{2}, \frac{ a _3}{2^2}, \ldots ., \frac{ a _{10}}{2^9}\) be a G.P. of common ratio \(\frac{1}{\sqrt{2}}\). If \(a _1+ a _2+\ldots+ a _{10}=62\), then \(a _1\) is equal to \(:\)
- A \( 2(\sqrt{2}-1) \)
- B \( 2-\sqrt{2} \)
- C \( \sqrt{2}-1 \)
- D \( 2(2-\sqrt{2}) \)
Answer & Solution
Correct Answer
(A) \( 2(\sqrt{2}-1) \)
Step-by-step Solution
Detailed explanation
\( \frac{a_{2}}{2a_{1}}=\frac{a_{3}}{2a_{2}}=\frac{a_{4}}{2a_{3}}=...=\frac{a_{10}}{2a_{9}}=\frac{1}{\sqrt{2}} \) \(\therefore a_{1}, a_{2}, a_{3},.,a_{10} \) are in G.P. with common ratio \( \sqrt{2} \) \( \sum_{i=1}^{10}a_{i}=\frac{a_{i}((\sqrt{2})^{10}-1)}{\sqrt{2}-1}=62 \)…
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