JEE Mains · Maths · STD 11 - 8. sequence and series
Let \(G\) be the geometric mean of two positive numbers \(a\) and \(b,\) and \(M\) be the arithmetic mean of \(\frac {1}{a}\) and \(\frac {1}{b}\). If \(\frac {1}{M}\,:\,G\) is \(4:5,\) then \(a:b\) can be
- A \(1:4\)
- B \(1:2\)
- C \(2:3\)
- D \(3:4\)
Answer & Solution
Correct Answer
(A) \(1:4\)
Step-by-step Solution
Detailed explanation
\(G=\sqrt {ab} \) \(M = \frac{{\frac{1}{a} + \frac{1}{b}}}{2}\) \(M = \frac{{a + b}}{{2ab}}\) Given that \(\frac{1}{M}:G = 4:5\) \(\frac{{2ab}}{{\left( {a + b} \right)\sqrt {ab} }} = \frac{4}{5}\) \( \Rightarrow \frac{{a + b}}{{2\sqrt {ab} }} = \frac{5}{4}\)…
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