JEE Mains · Maths · STD 11 - 8. sequence and series
If \(m\) arithmetic means \(( A . Ms )\) and three geometric means \((G.Ms)\) are inserted between \(3\) and \(243\) such that \(4^{\text {th }}\) \(A.M.\) is equal to \(2^{\text {nd }}\) \(G.M.\), then \(m\) is equal to
- A \(39\)
- B \(40\)
- C \(38\)
- D \(41\)
Answer & Solution
Correct Answer
(A) \(39\)
Step-by-step Solution
Detailed explanation
\(3, A _{1}, A _{2} \ldots \ldots \ldots . A _{ m }, 243\) \(d =\frac{243-3}{ m +1}=\frac{240}{ m +1}\) Now \(3, G _{1}, G _{2}, G _{3}, 243\) \(r=\left(\frac{243}{3}\right)^{\frac{1}{3+1}}=3\) \(\therefore \quad A_{4}=G_{2}\) \(\Rightarrow \quad a +4 d = ar ^{2}\)…
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