Question:
If m arithmetic means (A.Ms) and three geometric means (G.Ms) are inserted between 3 and 243 such that 4th A.M. is equal to 2nd G.M., then m is equal to ______ .
If m arithmetic means (A.Ms) and three geometric means (G.Ms) are inserted between 3 and 243 such that 4th A.M. is equal to 2nd G.M., then m is equal to ______ .
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Find the common ratio of the GP \(3, G_1, G_2, G_3, 243\) to get \(G_2\), then express the 4th A.M. using the common difference \(d = \frac{240}{m+1}\).
Updated On: Nov 5, 2026
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Correct Answer: 39
Solution and Explanation
GP: 3, G1, G2, G3, 243 has 5 terms, so \(3r^4 = 243\), \(r^4 = 81\), \(r = 3\).
\(G_2 = 3r^2 = 3 \times 9 = 27\).
AP: 3, A1, ..., Am, 243 has m + 2 terms, so the common difference is \(d = \frac{243-3}{m+1} = \frac{240}{m+1}\).
4th A.M. \(A_4 = 3 + 4d = 3 + \frac{960}{m+1}\).
Given \(A_4 = G_2\): \(3 + \frac{960}{m+1} = 27\), so \(\frac{960}{m+1} = 24\), \(m + 1 = 40\).
\(m = 39\).
Answer: 39
\(G_2 = 3r^2 = 3 \times 9 = 27\).
AP: 3, A1, ..., Am, 243 has m + 2 terms, so the common difference is \(d = \frac{243-3}{m+1} = \frac{240}{m+1}\).
4th A.M. \(A_4 = 3 + 4d = 3 + \frac{960}{m+1}\).
Given \(A_4 = G_2\): \(3 + \frac{960}{m+1} = 27\), so \(\frac{960}{m+1} = 24\), \(m + 1 = 40\).
\(m = 39\).
Answer: 39
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