If $ \mathrm{m} $ arithmetic means (A.Ms) and three geometric means (G.Ms) are inserted between $ 3 $ and $ 243 $ such that $ 4^{t h} $ A.M. is equal to $ 2^{n d} $ G.M., then $ m $ is equal to:

Question

If $ \mathrm{m} $ arithmetic means (A.Ms) and three geometric means (G.Ms) are inserted between $ 3 $ and $ 243 $ such that $ 4^{t h} $ A.M. is equal to $ 2^{n d} $ G.M., then $ m $ is equal to:,

MARKS App · Accepted Answer

Let 3 , A 1 , A 2 , A 3 , . . . . . . A m , 243 are in arithmetic progression with m arithmetic means. Common difference d = 243 - 3 m + 1 = 240 m + 1 Let 3 , G 1 , G 2 , G 3 , 243 are in geometric progression with 3 geometric means. Common ratio r = 243 3 1 3 + 1 = 81 1 4 = 3 Given G 2 = A 4 ⇒ 3 3 2 = 3 + 4 240 m + 1 ⇒ 27 = 3 + 960 m + 1 ⇒ m + 1 = 40 ⇒ m = 39

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