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in Geometric Progressions by (51.9k points)
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If a, b, c are in G.P., prove that 1/loga m , 1/logb m, 1/logc m are in A.P.

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Best answer

Given:

a, b and c are in GP

b2 = ac {property of geometric mean}

Apply log on both sides with base m

logm b2 = logm ac

logm b2 = logm a + logm c {using property of log}

2logm b = logm a + logm c

2/logb m = 1/loga m + 1/logc m

∴ 1/loga m , 1/logb m, 1/logc m are in A.P.

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