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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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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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