If a, b, c are in G.P., prove that $$\dfrac{1}{log_a} m , \dfrac{1}{log_b} m, \dfrac{1}{log_c} m$$ are in A.P.

Asked by Aaryan | 1 year ago |  38

#### 1 Answer

##### Solution :-

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

$$\dfrac{2}{log_b} m = \dfrac{1}{log_a} m + \dfrac{1}{log_c} m$$

Answered by Aaryan | 1 year ago

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