First slide

Geometric progression

Question

lf a, b, c are pth, qth and rth terms of a GP, then (q - r) log a + (r - p) log b + (p - q) log c is equal to

NA

A

p+q+r
B

1
C

-pqr
D

0

Solution

Let A and R be the first term and common ratio of the given GP. Then, a = AR^P-1
$\begin{array}{l} \Rightarrow \log a = \log A + (p - 1) \log R - - - - - (i) \\ similarly, \log b = \log A + (q - 1) \log R - - - - (ii) \\ and \log c = \log A + (r - 1) \log R - - - - - (iii) \end{array}$
now , (q - r) log a + (r - p) log b + (p - q) log c
$\begin{array}{l} = (q - r) {\log A + (p - 1) \log R} + (r - p) {\log A + (q - 1) \log R} + (p - q) {\log A + (r - 1) \log R} \\ = \log A [q - r + r - p + p - q] + \log R [p (q - r) + q (r - p) + r (p - q) - (q - r) - (r - p) - (p - q)] \\ = \log A \cdot 0 + \log R \cdot 0 = 0 \end{array}$

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