## A, B, C are in GP and M,N,O are in AP, then (O-N)log A + (M-O)log B + (N-M)log C is equal to ?

## Options :

## A) (O-N)log ABC

## B) 0

## C) 1

## D) none

## Answer is : B

## Solution :

**Given :-**

1. A, B, C are in GP. So B

^{2}= A C

2. M,N,O are in AP. So

=> O-N = N-M = x (assume),

then => (O–M) = 2(O–N) = 2x.

**What To Find:-**

value of (O-N)log A + (M-O)log B + (N-M)log C

**Procedure :-**

Remind that (O-M) = 2x,so (M-O) = -2x

Now, find value

(O-N)log A + (M-O)log B + (N-M)log C

=> X
log A - 2X log B + X log
C

=> X
(log A - 2 log B + log C)

=> X
(log A - log B

^{2}+ log C)
=> X(log
AC – log B

^{2})
=> X(log
B

^{2}– log B^{2})
=>
X(0)

=>
0.

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