Sequence And Series Question 83
Question: If x, y, z are real and $ 4x^{2}+9y^{2}+16z^{2}-6xy-12yz-8zx=0 $ , then x, y, z are in
Options:
A) A.P
B) G.P.
C) H.P.
D) none of these
Show Answer
Answer:
Correct Answer: C
Solution:
[c] Multiplying the given expression by 2 and rewriting it, we have $ {{(2x-3y)}^{2}}+{{(3y-4z)}^{2}}+{{(4z-2x)}^{2}}=0 $
$ \Rightarrow 2x=3y=4z $
$ \Rightarrow \frac{1}{x},\frac{1}{y},\frac{1}{z} $ are in A.P
$ \Rightarrow x,y,z $ are in H.P.
BETA
