A) A.P
B) G.P.
C) H.P.
D) none of these
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.You need to login to perform this action.
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