A) A.P.
B) G.P.
C) H.P.
D) none of these
Correct Answer: C
Solution :
\[x,1,z\]are A.P \[\Rightarrow 2=x+z\]\[x,2,z\]are in G.P \[\Rightarrow 4=xz\] ?(i) Since (i) does not satisfy \[8=x+z\]and \[16=xz\]but it satisfies \[8=\frac{2xz}{z+x}.\] Therefore \[x,4,z\]are in H.PYou need to login to perform this action.
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