Question

If `x,y,z` be three numbers in `G.P.` such that `4` is the `A.M.` between `x` and `y` and `9` is the `H.M.` between `y` and `z` , then `y` is
A. `4`
B. `6`
C. `8`
D. `12`

Answer 1

Correct Answer - B
`(b)` Let `x=(k)/(r )`, `y=k`, `z=kr`
`implies((k)/(r )+k)/(2)=4` ……..`(i)`
and `(2k*kr)/(k+kr)=9` ………`(ii)`
Solving we get `k=6`