If `a,b,c` are in `H.P`, `b,c,d` are in `G.P` and `c,d,e` are in `A.P.` , then the value of `e` is

Options

A. `(ab^(2))/((2a-b)^(2))`
B. `(a^(2)b)/((2a-b)^(2))`
C. `(a^(2)b^(2))/((2a-b)^(2))`
D. None of these

Answer

A

Explanation

`(a)` `a,b,c` are in `H.P.impliesb=(2ac)/(a+c)`….`(i)`
`b,c,d` are in `G.P.impliesc^(2)=bd`..`(ii)`
`c,d,e` are in `A.P.impliesd=(c+e)/(2)`…..`(iii)`
From `(i)`, `ab+bc=2ac`
`impliesc=(ab)/(2a-b)`….`(iv)`
From `(iii)` and `(iv)`
`d=(1)/(2)[(ab)/(2a-b)+e]` ......`(v)`
From `(ii)`, `(iv)` and `(v)`
`(a^(2)b^(2))/((2a-b)^(2))=(b)/(2)[(ab)/(2a-b)+e]`
`impliese=(ab^(2))/((2a-b)^(2))`