If `a,b,c` are in HP,b,c,d are in GP and `c,d,e` are in AP, than show that `e=(ab^(2))/(2a-b)^(2)`.

Explanation

Given, `a,b,c` are in HP.
`therefore b=(2ac)/(a+c)" or "c=(ab)/(2a-b) "……..(i)"`
Given,` b,c,d` are in GP.
`therefore " " c^(2)=bd " " "……..(ii)"`
and given, `c,d,e` are in AP.
`therefore " " d=(c+e)/(2)`
`implies e=2d-c`
`e=((2c^(2))/(b)-c) " " [" from Eq. (ii) "] "........(iii)"`
From Eqs. (i) and (iii), `e=(2)/(b)((ab)/(2a-b))^(2)-((ab)/(2a-b))`
`=(ab)/(2a-b)^(2){(2a-(2a-b)}`
`=(ab^(2))/(2a-b)^(2)`