In a GP of 6 terms, the first and last terms are `(x^(3))/(y^(2)) and (y^(3))/(x^(2))` respectively. Find the ratio of 3rd and 4th terms of that GP.

Options

A. `x^(2) : 1`
B. `y^(2) :x`
C. `y :x`
D. `x:y`

Answer

D

Explanation

`a = (x^(3))/(y^(2)) = t_1 , b= (y^(3))/(x^(2))= t_6`
There are 4 GM's between a and b.
`therefore r = ((b)/(a))^(1//n+1)`
`r=[(((y^(3))/(x^(2))))/(((x^(3))/(y^(2))))]^(1//4+1) rArr [((y)/(x))^(5)]^(1//5)`
`rArr r = (y)/(x)`
But the ratio of 3 rd and 4th term is `(1)/(r )`.
`rArr (1)/(r ) = (x)/(y)`.
Therefore, the required ratio is `x :y`.