There are two sets A and B each of which consists of three numbers in GP whose product is 64 and R and r are the common ratios sich that `R=r+2`. If `(p)/(q)=(3)/(2)`, where p and q are sum of numbers taken two at a time respectively in the two sets.
The value of `r^(R)+R^(r)` is

Options

A. 66
B. 72
C. 78
D. 84

Answer

D

Explanation

Let `A={(A)/(R),A,AR}`
`B={(a)/(r ),a,ar}`
According to the question, `(A)/(R )*A*AR=64`
`implies A^(3)=64 " "implies A=4" " "……..(i)"`
`(a)/(r )*a*ar=64" " implies a^(3)=64 " " implies a=4 " " "……(ii)"`
and `R=r+2" " "……..(iii)"`
`p=(A)/(R)*A*AR+AR*(A)/(R)`
`=(A^(2))/(R)+A^(2)R+A^(2)=(16)/(R )+16R+16`
`q=(a)/(r )*a+a*ar+ar*(a)/( r)`
`=(a^(2))/(r )+a^(2)r+a(^2)=(16)/(r )=(16)/(r )+16r+16`
Give that, `(p)/(q)=(3)/(2)`
So, `((16+16R^(2)+16R)r)/((16+16r^(2)+16r)R)=(3)/(2)`
`((1+R^(2)+R)r)/((1+r^(2)+r)R)=(3)/(2)`
From Eq. (iii), `R=r+2`
`implies ((1+r^(2)+4+4r+r+2)r)/((1+r+r^(2))(r+2))=(3)/(2)`
`implies (r^(3)+5r^(2)+7r)/(r^(3)+3r^(2)+3r+2)=(3)/(2)`
`implies r^(3)-r^(2)-5r+6=0`
`implies (r-2)(r^(2)+r-3)=0`
`impliesr=2" or " r=(-1pmsqrt(13))/(2)`
So, `R=4`
`p=16((1)/(R)+R+1)=16((1)/(4)+4+1)=(16)/(4)(21)=84`