If a is the A.M. of b and c and the two geometric means are `G_(1)` and `G_(2)`, then prove that `G_(1)^(3)+G_(2)^(3)`

Options

A. 1
B. 2
C. `(1)/(2)`
D. 3

Answer

B

Explanation

It is given that a is the A.M. of and c.
`:." "a=(b+c)/(2)rArrb+c=2a` . . .(i)
Since `G_(1)andG_(2)` are two geometric means between b and c.
Therefore, b, `G_(1),G_(2),` c is a G.P. with common ratio r given by
`r=((c)/(b))^(1//3)`
`:." "G_(1)=br=b((c)/(b))^(1//3)=c^(1//3)b^(2//3)`
`G_(2)=br^(2)=b((c)/(b))^(2//3)=b^(1//3)c^(2//3)`
`rArr" "G_(1)^(3)=b^(2)candG_(2)^(3)=bc^(2)`
`rArr" "G_(1)^(3)+G_(2)^(3)=b^(2)c+bc^(2)=bc(b+c)=2abc" [Using (i)]"`
`rArr" "G_(1)^(3)+G_(2)^(3)=lamdaabcrArrlamda-2`