Let the given numbers be a and b. Then,
`A=(a+b)/2 rArr a+b=2A` ...(i)
And, `a, G_(1), G_(2), b` are in GP.
`:. G_(1)/a=G_(2)/G_(1)=b/G_(2)`
`rArr a=G_(1)^(2)/G_(2)` and `b=G_(2)^(2)/G_(1)`
`rArr a+b=G_(1)^(2)/G_(2)+G_(2)^(2)/G_(1)`
`rArr 2A=G_(1)^(2)/G_(1)+G_(2)^(2)/G_(1)` [using (i)].
Hence, `2A=G_(1)^(2)/G_(2)+G_(2)^(2)/G_(1)`.