Three numbers are in G.P. whose sum is 70. If the extremes be each multiplied by 4 and the means by 5, they will be in A.P. Find the numbers.

Explanation

Let the three numbers in GP be `(a)/(r ),a,ar`.
Given, `(a)/( r)+a+ar=70`
and `(4a)/(r ),5a,4ar` are in AP.
`:. 10a=(4a)/(r )+4ar` or `(10a)/(4)=(a)/(r )+ar`
or `(5a)/(2)=70-a " " [" from Eq. (i) "]`
or`5a=140-2a` or `7a=140`
`:. a=20`
From Eq.(i), we get
`(20)/(r )+20+20r=70`
or `(20)/(r )+20r=50`
or `2+2r^(2)=5r` or `2r^2)-5r+2=0`
or `(r-2)(2r-1)=0 :. r=2` or `(10)/(2)`
Hence, the three numbers are `10,20,40,20,10.`