askiitiansexpert soumyajit_iitkanpur
Last Activity: 15 Years ago
Hi,
Let the first three terms of the GP be 4, 4r, 4r^2 .
According to the problem, 4 x 4r x 4r^2 = 8000
(4r)^3 - (20)^3 = 0
(4r - 20) (16r^2 + 80r + 400) = 0
either 4r - 20 = 0 => r = 5
or, 16r^2 + 80r + 400 = 0 => r^2 + 5r + 25 = 0
r = (-5 ± i 5√3)/2
If imaginary value of r is considered then second and third term of the GP will be imaginary also. Hence product of first three terms will also be imaginary, which is not true. Hence acceptable soln is r = 5.
Therefore three terms are 4, 20, 100 .