If the continued product of three numbes in g.p. is 216 and the sum of their products in pair is 156. find the numbers.

Answer 1

apparently three numbers are in a GP
then the 3 numbers are: a, ar, and ar^2

"the continued product of three numbes in g.p. is 216"
I am guessing you mean:
a(ar)(ar^2) = 216
a^3 r^3 = 216
ar = 6
a = 6/r

"the sum of their products in pair is 156"
a(ar) + a(ar^2) + ar(ar^2) = 156 ???
a^2 r + a^2 r^2 + a^2 r^3) = 156
(ar)(a + ar + ar^2) = 156
6a(1 + r + r^2) = 156
a(1+r+r^2) = 26
(6/r)(1 + r + r^2) = 26
6/r + 6 + 6r = 26
times r
6 + 6r + 6r^2 = 26r
6r^2 - 20r + 6 = 0
(r - 3)(6r - 2) = 0
r = 3 or r = 1/3

carry on