The sum of three numbers in
g. p. is 21 and the sum of their squares is 189. find the numbers.

Let the three gp be a, ar and ar^2
a + ar + ar^2 = 21 => a(1 + r + r^2) = 21 . . . (1)
a^2 + a^2r^2 + a^2r^4 = 189 => a^2(1 + r^2 + r^4) = 189 . . . (2)
squaring (1) gives
a^2(1 + r + r^2)^2 = 441 . . . (3)
(3) ÷ (2) => (1 + r + r^2)^2 / (1 + r^2 + r^4) = 441/189 = 7/3
3(1 + r + r^2)^2 = 7(1 + r^2 + r^4)
3(r^4 + 2r^3 + 3r^2 + 2r + 1) = 7(1 + r^2 + r^4)
3r^4 + 6r^3 + 9r^2 + 6r + 3 = 7 + 7r^2 + 7r^4
4r^4 - 6r^3 - 2r^2 - 6r + 4 = 0
r = 1/2 or r = 2
From (1), a = 21/(1 + r + r^2)
When r = 2:
a = 21/(1 + 2 + 4) = 21/7 = 3
Therefore, the numbers are 3, 6 and 12.

Option a : week- amuont paid