Question

Find three numbers in GP whose product is 216 and sum of their product in pairs in 156.Plz do the sum urgent

Answer 1

Given that 3 three numbers are n GP.

Let the three numbers in GP be a/r, a, ar.   ------ (1)

Given that their product = 216.

a/r * a * ar = 216

a^3 = 216

a^3 = (6)^3

a = 6.

Given that the sum of their product in pairs= 156.

a/r * a + a * ar + ar * a/r = 156

a^2(1/r + r + 1)  = 156

a^2(1 + r^2 + r/r) = 156

From (1), the value of a = 6.

6^2(1 + r^2 + r/r) = 156

36(1 + r^2 + r/r) = 156

3(1 + r^2 + r/r) = 13

3r^2 + 3r + 3 = 13r

3r^2 - 10r + 3 = 0

(3r - 1)(r - 3) = 0

3r - 1 = 0         (or)   r - 3= 0

r = 1/3                      r = 3.


Substitute r = 1/3 in (1), we get

a/r = 6(3/1)               a = 6        ar = 6(1/3)

     = 18.                                        = 2.


Substitute r = 3 in (1), we get

a/r = 6/3 = 2

a = 6

ar = 6 * 3 = 18.


Their values are 18,6,2 and 2,6,18.

Hope this helps!