The sum of three numbers in GP is 31 and sum of their square is 651 find numbers?

Question

Answer

Given:

The sum of three numbers in GP is 31

The sum of their square is 651

To find: The three numbers

Solution:
Let the three numbers be a/r, a, and ar where r is the common ratio.

Step 1: Find the sum of the three numbers

a/r + a + ar = 31

a(1/r + 1 + r) = 31

a = 31/(1/r + 1 + r)

Step 2: Find the sum of their square

(a/r)^2 + a^2 + (ar)^2 = 651

a^2/r^2 + a^2 + a^2r^2 = 651

a^2(1/r^2 + 1 + r^2) = 651

a^2 = 651/(1/r^2 + 1 + r^2)

Step 3: Substitute the value of a from step 1 into step 2

(31/(1/r + 1 + r))^2 = 651/(1/r^2 + 1 + r^2)

961/(1/r^2 + 1 + r^2)^2 = 651/(1/r^2 + 1 + r^2)

(1/r^2 + 1 + r^2) = 651/961

(1/r^2 + 1 + r^2) = 0.6774

r^4 + r^2 - 0.6774 = 0

solve for r using quadratic formula

r = 0.9, -0.9, 0.3i, -0.3i

Step 4: Discard the negative values of r as the common ratio cannot be negative

r = 0.9, 0.3i, -0.3i

Step 5: Substitute the value of r into step 1 to find a

r = 0.9: a = 8, a/r = 9, ar = 14

r = 0.3i: a = 12i, a/r = -4i, ar = 4i

r = -0.3i: a = -12i, a/r = 4i, ar = -4i

Answer: The three numbers in GP are 8, 9, and 14 or 12i, -4i, and 4i or -12i, 4i, and -4i.

Answer

The numbers are 1,5,25

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