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The sum of three numbers in a geometric progression is 13 and their product is -64.
Find the numbers.
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So, in standard designations, the terms are a, ar and ar^2.
From the problem, we have two equations
a + ar + ar^2 = 13, (1) (the sum)
a*(ar)*(ar^2) = -64. (2) (the product).
Equation (2) is the same as
(ar)^3 = -64,
which gives us
ar = -4. (3)
We substitute ar = -4 into equation (1). We get then
a - 4 + ar^2 = 13,
a + ar^2 = 13 + 4 = 17. (4)
We also represent ar^2 as (ar)*r = -4r.
Then equation (4) takes the form
a - 4r = 17. (5)
Now we have a system of two equations (3) and (5)
ar = -4, (3)
a - 4r = 17. (5)
From (3), express r = and substitute it into (5). You will get
= 17,
a^2 + 16 = 17a,
a^2 - 17a + 16 = 0,
(a-16)*(a-1) = 0.
It gives two possible solutions for 'a', 16 and 1.
If a = 16, then r = = = .
If a = 1, then r = = = -4.
Thus, two progressions are possible.
One progression is 16, -4, 1. (*)
The other progression is 1, -4, 16 (the same as (*), but reversed)
Both progressions satisfy the imposed conditions.
Solved.
Enjoy !