Question

Find 3 numbers in G.P whose sum is 28 and whose product is 512

Answer 1

Answer:

4, 8, 16.

Step-by-step explanation:

Let the three numbers in GP be $r, ra, ra^2$, where $a \ne 0$. We are given that $r + ra + ra^2 = 28$ and $r \cdot ra \cdot ra^2 = r^3a^3 = 512$, the latter of which gives $ra = \sqrt[3]{512} = 8$. So substituting this into the former equation gives $\frac{8}{a} + 8 + 8a = 28$ or

$8 + 8a + 8a^2 = 28a\\2a^2 + 2a + 2 = 7a\\2a^2 - 5a + 2 = 0\\(2a-1)(a-2) = 0\\a = \frac{1}{2} \vee a = 2$

And this gives our answer directly.