Question

The sum of three numbers in GP is 38 and their product is 1728. Find the GP.

Answer 1

Let the numbers be \(\frac{a}{r}\), a, ar, the given;

\(\frac{a}{r} \times a \times ar = 1728\) 

\(\Rightarrow\) a³ = 1728 

\(\Rightarrow\)a = 12

\(\frac{12}{r}+12+12r\) = 38 

\(\Rightarrow\) \(12(\frac{1}{r} + 1+r) =38\)

⇒ 6(1 + r + r²) = 19r 

⇒ 6 + 6r + 6r² = 19r

⇒ 6r² – 13r + 6 = 0

⇒ 6r² – 9r – 4r + 6 = 0

⇒ 3r(2r – 3) – 2(2r – 3) = 0

⇒ (3r – 2)(2r – 3) = 0

⇒ r = \(\frac{2}{3}, \frac{3}{2}\)

Therefore GP is 8, 12, 18 or 18, 12, 8.