Question

Find three numbers in G.P. whose sum is 38 and their product is 1728.

Answer 1

Let the terms of the the given G.P. be

\[\frac{a}{r}, \text { a and ar } .\]

Then, product of the G.P. = 1728

\[\Rightarrow\] a³ = 1728

\[\Rightarrow\] a = 12

Similarly, sum of the G.P. = 38

\[\Rightarrow \frac{a}{r} + a + ar = 38\]

Substituting the value of a

\[\frac{12}{r} + 12 + 12r = 38\]

\[ \Rightarrow 12 r^2 + 12r + 12 = 38r\]

\[ \Rightarrow 12 r^2 - 26r + 12 = 0\]

\[ \Rightarrow 2\left( 6 r^2 - 13r + 6 \right) = 0\]

\[ \Rightarrow 6 r^2 - 13r + 6 = 0\]

\[ \Rightarrow \left( 3r - 2 \right)\left( 2r - 3 \right) = 0\]

\[ \Rightarrow r = \frac{2}{3}, \frac{3}{2}\]

Hence, the G.P. for a = 12 and r =  \[\frac{2}{3}\] is 18, 12 and 8.

And, the G.P. for  a = 12 and r = \[\frac{3}{2}\] is 8, 12 and 18.

Hence, the three numbers are 8, 12 and 18.