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Question
Find three numbers in G.P. whose sum is 38 and their product is 1728.
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Solution
Let the terms of the the given G.P. be
\[\frac{12}{r} + 12 + 12r = 38\]
\[ \Rightarrow 12 r^2 + 12r + 12 = 38r\]
\[ \Rightarrow 12 r^2 - 26r + 12 = 0\]
\[ \Rightarrow 2\left( 6 r^2 - 13r + 6 \right) = 0\]
\[ \Rightarrow 6 r^2 - 13r + 6 = 0\]
\[ \Rightarrow \left( 3r - 2 \right)\left( 2r - 3 \right) = 0\]
\[ \Rightarrow r = \frac{2}{3}, \frac{3}{2}\]
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.
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