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Question
Find k such that k + 9, k − 6 and 4 form three consecutive terms of a G.P.
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Solution
k, k + 9, k−6 are in G.P.
\[\therefore \left( k - 6 \right)^2 = 4\left( k + 9 \right)\]
\[ \Rightarrow k^2 + 36 - 12k = 4k + 36\]
\[ \Rightarrow k^2 - 16k = 0\]
\[ \Rightarrow k \left( k - 16 \right) = 0\]
\[ \Rightarrow k = 0, 16\]
But, k = 0 is not possible.
∴ k = 16
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