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Find K Such that K + 9, K − 6 and 4 Form Three Consecutive Terms of a G.P.

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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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Chapter 20: Geometric Progression - Exercise 20.5 [Page 45]

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R.D. Sharma Mathematics [English] Class 11
Chapter 20 Geometric Progression
Exercise 20.5 | Q 3 | Page 45

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