Find k such that k + 9, k − 6 and 4


Find k such that k + 9, k − 6 and 4 form three consecutive terms of a G.P.


k, k + 9k−6 are in G.P.


$\Rightarrow k^{2}+36-12 k=4 k+36$

$\Rightarrow k^{2}-16 k=0$

$\Rightarrow k(k-16)=0$

$\Rightarrow k=0,16$

But, k = 0 is not possible.

∴ k = 16



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