Find k such that k + 9, k − 6 and 4

k, k + 9, k−6 are in G.P.

$\therefore(k-6)^{2}=4(k+9)$

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

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

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

$\Rightarrow k=0,16$

But, k = 0 is not possible.

∴ k = 16