Question
Find k such that $\text{k}+9,\text{k}-6$ and 4 from three consecutive terms of a G.P.
✓
Answer
$\text{k}+9,\text{k}-6,4 \text{ are in G.P.}$ $(\text{k}-6)^2=(\text{k}+9)4$ $\text{k}^2+36-12\text{k}=4\text{k}+36$ $\text{k}^2-16\text{k}=0$ $\text{k}(\text{k}-16)=0$ $\text{k}=0,\text{k}=16$
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