If $k - 1$ , $k + 2$ and $3 k$ are in $G P$ , find the value of $k$ .

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Solution

$k - 1, k + 2, 3 k$ are in GP
therefore we have $\frac{k + 2}{k - 1} = \frac{3 k}{k + 2}$
$(k + 2)^{2} = 3 k (k - 1) k^{2} + 4 + 4 k = 3 k^{2} - 3 k 2 k^{2} - 8 k - 4 = 0 2 k (k - 4) + 1 (k - 4) = 0 k = 4$

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