Find $k$ given the consecutive arithmetic terms:
$k, k^{2}, k^{2} + 6$

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Solution

Since, $k, k^{2}, k^{2} + 6$ are in A.P

$\Rightarrow k^{2} - k = k^{2} + 6 - k^{2}$

$k^{2} - k = 6$

$k^{2} - k - 6 = 0$

$(k + 2) (k - 3) = 0$

$∴ k = - 2, 3$

when $k = - 2$ series is $- 2, 4, 10$

when $k = 3$ series is $3, 9, 15$

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