Find $k$ given the consecutive arithmetic terms:
$5, k, k^{2} - 8$

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Solution

Since, $5, k, k^{2} - 8$ are in A.P

$\Rightarrow k - 5 = k^{2} - 8 - k$

$2 k = k^{2} - 3$

$k^{2} - 2 k - 3 = 0$

$(k + 1) (k - 3) = 0$

$∴ k = - 1, 3$

when $k = - 1$ series is $5, - 1, - 7$

when $k = 3$ series is $5, 3, 1$

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