If $2 k, 4 k (3)$ , and $(4 k + 4)$ are in AP. Find k.

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Solution

$2 k, 4 k (3), (4 k + 4)$ are in $A P$
$=> (4 k + 4) - 12 k = 12 k - 2 k$
$=> - 8 k + 4 = 10 k$
$=> 18 k = 4$
$=> k = \frac{4}{18}$

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