Determine value of $k$ so that $3 k - 2$ , $4 k - 6$ and $k + 2$ are three consecutive terms of an $A P .$

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Solution

Given,

$(3 k - 2), (4 k - 6)$ and $(k + 2)$ are in $A . P$

So,

$2 (4 k - 6) = 3 k - 2 + k + 2$

$8 k - 12 = 4 k$

$4 k = 12$

$k = 3$

Hence, this is the answer.

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