If $k+2,\ 4k−6,\ 3k−2$ are the three consecutive terms of an A.P., then find the value of $k$.

Given: $k+2,\ 4k-6,\ 3k-2$ are the three consecutive terms of an A.P.

To do: To find the value of $k$.

It is given that $k+2,\ 4k-6,\ 3k-2$ are the consecutive terms of A.P, therefore, by arithmetic mean property it's well known that, first term + third term is equal to twice of second term that is:

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

$\Rightarrow 4k=8k-12$

$\Rightarrow 4k−8k=-12$

$\Rightarrow -4k=-12$

$\Rightarrow 4k=12$

$\Rightarrow k=\frac{12}{4}=3$

Hence $k=3$.