Determine $\text{'}k\text{'}$ so that $8k+4,6k-2$ and $2k-7$ are the three consecutive terms of an A.P.

$8k+4,6k-2$, $2k-7$ are three consecutive terms then

$a_2-a_1=a_3-a_2$

$\Rightarrow (6k-2)-(8k+4)=(2k-7)-(6k-2)$

$\Rightarrow 6k-2-8k-4=2k-7-6k+2$

$\Rightarrow -2k-6=-4k-5$

$\Rightarrow 4k-2k=6-5$

$\Rightarrow 2k=1$

$\therefore k=\frac{1}{2}$