An algebra problem by Shriniketan Ruppa

• For any three terms $a, b, c$ in A.P. we have $2b = a + c$ . Hence:

\[\begin{align} 2(2k^2 + 3k + 6) &= (k^2 + 4k + 8) + (3k^2 +4k +4) \\ \cancel{4k^2} + 6k + \cancel{12} &= \cancel{4k^2} + 8k + \cancel{12} \\ 6k &= 8k \\ k&= \boxed{0}

\end{align}\]

For three terms $a,b,c$ to be in A.P., $a+b=2b$ $\implies 4k^2+8k+12=4k^2+6k+12$ $\implies 8k=6k \implies \boxed{k=0}$