. $\left| {\begin{array}{llllllllllllllllllll}{x + 1}&{x + 2}&{x + a}\\{x + 2}&{x + 3}&{x + b}\\{x + 3}&{x + 4}&{x + c}\end{array}} \right| = 0$, where a, b and c are in AP.

Correct Answer True

Since, a, b and $c$ are in AP, then $2b = a + c$

$\therefore$ $\left| {\begin{array}{llllllllllllllllllll}{x + 1}&{x + 2}&{x + a}\\{x + 2}&{x + 3}&{x + b}\\{x + 3}&{x + 4}&{x + c}\end{array}} \right| = 0$

$\Rightarrow$ $\left| {\begin{array}{cccccccccccccccccccc}{2x + 4}&{2x + 6}&{2x + a + c}\\{x + 2}&{x + 3}&{x + b}\\{x + 3}&{x + 4}&{x + c}\end{array}} \right| = 0$

$\Rightarrow$ $\left| {\begin{array}{cccccccccccccccccccc}{2(x + 2)}&{2(x + 3)}&{2(x + b)}\\{x + 2}&{x + 3}&{x + b}\\{x + 3}&{x + 4}&{x + c}\end{array}} \right| = 0$

$\Rightarrow$ $0 = 0$
[since, ${R_1}$ and ${R_2}$ are in proportional to each other]

Hence, statement is true.