# Choose the correct answer. If a, b, c, are in A.P., then the determinant

Question:

If abc, are in A.P., then the determinant

$\left|\begin{array}{lll}x+2 & x+3 & x+2 a \\ x+3 & x+4 & x+2 b \\ x+4 & x+5 & x+2 c\end{array}\right|$

A. 0 B. 1 C. $x$ D. $2 x$

Solution:

\begin{aligned} \Delta &=\left|\begin{array}{lll}x+2 & x+3 & x+2 a \\ x+3 & x+4 & x+2 b \\ x+4 & x+5 & x+2 c\end{array}\right| \\ &=\left|\begin{array}{lll}x+2 & x+3 & x+2 a \\ x+3 & x+4 & x+(a+c) \\ x+4 & x+5 & x+2 c\end{array}\right| \quad(2 b=a+c \text { as } a, b, \text { and } c \text { are in A.P. }) \end{aligned}

Applying $R_{1} \rightarrow R_{1}-R_{2}$ and $R_{3} \rightarrow R_{3}-R_{2}$, we have:

$\Delta=\left|\begin{array}{lll}-1 & -1 & a-c \\ x+3 & x+4 & x+(a+c) \\ 1 & 1 & c-a\end{array}\right|$

Applying $R_{1} \rightarrow R_{1}+R_{3}$, we have:

$\Delta=\left|\begin{array}{ccc}0 & 0 & 0 \\ x+3 & x+4 & x+a+c \\ 1 & 1 & c-a\end{array}\right|$

Here, all the elements of the first row $\left(\mathrm{R}_{1}\right)$ are zero.

Hence, we have $\Delta=0$.