Question

If$\left| \begin{matrix} x + 1 & x + 2 & x + 3 \\ x + 2 & x + 3 & x + 4 \\ x + a & x + b & x + c \end{matrix} \right| = 0$, then $a,b,c$ are in.

• A. A. P.
• B. G. P.
• C. H. P.
• D. None of these

Answer 1

Answer: (A)

A. P.

Answer 2

As given $\left| \begin{matrix} x + 1 & x + 2 & x + 3 \\ x + 2 & x + 3 & x + 4 \\ x + a & x + b & x + c \end{matrix} \right| = 0$

= $\left| \begin{matrix}

• 1 & - 1 & x + 3 \
• 1 & - 1 & x + 4 \ a - b & b - c & x + c \end{matrix} \right| = 0$, by$C_{1} \rightarrow C_{1} - C_{2} $$C_{2} \rightarrow C_{2} - C_{3} $

$\Rightarrow$ $\left| \begin{matrix} 0 & 0 & - 1 \

• 1 & - 1 & x + 4 \ a - b & b - c & x + c \end{matrix} \right| = 0$, by$R_{1} \rightarrow R_{1} - R_{2}$

$$\Rightarrow ( - 1)( - b + c + a - b) = 0$$

$\Rightarrow$ $2b - a - c = 0 \Rightarrow a + c = 2b$ i.e., $a,b,c$ are in A.P.

Trick : In such type of problem, put any suitable value of x i.e. 0, so that the determinant.

$$\left| \begin{matrix} 1 & 2 & 3 \ 2 & 3 & 4 \ a & b & c \end{matrix} \right| = 0$$

$$\Rightarrow 1(3c - 4b) - 2(2c - 4a) + 3(2b - 3a) = 0$$

$\Rightarrow - c + 2b - a = 0 \Rightarrow 2b = a + c$.

Hence the result.