Question

Show that

\[\begin{vmatrix}x + 1 & x + 2 & x + a \\ x + 2 & x + 3 & x + b \\ x + 3 & x + 4 & x + c\end{vmatrix} =\text{ 0 where a, b, c are in A . P .}\]

 

Answer 1

Given: a, b, c are in A.P.

\[2b = a + c\]

\[∆ = \begin{vmatrix}x + 1 & x + 2 & x + a \\ x + 2 & x + 3 & x + b \\ x + 3 & x + 4 & x + c\end{vmatrix} \left[\text{ Applying }R_2 = 2 R_2 \right]\] 

\[ ∆ = \frac{1}{2}\begin{vmatrix}x + 1 & x + 2 & x + a \\ 2x + 4 & 2x + 6 & 2x + 2b \\ x + 3 & x + 4 & x + c\end{vmatrix} \] 

\[ ∆ = \frac{1}{2}\begin{vmatrix}x + 1 & x + 2 & x + a \\ 0 & 0 & 0 \\ x + 3 & x + 4 & x + c\end{vmatrix} \left[ \because 2b = a + c \right] \left[\text{ Applying }R_2 \to R_2 - \left( R_1 + R_3 \right) \right]\] 
\[ ∆ = 0\]