Question

Choose the correct answer.
If $a,b,c$,are in A.P.,then the determinant
$\begin{vmatrix}x+2& x+3& x+2a\\\ x+3& x+4& x+2b\\\ x+4& x+5& x+2c\end{vmatrix}$ is:

• A. 0
• B. 1
• C. x
• D. 2x

Answer 1

Answer: (A)

0

Answer 2

The correct answer is A:0
$\triangle=\begin{vmatrix}x+2& x+3& x+2a\\\ x+3& x+4& x+2b\\\ x+4& x+5& x+2c\end{vmatrix}$
$\begin{vmatrix}x+2& x+3& x+2a\\\ x+3& x+4& x+(a+c)\\\ x+4& x+5& x+2c\end{vmatrix}$ ($2b=a+c$ as $a,b$ and $c$ are in AP)
Applying $R_1→R_1-R_2$ and $R_3→R_3-R_2$,we have:
$Δ=\begin{vmatrix}-1& -1& a-c\\\ x+3& x+4& x+(a+c)\\\ 1& 1& c-a\end{vmatrix}$
Applying $R_1→R_1+R_3$ we have:
$Δ=\begin{vmatrix}0&0&0\\\ x+3& x+4& x+a+c\\\ 1& 1& c-a\end{vmatrix}$
Here, all the elements of the first row $(R_1)$ are zero.
Hence, we have ∆ = 0.
The correct answer is A