Question

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

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

Answer 1

The correct option is A A.P.

As given $\left|\begin{array}{ccc}x+1& x+2& x+3\\ x+2& x+3& x+4\\ x+a& x+b& x+c\end{array}\right|=0=\left|\begin{array}{ccc}-1& -1& x+3\\ -1& -1& x+4\\ a-b& b-c& x+c\end{array}\right|=0,by\begin{array}{c}{C}_1\to C_1-C_2\\ C_2\to C_2-C_3\end{array}\Rightarrow \left|\begin{array}{ccc}0& 0& -1\\ -1& -1& x+4\\ a-b& b-c& x+c\end{array}\right|=0,by{R}_1\to R_1-R_2\Rightarrow (-1)(-b+c+a-b)=0\Rightarrow 2b-a-c=0\Rightarrow a+c=2b$ 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.
$\left|\begin{array}{ccc}1& 2& 3\\ 2& 3& 4\\ a& b& c\end{array}\right|=0$
$\Rightarrow$ 1(3c-4b) -2(2c-4a) +3(2b-3a)=0
$\Rightarrow$ -c+2b-a=0 $\Rightarrow$ 2b =a+c. Hence the result.