Question

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

Answer 1

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

Using the row transformation, $R_1\to R_2-R_1$ and $R_2\to R_3-R_2,$ we get

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

and then using $R_1\to R_2-R_1$ we get,

$\left|\begin{array}{ccc}0& 0& 2b-a-c\\ 1& 1& c-b\\ x+3& x+4& x+c\end{array}\right|.$

Now expanding the determinant by $R_1$ we get $(2b-a-c)\times 1=0.$

So, $2b=a+c$.

Hence, $a,b,c$ are in A.P