Question

If $a,b,c$ are in AP then Prove that $\left|\begin{array}{ccc}x+2& x+3& x+2a\\ x+3& x+4& x+2b\\ x+4& x+5& x+2c\end{array}\right|=0$

Answer 1

$△=\left|\begin{array}{ccc}x+2& x+3& x+2a\\ x+3& x+4& x+2b\\ x+4& x+5& x+2c\end{array}\right|$

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

$\Rightarrow 2b=a+c$

$△=\left|\begin{array}{ccc}x+2& x+3& x+2a\\ x+3& x+4& x+a+c\\ x+4& x+5& x+2c\end{array}\right|$

Applying $R_1\to R_1-R_2$ and $R_3\to R_3-R_2$ we have

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

Applying $R_1\to R_1+R_3$

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

Here all the elements of the first row $R_1$ are zero.

Hence, we have $\Delta =0$