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 a, b, c are in ____________.

Answer 1

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


$$\begin{gathered} \left|\begin{array}{ccc}x+1& x+2& x+a\\ x+2& x+3& x+b\\ x+3& x+4& x+c\end{array}\right|=0 \\ \mathrm{Applying}{C}_1\to C_1-C_2 \\ \Rightarrow \left|\begin{array}{ccc}x+1-x-2& x+2& x+a\\ x+2-x-3& x+3& x+b\\ x+3-x-4& x+4& x+c\end{array}\right|=0 \\ \Rightarrow \left|\begin{array}{ccc}-1& x+2& x+a\\ -1& x+3& x+b\\ -1& x+4& x+c\end{array}\right|=0 \\ \mathrm{Taking}\mathrm{out}\left(-1\right)\mathrm{common}\mathrm{from}{C}_1 \\ \Rightarrow \left(-1\right)\left|\begin{array}{ccc}1& x+2& x+a\\ 1& x+3& x+b\\ 1& x+4& x+c\end{array}\right|=0 \\ \mathrm{Applying}{R}_2\to R_2-R_1\mathrm{and}{R}_3\to R_3-R_1 \\ \Rightarrow \left(-1\right)\left|\begin{array}{ccc}1& x+2& x+a\\ 1-1& x+3-x-2& x+b-x-a\\ 1-1& x+4-x-2& x+c-x-a\end{array}\right|=0 \\ \Rightarrow \left(-1\right)\left|\begin{array}{ccc}1& x+2& x+a\\ 0& 1& b-a\\ 0& 2& c-a\end{array}\right|=0 \\ \mathrm{Expanding}\mathrm{through}{C}_1 \\ \Rightarrow \left(-1\right)\left(1\left(c-a\right)-2\left(b-a\right)\right)=0 \\ \Rightarrow c-a-2b+2a=0 \\ \Rightarrow c+a-2b=0 \\ \Rightarrow c+a=2b \\ \Rightarrow a,b,c\mathrm{are}\mathrm{in}\mathrm{A}.\mathrm{P}. \end{gathered}$$

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