A) A. P.
B) G. P.
C) H. P.
D) None of these
Correct Answer: A
Solution :
As given \[\left| \,\begin{matrix} x+1 & x+2 & x+3 \\ x+2 & x+3 & x+4 \\ x+a & x+b & x+c \\ \end{matrix}\, \right|\,=\,0\] = \[\left| \,\begin{matrix} -1 & -1 & x+3 \\ -1 & -1 & x+4 \\ a-b & b-c & x+c \\ \end{matrix}\, \right|\,=0\], by \[\begin{align} & {{C}_{1}}\to {{C}_{1}}-{{C}_{2}} \\ & {{C}_{2}}\to {{C}_{2}}-{{C}_{3}} \\ \end{align}\] \[\Rightarrow \] \[\left| \,\begin{matrix} 0 & 0 & -1 \\ -1 & -1 & x+4 \\ a-b & b-c & x+c \\ \end{matrix}\, \right|\,=0\], 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. \[\left| \,\begin{matrix} 1 & 2 & 3 \\ 2 & 3 & 4 \\ a & b & c \\ \end{matrix}\, \right|=0\] \[\Rightarrow 1\,(3c-4b)-2(2c-4a)+3(2b-3a)=0\] \[\Rightarrow \]\[-c+2b-a=0\Rightarrow 2b=a+c\]. Hence the result.
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