A) a, b, c are in GP
B) a, b, care in HP
C) a, b, c are in AP
D) \[{{(a+b)}^{2}}=c\]
E) \[a+b=c\]
Correct Answer: C
Solution :
Since, the given lines are concurrent. \[\therefore \] \[\left| \begin{matrix} a & k & 10 \\ b & k+1 & 10 \\ c & k+2 & 10 \\ \end{matrix} \right|=0\] \[\Rightarrow \] \[10\left| \begin{matrix} a & k & 1 \\ b & k+1 & 1 \\ c & k+2 & 1 \\ \end{matrix} \right|=0\] Applying\[{{R}_{2}}\to {{R}_{2}}-{{R}_{1}}\]and \[{{R}_{3}}\to {{R}_{3}}-{{R}_{1}}\] \[\Rightarrow \] \[10\left| \begin{matrix} a & k & 1 \\ b-a & 1 & 0 \\ c-a & 2 & 0 \\ \end{matrix} \right|=0\] \[\Rightarrow \] \[10[1(2b-2a-c+a)]=0\] \[\Rightarrow \] \[2b=a+c\] Hence, a, b and c are in AP.
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