A) in GP
B) in HP
C) equal
D) in AP
Correct Answer: D
Solution :
Given, \[\left| \begin{matrix} x+1 & x+2 & x+a \\ x+2 & x+3 & x+b \\ x+3 & x+4 & x+c \\ \end{matrix} \right|=0\] Applying \[{{R}_{1}}\to {{R}_{1}}+{{R}_{3}}-2{{R}_{2}},\] we get \[\left| \begin{matrix} 0 & 0 & a+c-2b \\ x+2 & x+3 & x+b \\ x+3 & x+4 & x+c \\ \end{matrix} \right|=0\] \[\Rightarrow \] \[(a+c-2b)(-1)=0\] \[\Rightarrow \] \[2b=a+c\] \[\Rightarrow \] \[a,b,c\] are AP.
You need to login to perform this action.
You will be redirected in
3 sec
