A) \[1\]
B) \[-1\]
C) \[0\]
D) \[pqr\]
Correct Answer: C
Solution :
Let the first term be \[{{a}_{1}}\] and common difference be\[d.\] According to the condition \[{{T}_{p}}={{a}_{1}}+(p-1)d\] \[\Rightarrow \] \[a={{a}_{1}}+(p-1)d\] ... (i) \[{{T}_{q}}={{a}_{1}}+(q-1)d\] \[\Rightarrow \] \[b={{a}_{1}}+(q-1)d\] ... (ii) and \[{{T}_{r}}={{a}_{1}}+(r-1)d\] \[\Rightarrow \] \[c={{a}_{1}}+(r-1)d\] ... (iii) \[\therefore \]\[\left| \begin{matrix} a & p & 1 \\ b & q & 1 \\ c & r & 1 \\ \end{matrix} \right|=\left| \begin{matrix} {{a}_{1}} & +(p-1)d & p\,\,\,1 \\ {{a}_{1}} & +(q-1)d & p\,\,\,1 \\ {{a}_{1}} & +(r-1)d & r\,\,\,1 \\ \end{matrix} \right|\] Applying \[{{C}_{1}}\to {{C}_{1}}-({{C}_{2}}-{{C}_{1}})d\] \[=\left| \begin{matrix} {{a}_{1}} & p & 1 \\ {{a}_{1}} & q & 1 \\ {{a}_{1}} & r & 1 \\ \end{matrix} \right|\] \[={{a}_{1}}\left| \begin{matrix} 1 & p & 1 \\ 1 & q & 1 \\ 1 & r & 1 \\ \end{matrix} \right|\] \[=0[\because \]two columns are identical] Note: If any two rows or columns are identical or proportional, then the value of determinant will be zero:
You need to login to perform this action.
You will be redirected in
3 sec
