A) 1
B) \[-1\]
C) 0
D) \[pqr\]
E) \[p+q+r\]
Correct Answer: C
Solution :
\[\because \]a, b, c are respectively p th, q th and r th terms of an AP, then \[a=A+(p-1)D\] ...(i) \[b=A+(q-1)D\] ...(ii) and \[c=A+(r-1)D\] ...(iii) Now,\[\left| \begin{matrix} a & p & 1 \\ b & q & 1 \\ c & r & 1 \\ \end{matrix} \right|=a(q-r)+b(r-p)+c(p-q)\] \[=A(q-r+r-p+p-q)\] \[+D\{(p-1)(q-r)+(q-1)(r-p)+(r-1)\] \[(p-q)\}\] \[=A0+D0=0\]You need to login to perform this action.
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