A) 0
B) 1
C) \[abc\]
D) \[pqr\]
Correct Answer: A
Solution :
\[{{T}_{p}}={{a}_{1}}+(p-1)d\] \[{{T}_{q}}={{q}_{a}}+(q-1)d\] \[\therefore \] \[LHS=a(q-r)+b(r-p)+c(p-q)\] \[=\{{{a}_{1}}+(p-1)d\}\,(q-r)+\{{{a}_{1}}+(q-1)d\}\] \[(r-p)+\{{{a}_{1}}+(r-1)d\}(p-q)\] \[={{a}_{1}}\cdot (q-r+r-p+p-q)+d\,[(p-1)\] \[=(q-r)+(q-1)\,(r-p)+(r-1)\,(p-q)]=0\]You need to login to perform this action.
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