A) 0
B) 1
C) 2
D) 3
Correct Answer: A
Solution :
\[p\{a+(p-1)\,d\}=q\{a+(q-1)\,d\}\] \[\Rightarrow \] \[a(p-q)+({{p}^{2}}-{{q}^{2}})\,d+(q-p)\,d=0\] \[\Rightarrow \] \[(p-q)\{a+(p+q-1)\,d\}=0\] \[\Rightarrow \] \[a+(p+q-1)\,d=0\]\[\Rightarrow \]\[{{T}_{p+q}}=0\], \[\{\because \ p\ne q\}\].
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