If the sum of an A.P. is the same for p terms and q terms, then the sum of its (p+q) terms is:
Answer: B
Given \(S_p = S_q\). \(\frac{p}{2}[2a+(p-1)d] = \frac{q}{2}[2a+(q-1)d]\). After simplification, we get \(2a(p-q) + d(p^2-p-q^2+q)=0 \Rightarrow (p-q)[2a+d(p+q-1)]=0\). Since p is not equal to q, \(2a+d(p+q-1)=0\). The sum of (p+q) terms is \(S_{p+q} = \frac{p+q}{2}[2a+(p+q-1)d]\). Since the term in the bracket is 0, the sum is 0.