### If the sum of p terms of an A.P. is q and the sum of q terms is p, then the sum of (p + q) terms will be

A. 0 B. p – q C. p + q D. -(p + q) Answer: Option D
$\begin{array}{l}\text{ Sum of }p\text{ terms }=q\\ \text{ i.e., }\\ S_p=\frac{p}{2}[2a+(p-1)d]=q\\ \Rightarrow p[2a+(p-1)d]=2q\\ \Rightarrow2ap+p(p-1)d=2q\ldots.\text{ (1) }\\ \begin{array}{l} \text { and sum of } q \text { terms }=p \\ \text { i. } e . \\ S_{q}=\frac{q}{2}[2 a+(q-1) d]=p \\ \Rightarrow q[2 a+(q-1) d]=2 p \\ \Rightarrow 2 a q+q(q-1) d=2 p \ldots . .(2) \end{array}\end{array}$ $\begin{array}{l} \text { Subtracting (2) from (1) }\\ \begin{array}{l} \Rightarrow 2 a(p-q)+\left\{p^{2}-p-q^{2}+q\right\} d=2 q-2 p \\ \Rightarrow 2 a(p-q)+\left\{p^{2}-q^{2}-(p-q)\right\} d=-2(p-q) \\ \Rightarrow 2 a(p-q)+\{(p+q)(p-q)-(p-q)\} d=-2(p-q) \\ \Rightarrow 2 a(p-q)+(p-q)[p+q-1] d=-2(p-q) \text { Dividing by }(p-q) \\ \Rightarrow 2 a+(p+q-1) d=-2 \ldots(3) \\ S_{p+q}=\frac{p+q}{2}[2 a+(p+q-1) d] \\ \quad=\frac{p+q}{2}[-2] \quad[\text { From }(3)] \\ \quad=-(p+q) \end{array} \end{array}$