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 DShow Answer
Solution(By Apex Team)
$\begin{array}{l}\text{ Sum of }p\text{ terms }=q\\
\text{ i.e., }\\
S_p=\frac{p}{2}[2a+(p-1)d]=q\\
\Rightarrow p[2a+(p-1)d]=2q\\
\Rightarrow2ap+p(p-1)d=2q\ldots.\text{ (1) }\\
\begin{array}{l}
\text { and sum of } q \text { terms }=p \\
\text { i. } e . \\
S_{q}=\frac{q}{2}[2 a+(q-1) d]=p \\
\Rightarrow q[2 a+(q-1) d]=2 p \\
\Rightarrow 2 a q+q(q-1) d=2 p \ldots . .(2)
\end{array}\end{array}$
$\begin{array}{l}
\text { Subtracting (2) from (1) }\\
\begin{array}{l}
\Rightarrow 2 a(p-q)+\left\{p^{2}-p-q^{2}+q\right\} d=2 q-2 p \\
\Rightarrow 2 a(p-q)+\left\{p^{2}-q^{2}-(p-q)\right\} d=-2(p-q) \\
\Rightarrow 2 a(p-q)+\{(p+q)(p-q)-(p-q)\} d=-2(p-q) \\
\Rightarrow 2 a(p-q)+(p-q)[p+q-1] d=-2(p-q) \text { Dividing by }(p-q) \\
\Rightarrow 2 a+(p+q-1) d=-2 \ldots(3) \\
S_{p+q}=\frac{p+q}{2}[2 a+(p+q-1) d] \\
\quad=\frac{p+q}{2}[-2] \quad[\text { From }(3)] \\
\quad=-(p+q)
\end{array}
\end{array}$
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