Question

If the sum of the first p terms of an AP is q and that of q terms is p, then the sum of the first (p + q) terms will be:

• A. p - q
• B. p + q
• C. -(p + q)

Answer 1

The correct option is C -(p + q)

Let the first term and common difference be a and d respectively and let $T_n$ denote the $n^{th}$ term.
$\therefore T_p=a+(p-1)d$
Sum of the first p terms of this A.P = $\left(\frac{p}{2}\right)(2a+(p-1\left)d\right)=q$
$\Rightarrow (2a+(p-1\left)d\right)=\frac{2q}{p}\dots \left(1\right)$

Similarly,
Sum of the first q terms of this A.P = $\left(\frac{q}{2}\right)(2a+(q-1\left)d\right)=p$
$\Rightarrow (2a+(q-1\left)d\right)=\frac{2p}{q}\dots \left(2\right)$

Subtracting (1) and (2),
$(p-q)d=2(\frac{q}{p}-\frac{p}{q})$
$(p-q)d=2\left(\frac{q^2-p^2}{pq}\right)$
$(p-q)d=2\left(\frac{(q-p)(p-q)}{pq}\right)$
$\Rightarrow d=-2\left(\frac{p+q}{pq}\right)\dots \left(3\right)$

$\therefore$ Sum of the first (p+q) terms
= $\left(\frac{p+q}{2}\right)(2a+(p+q-1\left)\right(-2\left(\frac{p+q}{pq}\right))$
= $\left(\frac{p+q}{2}\right)(2a+(q-1)d+pd)$
= $\left(\frac{p+q}{2}\right)(\frac{2p}{q}+pd)$ (from (2))
= $\left(\frac{p+q}{2}\right)(\frac{2p}{q}+(-2\left)\right(\frac{p+q}{q}\left)\right)$ (from (3))
= $\left(\frac{p+q}{2}\right)(\frac{2p}{q}-2\frac{p}{q}-2)$
= $-(p+q)$