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

#### Options

0

p − q

p + q

− (p + q)

#### Solution

− (p + q)

\[S_p = q\]

\[ \Rightarrow \frac{p}{2}\left\{ 2a + \left( p - 1 \right)d \right\} = q\]

\[ \Rightarrow 2ap + \left( p - 1 \right)pd = 2q . . . . . \left( 1 \right)\]

\[ S_q = p\]

\[ \Rightarrow \frac{q}{2}\left\{ 2a + \left( q - 1 \right)d \right\} = p\]

\[ \Rightarrow 2aq + \left( q - 1 \right)qd = 2p . . . . . \left( 2 \right)\]

\[\text { Multiplying equation } \left( 1 \right) \text { by q and equation } \left( 2 \right) \text { by p and then solving, we get }: \]

\[d = \frac{- 2\left( p + q \right)}{pq}\]

\[\text { Now }, S_{p + q} = \frac{\left( p + q \right)}{2}\left[ 2a + \left( p + q - 1 \right)d \right]\]

\[ = \frac{p}{2}\left[ 2a + \left( p - 1 \right)d + qd \right] + \frac{q}{2}\left[ 2a + \left( q - 1 \right)d + pd \right]\]

\[ = S_p + \frac{pqd}{2} + S_q + \frac{pqd}{2}\]

\[ = p + q + pqd\]

\[ = p + q - \frac{2\left( p + q \right)pq}{pq}\]

\[ = - (p + q)\]