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Question
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)
\[S_p = q\]
\[ \Rightarrow \frac{p}{2}\left\{ 2a + \left( p - 1 \right)d \right\} = q\]
\[ \Rightarrow 2ap + \left( p - 1 \right)pd = 2q . . . . . \left( 1 \right)\]
\[ S_q = p\]
\[ \Rightarrow \frac{q}{2}\left\{ 2a + \left( q - 1 \right)d \right\} = p\]
\[ \Rightarrow 2aq + \left( q - 1 \right)qd = 2p . . . . . \left( 2 \right)\]
\[\text { Multiplying equation } \left( 1 \right) \text { by q and equation } \left( 2 \right) \text { by p and then solving, we get }: \]
\[d = \frac{- 2\left( p + q \right)}{pq}\]
\[\text { Now }, S_{p + q} = \frac{\left( p + q \right)}{2}\left[ 2a + \left( p + q - 1 \right)d \right]\]
\[ = \frac{p}{2}\left[ 2a + \left( p - 1 \right)d + qd \right] + \frac{q}{2}\left[ 2a + \left( q - 1 \right)d + pd \right]\]
\[ = S_p + \frac{pqd}{2} + S_q + \frac{pqd}{2}\]
\[ = p + q + pqd\]
\[ = p + q - \frac{2\left( p + q \right)pq}{pq}\]
\[ = - (p + q)\]
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