Answer & Solution

Answer: Option D

Solution:

$$\eqalign{ & {\text{Sum}}\,{\text{of}}\,p\,{\text{terms}} = q \cr & i.e., \cr & {S_p} = \frac{p}{2}\left[ {2a + \left( {p - 1} \right)d} \right] = q \cr & \Rightarrow p\left[ {2a + \left( {p - 1} \right)d} \right] = 2q \cr & \Rightarrow 2ap + p\left( {p - 1} \right)d = 2q\,.....\left( 1 \right) \cr & {\text{and}}\,{\text{sum}}\,{\text{of}}\,q\,{\text{terms}} = p \cr & i.e., \cr & {S_q} = \frac{q}{2}\left[ {2a + \left( {q - 1} \right)d} \right] = p \cr & \Rightarrow q\left[ {2a + \left( {q - 1} \right)d} \right] = 2p \cr & \Rightarrow 2aq + q\left( {q - 1} \right)d = 2p\,.....\left( 2 \right) \cr & {\text{Subtracting}}\,\left( 2 \right)\,{\text{from}}\,\left( 1 \right) \cr} $$
$$ \Rightarrow 2a\left( {p - q} \right) + \left\{ {{p^2} - p - {q^2} + q} \right\}d$$ $$ = 2q - 2p$$
$$ \Rightarrow 2a\left( {p - q} \right) + \left\{ {{p^2} - {q^2} - \left( {p - q} \right)} \right\}d$$ $$ = - 2\left( {p - q} \right)$$
$$ \Rightarrow 2a\left( {p - q} \right) + \left\{ {\left( {p + q} \right)\left( {p - q} \right) - \left( {p - q} \right)} \right\}d$$ $$ = - 2\left( {p - q} \right)$$
$$ \Rightarrow 2a\left( {p - q} \right) + \left( {p - q} \right)\left[ {p + q - 1} \right]d$$ $$ = - 2\left( {p - q} \right)$$ $${\text{Dividing}}\,{\text{by}}$$ $$\,\left( {p - q} \right)$$
$$\eqalign{ & \Rightarrow 2a + \left( {p + q - 1} \right)d = - 2\,.....\left( 3 \right) \cr & {S_{p + q}} = \frac{{p + q}}{2}\left[ {2a + \left( {p + q - 1} \right)d} \right] \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{p + q}}{2}\left[ { - 2} \right]\,\,\,\left[ {{\text{From}}\,\left( 3 \right)} \right] \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = - \left( {p + q} \right) \cr} $$

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

Solution