The sum of p terms of …

Let a be the first term and d the common difference of the given A.P.
{tex}\therefore S_{p}=\frac{p}{2}{/tex} [2a + (p - 1)d] = q 
{tex}\Rightarrow{/tex} 2a + (p - 1)d {tex}=\frac{2 q}{p}{/tex}  ….(i)
And {tex}S_{q}=\frac{q}{2}{/tex} [2a + (q - 1)d] = p
{tex}\Rightarrow{/tex} 2a + (q - 1)d {tex}=\frac{2 p}{q}{/tex} ….(ii)
Subtracting eq. (ii) from eq. (i) we get
(p - q)d = {tex}\frac{2 q}{p}-\frac{2 p}{q}{/tex} {tex}\Rightarrow{/tex} (p - q)d {tex}=\frac{2\left(q^{2}-p^{2}\right)}{p q}{/tex}
{tex}\Rightarrow{/tex} (p - q)d {tex}=\frac{-2}{p q}{/tex}(p² - q²)
{tex}\Rightarrow{/tex} (p - q)d {tex}=\frac{-2}{p q}{/tex} (p + q)(p - q) {tex}\Rightarrow d=\frac{-2}{p q}{/tex} (p + q)
Substituting the value of d in eq. (i) we get
2a + (p - 1) {tex}\left[\frac{-2(p+q)}{p q}\right]=\frac{2 q}{p}{/tex}
{tex}\Rightarrow 2 a=\frac{2 q}{p}+\frac{2(p-1)(p+q)}{p q}{/tex}
{tex}\Rightarrow a=\frac{q}{p}+\frac{(p-1)(p+q)}{p q}{/tex}
{tex}a=\frac{q^{2}+p^{2}+p q-p-q}{p q}{/tex}
Now S_p+q {tex}=\frac{p+q}{2}{/tex} [2a + (p + q - 1)d
{tex}=\frac{p+q}{2}\left[\frac{2 q^{2}+2 p^{2}+2 p q-2 q-2 q}{p q}+\frac{(p+q-1)[-2(p+q)}{p q}\right]{/tex}
{tex}=\frac{p+q}{2}\left[\frac{2q^{2} + 2p^{2} + 2pq - 2p - 2q -2p^{2} -2 p q+2 p-2 p q-2 q^{2}+2 q}{p q}\right]{/tex}
{tex}=\frac{p+q}{2}\left[\frac{-2 p q}{p q}\right]{/tex} = -(p + q) hence proved.