Question

The sum of $p$ terms of an A.P. is $q$, and the sum of $q$ terms is $p$; find the sum of $p+q$ terms.

Answer 1

Let the first term of the AP be $a$ and the common difference be $d$

Sum of $p$ terms $=q$

$\therefore q=pa+\frac{p(p-1)}{2}d$ ........... Since, sum of n terms $=na+\frac{n(n-1)}{2}d$

$\Rightarrow \frac{2q}{p}=2a+(p-1)d$ ...... $\left(i\right)$

Similarly, sum of $q$ terms $=p=qa+\frac{q(q-1)}{2}d$

$\Rightarrow \frac{2p}{q}=2a+(q-1)d$ ....... $\left(ii\right)$

Subtract $\left(i\right)-\left(ii\right)$

$2\left(\frac{q^2-p^2}{pq}\right)=(p-q)d$

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

$d=-2\frac{p+q}{pq}$ ...... $\left(iii\right)$

Sum of first $p+q$ terms = $(p+q)a+\frac{(p+q)(p+q-1)}{2}d$

= $\frac{(p+q)}{2}(2a+(p-1)d+qd)$

Using $\left(i\right)$ and $\left(iii\right)$,

$=\frac{(p+q)}{2}(\frac{2q}{p}+q(-2\frac{p+q}{pq}))$

$=(p+q)\times (\frac{q}{p}-\frac{p+q}{p})$

$=(p+q)\times (-1)$

$=-(p+q)$