Question

If the sum of the first p terms of an AP is the same as the sum of its first q terms ( where $p\ne q$ ) then show that the sum of its first $(p+q)$ terms is zero.

Answer 1

Let a be the first term and d be the common difference of the given AP. Then,

$S_p=S_q$

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

$(p-q)\left(2a\right)=(q-p)(q+p-1)d$

$2a=(1-p-q)d$ ......(1)

Sum of the first $(p+q)$ terms of the given AP

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

$=\frac{p+q}{2}\left[\right(1-p-q)d+(p+q-1\left)d\right]=0$