The sum of the first p terms of an AP in equal to the sum of the
first q terms.Prove that sum of the first p+q terms of the same
AP is zero.
$\displaystyle S_p=S_q$
$\displaystyle \frac{(2a_1 +(p-1)d)p}{2}=\frac{(2a_1+(q-1)d)q}{2}$
$\displaystyle p(2a_1 +pd-d)=q(2a_1+qd-d)$
$\displaystyle 2a_1p+p^2d-pd =2a_1q+q^2d-qd$
$\displaystyle 2a_1p-2a_1q +p^2d -q^2d -pd +qd =0$
$\displaystyle 2a_1(p-q)+d(p+q)(p-q)-d(p-q)=0$
$\displaystyle (p-q)(2a_1+(p+q)d-d)=0$
$\displaystyle 2a_1+(p+q)d-d=0$
$\displaystyle 2a_1+((p+q)-1)d=0$
$\displaystyle \frac{(p+q)(2a_1+((p+q)-1)d)}{2}=0$
$\displaystyle S_{p+q}=0$
You could also use
$\displaystyle a+(a+d)+(a+2d)+....+[a+(p-1)d]$
$\displaystyle =a+(a+d)+...+[a+(p-1)d]+[a+pd]+[a+(p+1)d]+....+[a+(q-1)d]$
for q>p
Then
$\displaystyle [a+pd]+[a+(p+1)d]+....+[a+(q-1)d]=0$
Therefore the average must be zero, which is also the average of the first and last terms of this progression.
$\displaystyle 0.5[a+pd+a+(q-1)d]=0$
$\displaystyle 2a+(p+q-1)d=0$
This causes the sum of p+q terms to be zero, since
$\displaystyle S_{p+q}=\frac{2a+(p+q-1)d}{2}(p+q)$