Question

If sum of first $p$ terms of an AP is equal to the sum of first $q$ terms, then find the sum of the first $(p+q)$ terms.

Answer 1

Formula:

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

Sum of first p terms,

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

Sum of first q terms,

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

given,

$S_p=S_q$

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

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

$2a(p-q)=-d(p-q)[p+q-1]$

$2a(p-q)+d(p-q)[p+q-1]=0$

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

$p-q\ne 0\Rightarrow [2a+d(p+q-1\left)\right]=0$

Now,

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

$=\frac{p+q}{2}\times 0$

$=0$

Therefore the sum of $(p+q)$ terms is $0$.