Question

If the sum of first $p$ terms of an $A.P.$ is equal to the sum of first $(p+q)$ terms is zero. Where $p\ne q$

Answer 1

Let sum of $1^{st}$ p terms be $s_p$ & sum

of $1^{st}$ 2 terms be $s_q$ and sum of

$(p+q)$ terms be $s_{(p+q)}$

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

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

According to question,

$s_p=s_q$

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

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

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

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

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

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

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

$\therefore s_{(p+q)}=0$