Messed up!

Since $0$ is a term of the AP, we can assume the AP either starts with a negative first term $a_1$ with a positive common difference $d$ or with a positive $a_1$ and negative $d$ . Either way we get the same result for this problem. Let us take the former assumption of $a_1 < 0$ and $d>0$ . Also let the $n$ th term $a_n = 0$ . Then any term higher than $n$ is positive. Assuming $p > r > n$ (again, the result of the problem is the same if we assume $n>r>p$ ), then $a_p = (p-n)d$ , $a_r = (r-n)d$ , and

$\begin{aligned} (p-n)d & = q(r-n)d \\ qn-n & = qr-p \\ \implies n & = \frac {qr-p}{q-1} = \frac {qr-r+r-p}{q-1} = r + \frac {r-p}{q-1} \end{aligned}$

Since $n$ is an integer, $\dfrac {r-p}{q-1}$ must be an integer, that is $r-p$ is divisible by $q-1$ .