The numbers $p, 10, q$ are the consecutive terms of an arithmetic series. The numbers $p, 6, q$ are from a geometric series. Show that $p^2-20p+36=0$ and hence find the values of $p$ and $q$ for which the geometric series converges.
What I have tried:
$p^2-20p+36 = 0 \implies (p-2)(p-18)=0$
Therefore, $p=2$ or $p=18$.
However, I'm not sure how to approach the part of the question which calculates for q
Let $d$ be the common difference and $r$ be the common ratio of the arithmetic series and geometric series respectively. As $p,10,q$ forms an arithmetic series, we have $p+q=(10-d)+(10+d)=20$. As $p,6,q$ forms an geometric series, we have $pq=(\frac 6 r)(6r)=36$, then by vieta theorem, $p$, $q$ is the solution of the equation $x^2-(p+q)x+pq=0$ which is $x^2-20x+36=0$. This means that $p^2-20p+36=0$ so $p=2$ or $p=18$, the same for $q$. Now the geometric series $p,6,q$ converges when $p\geq6\geq q$, this means that $p=18$ and $q=2$