Solution 133008a1-b165-4079-aa13-b8af5a6b47b4

Points: 3 pts

Lemma 1: $1+3d=r^3$

Let $d$ be the common difference and $r$ be the common ratio such that:
\[
\begin{aligned}
d&=x-1=y-x=z-y,\\[2mm]
r&=\frac{p}{1}=\frac{q}{p}=\frac{z}{q}.
\end{aligned}
\]
We then have
\[
1+3d=1\cdot r^3=z,
\]
giving the desired result.

Lemma 2: $(d,r)=(21,4)$

Working modulo $3$, we see
\[
r^3\equiv 1\pmod{3}
\quad\Longrightarrow\quad
r\equiv 1\pmod{3}
\quad\Longrightarrow\quad
r=1+3i,
\]
where $i\in\mathbb{Z}$. Both sequences are strictly increasing, so
\[
r>1,\qquad d>0
\quad\Longrightarrow\quad
\min r=4.
\]
Since $z=r^3$, the smallest $r$ gives the smallest $z$. We then find
\[
d=\frac{64-1}{3}=21,
\]
which is a valid positive integer.

Lemma 3: $S=149$

Let
\[
S=x+y+z+p+q.
\]
The two sequences are
\[
\begin{aligned}
(1,x,y,z)&=(1,22,43,64),\\
(1,p,q,z)&=(1,4,16,64).
\end{aligned}
\]
Adding the required values gives
\[
S=22+43+64+4+16=149.
\]

Final answer:
149