Various progressions

We know that $b = ar$ and $c = ar^2$ , where $r$ is a rational number. Since $a$ , $b + 8$ , and $c + 64$ form a geometric sequence, $b + 8$ is equal to the geometric mean of $a$ and $c + 64$ :

$\begin{aligned} a(c + 64) &= (b + 8)^2 \\ a(ar^2 + 64) &= (ar + 8)^2 \\ a^2r^2 + 64a &= a^2r^2 + 16ar + 64 \\ 64a &= 16ar + 64 \\ 4a &= ar + 4. \end{aligned}$

Solving for $r$ , we get $r = \dfrac{4(a - 1)}{a}$ . Note that $a \neq 0$ , because this forces $b = 0$ and $c = 0$ , but $(a, b + 8, c) = (0, 8, 0)$ is not an arithmetic sequence.

Since $a$ , $b + 8$ , and $c$ form an arithmetic sequence, $b + 8$ is the arithmetic mean of $a$ and $c$ :

$\begin{aligned} \dfrac{a + c}{2} &= b + 8 \\ \dfrac{a + ar^2}{2} &= ar + 8 \\ a + ar^2 &= 2ar + 16. \end{aligned}$

Now, substitute our value for $r$ : $\begin{aligned} a + a \left ( \dfrac{4(a - 1)}{a} \right )^2 &= 2a \left ( \dfrac{4(a - 1)}{a} \right ) + 16 \\ a + \dfrac{16(a - 1)^2}{a} &= 8(a - 1) + 16 \\ \dfrac{16(a - 1)^2}{a} &= 7a + 8 \\ 16(a^2 - 2a + 1) &= 7a^2 + 8a \\ 9a^2 - 40a + 16 &= 0 \\ (9a - 4)(a - 4) &= 0 \\ a &= 4, \dfrac{9}{4}. \end{aligned}$

Since $a$ is an integer, $a = \boxed{4}.$ To check, we have $r = \dfrac{4(4 - 1)}{4} = 3$ , so $b = 4(3) = 12$ and $c = 12(3) = 36.$ Indeed, $(a, b + 8, c) = (4, 20, 36)$ is an arithmetic sequence, and $(a, b + 8, c + 64) = (4, 20, 100)$ is a geometric sequence.