Answer: 31

Solution:

Write the terms as $a - d,$ $a,$ $a + d$ with integer $d \ge 1.$ The condition is $$\begin{aligned} &(a-d)^2 + a^2 \\ &\quad {}+ (a+d)^2 = ad^2 \\ &\quad\Longleftrightarrow\quad 3a^2 + 2d^2 = ad^2, \end{aligned}$$ so $d^2(a - 2) = 3a^2$ and $d^2 = \frac{3a^2}{a - 2}.$ For $d^2$ to be positive we need $a \gt 2.$ Indeed, $a = 0$ forces $d = 0,$ while for any nonzero integer $a \lt 2$ the right side is negative; $a = 2$ makes the original equation impossible.

Substituting $t = a - 2 \ge 1$ gives $$d^2 = \frac{3(t+2)^2}{t} = 3t + 12 + \frac{12}{t},$$ so $t \mid 12.$ Testing $t = 1, 2, 3, 4, 6, 12$ gives $d^2 = 27, 24, 25, 27, 32, 49:$ only $t = 3$ and $t = 12$ yield perfect squares.

These give $(a, d) = (5, 5)$ with sequence $0, 5, 10,$ and $(a, d) = (14, 7)$ with sequence $7, 14, 21.$ The sum of the third terms is $10 + 21 = 31.$