Let the first term of the sequence be $a$.

The second term is $a + d$.
The fourth term is $a + 3d$.

Their product is $(a + d)(a + 3d) = -d^2$.

We are to find the product of the third and fifth terms:

Third term: $a + 2d$
Fifth term: $a + 4d$

Their product is $(a + 2d)(a + 4d)$.

Let us expand $(a + d)(a + 3d)$:

$(a + d)(a + 3d) = a^2 + 3ad + ad + 3d^2 = a^2 + 4ad + 3d^2 = -d^2$

So:
$a^2 + 4ad + 3d^2 = -d^2$
$a^2 + 4ad + 4d^2 = 0$

Now, $(a + 2d)(a + 4d) = a^2 + 4ad + 2ad + 8d^2 = a^2 + 6ad + 8d^2$

From above, $a^2 + 4ad + 4d^2 = 0$, so $a^2 + 4ad = -4d^2$

Therefore:
$(a + 2d)(a + 4d) = (a^2 + 4ad) + 2ad + 8d^2 = (-4d^2) + 2ad + 8d^2 = 2ad + 4d^2$

But $a^2 + 4ad + 4d^2 = 0$

Let us solve for $a$:
$a^2 + 4ad + 4d^2 = 0$
$a^2 + 4ad = -4d^2$

So $a^2 + 4ad = -4d^2$

Now, $2ad + 4d^2 = 2ad + 4d^2$

But we can express $a$ in terms of $d$:
$a^2 + 4ad + 4d^2 = 0$
$(a + 2d)^2 = 0$
So $a + 2d = 0$
Thus, $a = -2d$

Now, third term: $a + 2d = -2d + 2d = 0$
Fifth term: $a + 4d = -2d + 4d = 2d$

Their product: $0 \times 2d = 0$

Answer: $0$