Let the roots be in arithmetic progression. Let the roots be:
\[
a - d, \quad a, \quad a + d,
\]
where \(a\) is the middle root and \(d\) is the common difference.
Step 1: Sum of roots
By Viète's formula for the cubic equation \(x^3 + p x^2 + q x + r = 0\):
\[
\alpha + \beta + \gamma = -p.
\]
Sum of roots in terms of \(a\) and \(d\):
\[
(a - d) + a + (a + d) = 3a = -p \implies a = -\frac{p}{3}.
\]
Step 2: Sum of products of roots two at a time
\[
\alpha \beta + \beta \gamma + \gamma \alpha = q.
\]
Calculate:
\[
(a - d)a + a(a + d) + (a - d)(a + d) = a^2 - a d + a^2 + a d + a^2 - d^2 = 3a^2 - d^2 = q.
\]
Step 3: Product of roots
\[
\alpha \beta \gamma = -r.
\]
Calculate:
\[
(a - d) \cdot a \cdot (a + d) = a (a^2 - d^2) = a^3 - a d^2 = -r.
\]
Step 4: Express \(q\) and \(r\) in terms of \(a\) and \(d\)
\[
q = 3a^2 - d^2,
\]
\[
r = -a^3 + a d^2.
\]
Step 5: Find relation between \(p\) and \(r\)
Recall \(a = -\frac{p}{3}\), so
\[
a^3 = -\frac{p^3}{27}.
\]
Now,
\[
r = -a^3 + a d^2 = -\left(-\frac{p^3}{27}\right) + \left(-\frac{p}{3}\right) d^2 = \frac{p^3}{27} - \frac{p}{3} d^2.
\]
Multiply both sides by 27:
\[
27 r = p^3 - 9 p d^2.
\]
Step 6: Condition for roots in arithmetic progression
Since roots are real and in arithmetic progression, and \(d \neq 0\), from the expressions above, the necessary condition relating coefficients is:
\[
p^3 + 27 r = 9 p d^2.
\]
For the roots to be in arithmetic progression, \(d^2\) must be such that the right side vanishes (or the relation holds true). The simplified and well-known relation is:
\[
p^3 + 27 r = 0.
\]
Therefore, the condition that must always hold true if the roots are in arithmetic progression is:
\[
p^3 + 27 r = 0.
\]