First, expand \((a + x)^5\) using the binomial theorem:

\((a + x)^5 = a^5 + 5a^4x + 10a^3x^2 + \ldots\)

We are interested in the terms that will contribute to \(x^2\) in the product \(\left( 1 - \frac{2x}{a} \right)(a + x)^5\).

Consider the term \(-\frac{2x}{a} \cdot a^5\) which contributes \(-\frac{2a^5x}{a} = -2a^4x\).

Now consider the term \(1 \cdot 10a^3x^2 = 10a^3x^2\).

The coefficient of \(x^2\) is given by:

\(-\frac{2}{a} \times 5a^4 + 10a^3 = -10a^3 + 10a^3 = 0\).

Thus, the coefficient of \(x^2\) is zero.