To find the coefficients of x and x² in the expansion of (2 + ax)⁷, we use the binomial theorem.

The general term in the expansion is given by:

\(\binom{7}{r} (2)^{7-r} (ax)^r\)

\(For the coefficient of x, set r = 1:\)

\(\binom{7}{1} (2)^6 (ax) = 7 \times 2^6 \times a \times x\)

The coefficient is \(7 \times 2^6 \times a\).

\(For the coefficient of x², set r = 2:\)

\(\binom{7}{2} (2)^5 (ax)^2 = 21 \times 2^5 \times a^2 \times x^2\)

The coefficient is \(21 \times 2^5 \times a^2\).

Since the coefficients are equal, we have:

\(7 \times 2^6 \times a = 21 \times 2^5 \times a^2\)

Simplifying gives:

\(7 \times 2 = 21 \times a\)

\(a = \frac{2}{3}\)