To solve this problem, we need to find the values of \(a\) and \(n\) in the binomial expansion of \((1 + ax)^n\) which are consistent with the given first three terms \(1\), \(6x\), and \(16x^2\).
The general term in the expansion of \((1 + ax)^n\) is given by:
\(T_k = \binom{n}{k-1} (ax)^{k-1}\)
So, the first term (\(T_1\)) is:
\(T_1 = \binom{n}{0} (ax)^0 = 1\)
The second term (\(T_2\)) is:
\(T_2 = \binom{n}{1} (ax) = nax\)
It is given that \(T_2 = 6x\), which implies:
\(n \cdot a = 6 \quad \text{(Equation 1)}\)
The third term (\(T_3\)) is:
\(T_3 = \binom{n}{2} (ax)^2 = \frac{n(n-1)}{2} a^2 x^2\)
It is also given that \(T_3 = 16x^2\), which implies:
\(\frac{n(n-1)}{2} a^2 = 16 \quad \text{(Equation 2)}\)
We now have two equations:
From Equation 1, express \(a\) in terms of \(n\):
\(a = \frac{6}{n}\)
Substitute \(a = \frac{6}{n}\) into Equation 2:
\(\frac{n(n-1)}{2} \left(\frac{6}{n}\right)^2 = 16\)
Simplify this equation:
\(\frac{n(n-1)}{2} \cdot \frac{36}{n^2} = 16\)
\(\frac{36(n-1)}{2n} = 16\)
\(18(n-1) = 16n\)
\(18n - 18 = 16n\)
\(2n = 18\)
\(n = 9\)
Using \(n = 9\) in \(a = \frac{6}{n}\):
\(a = \frac{6}{9} = \frac{2}{3}\)
Thus, the values of \(a\) and \(n\) are \(\frac{2}{3}\) and 9, respectively.
Therefore, the correct answer is: \(\frac{2}{3}\) and 9.