7. Given a triangle with sides as three consecutive natural numbers, the largest angle is twice the smallest angle. Then the perimeter of the triangle is $\qquad$ .
Given $\triangle A B C$ with side lengths $A B=n+1, B C=n, A C=n-1$, the angle opposite side $A C$ is $\theta$, and the angle opposite side $A B$ is $2 \theta$.
According to the Law of Sines and the Law of Cosines, we have
$$
\begin{array}{l}
\frac{n-1}{\sin \theta}=\frac{n+1}{\sin 2 \theta} \Rightarrow \frac{n+1}{n-1}=2 \cos \theta, \\
\cos \theta=\frac{(n+1)^{2}+n^{2}-(n-1)^{2}}{2 n(n+1)} \\
=\frac{n+4}{2(n+1)} .
\end{array}
$$

From equations (1) and (2), we get
$$
\frac{n+1}{n-1}=\frac{n+4}{n+1} \Rightarrow n=5 \text {. }
$$

Therefore, the perimeter of the triangle is 15.