For 10% of marks, we can try every possible arrangement of French words and English words, making sure that no three consecutive words are in French. It takes $\mathcal{O}(N)$ time to check one arrangement, and there are $\mathcal{O}(2^N)$ such arrangements, for an $\mathcal{O}(N{2}^N)$ algorithm.

For full marks, we will employ a dynamic programming approach. Let $f(n,k)$ be the maximum mark attainable given that the first $n$ words have had their language determined, and the longest suffix of French words currently determined has length $k$. $f(n+1,0)$ is therefore the maximum of $f(n,0)$, $f(n,1)$ and $f(n,2)$, incremented by the mark of word $n+1$ in English. $f(n+1,1)=f(n,0)+a_{n+1}$, and $f(n+1,2)=f(n,1)+a_{n+1}$. The answer is therefore the maximum of $f(N,0)$, $f(N,1)$, and $f(N,2)$.