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By definition, $\left(\genfrac{}{}{0ex}{}{n}{k}\right)=\frac{n!}{k!(n-k)!}$. The ratio of the first two terms give us that$$\begin{array}{rl}\frac{1}{2}& =\frac{\frac{n!}{k!(n-k)!}}{\frac{n!}{(k+1)!(n-k-1)!}}=\frac{k+1}{n-k}\\ 2& =n-3k\end{array}$$The ratio of the second and third terms give us that$$\begin{array}{rl}\frac{2}{3}& =\frac{\frac{n!}{(k+1)!(n-k-1)!}}{\frac{n!}{(k+2)!(n-k-2)!}}=\frac{k+2}{n-k-1}\\ 8& =2n-5k\end{array}$$This is a linear system of two equations with two unknowns, indicating that there is a unique solution. Solving by substitution or multiplying the top equation and subtracting, we find $k=4,n=14$. Thus, $n+k=\boxed{18}$.