jmc

By the Binomial Theorem, the third, fourth, and fifth terms in the expansion of $(x+a{)}^n$ are $\left(\genfrac{}{}{0ex}{}{n}{2}\right)x^{n-2}{a}^2,$ $\left(\genfrac{}{}{0ex}{}{n}{3}\right)x^{n-3}{a}^3,$ and $\left(\genfrac{}{}{0ex}{}{n}{4}\right)x^{n-4}{a}^4,$ respectively. Then $$\frac{\left(\genfrac{}{}{0ex}{}{n}{2}\right)x^{n-2}{a}^2}{\left(\genfrac{}{}{0ex}{}{n}{3}\right)x^{n-3}{a}^3}=\frac{84}{280}.$$This simplifies to $$\frac{3x}{a(n-2)}=\frac{3}{10},$$so $10x=a(n-2).$

Also, $$\frac{\left(\genfrac{}{}{0ex}{}{n}{3}\right)x^{n-3}{a}^3}{\left(\genfrac{}{}{0ex}{}{n}{4}\right)x^{n-4}{a}^4}=\frac{280}{560}.$$This simplifies to $$\frac{4x}{a(n-3)}=\frac{1}{2},$$so $8x=a(n-3).$

Dividing the equations $10x=a(n-2)$ and $8x=a(n-3),$ we get $$\frac{n-3}{n-2}=\frac{4}{5}.$$Then $5n-15=4n-8,$ so $n=\boxed{7}.$