G1.1 Given that $n$ and $k$ are natural numbers and $1<k<n$. If $\frac{(1+2+3+\cdots+n)-k}{n-1}=10$ and $n+k=a$, find the value of $a$.
$$
\frac{n(n+1)}{2}-k=10 n-10 \Rightarrow n^{2}-19 n+2(10-k)=0
$$
$\Delta=281+8 k, n$ is an integer $\Rightarrow \Delta$ is a perfect square.
$$
281+8 k=289,361,441, \ldots \Rightarrow k=1,10,20, \ldots \text { Given } 1<k<n, \therefore k=10,20, \ldots
$$
when $k=10, n=19 ; a=n+k=29$; when $k=20, n=20$ rejected.