Given a positve integer n, we can find infinitely many positve integers $b$ such that the $n-1$ integers in the set $\{b+1,\,2b+1,\,3b+1,\,…,\,(n-1)b+1\}$ are pairwise relatively prime.
I assume that $b+1,\,2b+1,\,3b+1,\,…,\,(n-1)b+1$ are not r.p..
Let $1\le i<j\le n-1$ and $ib+1,jb+1$ are not r.p..
Let $p$ be a prime which is a factor of both $ib+1,\,jb+1$.
Question: Why should $p$ now $p\ge n-1$ ?
Proof finish:
$$
\begin{align*}
&p\mid (jb+1)-(ib+1)=(j-i)b\\
\Rightarrow& p\mid j-i\\
\Rightarrow& j-i<n-1\\
\Rightarrow& p<n-1,
\end{align*}
$$
which is a contradiction.
Best Answer
Well, you can choose $b$, so take $$b=c\prod_{p \text{prime} \atop p \leq n-2}{p}$$ where $c$ is any positive integer. This gives infinitely many $b$. The rest follows from what you have done, as the prime factor cannot divide $b$, it must be $\geq n-1$.