To find $n$ such that the expansion of $(1+x)^n$ has three consecutive coefficients $p,q,r$ that satisfy $$p:q:r = 1:7:35$$
My work:
Suppose the consecutive coefficients are $\binom{n}{k-1}, \binom{n}{k}$ and $\binom{n}{k+1}$. Then we have
$$\dfrac{\binom{n}{k-1}}{\binom{n}{k}} = \frac{k}{n-k+1}= \frac17$$ and
$$\dfrac{\binom{n}{k}}{\binom{n}{k+1}} = \frac{k+1}{n-k}= \frac{7}{35}= \frac15$$
Cross multiplying we have $7k = n-k+1 \implies n = 8k-1$ and $5k+5=n-k \implies n = 6k+5$.
Thus $8k-1=6k+5 \implies k = 3$ and hence $n = 23$.
Is my argument fine?