There are n A.M.s between 3 and 17. The ratio of the last mean to the first mean is 3 : 1.

Let $A_{1}, A_{2}, A_{3}, A_{4} \ldots A_{n}$ be the $n$ arithmetic means between 3 and 17 .

Let $d$ be the common difference of the A.P. $3, A_{1}, A_{2}, A_{3}, A_{4} \ldots A_{n}$ and17.

Then, we have:

$d=\frac{17-3}{n+1}=\frac{14}{n+1}$

Now, $A_{1}=3+d=3+\frac{14}{n+1}=\frac{3 n+17}{n+1}$

And, $A_{n}=3+n d=3+n\left(\frac{14}{n+1}\right)=\frac{17 n+3}{n+1}$

$\therefore \frac{A_{n}}{A_{1}}=\frac{3}{1}$

$\Rightarrow \frac{\left(\frac{17 n+3}{n+1}\right)}{\left(\frac{3 n+17}{n+1}\right)}=\frac{3}{1}$

$\Rightarrow \frac{17 n+3}{3 n+17}=\frac{3}{1}$

$\Rightarrow 17 n+3=9 n+51$

$\Rightarrow 8 n=48$

$\Rightarrow n=6$