n arithmetic means are inserted between 3 and 17. If the ratio of first to last mean is 1 : 3, then n is equal to:

Let the n arithmetic means between 3 and 

17 be $A_1, A_2,\dots , A_n$ If $d$ is the common difference, then,

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

Also, $\frac{A_1}{A_n}=\frac{1}{3}\Rightarrow \frac{3+d}{17-d}=\frac{1}{3}$

$\Rightarrow \frac{3+14/(n+1)}{17-14/(n+1)}=\frac{1}{3}\Rightarrow \frac{3n+17}{17n+3}=\frac{1}{3}\Rightarrow n=6$