Question

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

Answer 1

Formulae,

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

$A_1=a+d=3+\frac{14}{n+1}=\frac{3n+17}{n+1}$

$A_n=a+nd=3+\frac{14n}{n+1}=\frac{17n+3}{n+1}$

$\frac{A_n}{A_1}=\frac{3}{1}$

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

$8n=48\Rightarrow n=6$