Question

There are n arithmetic means between 3 and 17. The ratio of the last mean to the first mean is 3 : 1. Find the value of n.

Answer 1

Let , A1, A2,.......,An be nice A.M. between 3 and 17 and do be the common difference.

Answer 2

Answer:

$n=6$

Step-by-step explanation:

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

Then, the common difference is given by:

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

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

$d=\frac{14}{n+1}$

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

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

and $A_{n}=a+nd=3+\frac{14n}{n+1}$

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

Now, $\frac{A_{n}}{A_{1}}=\frac{3}{1}$

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

$17n+3=9n+51$

$n=6$