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:

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:

A
4
B
5
C
6
D
8

Solution:

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, $\frac{A_{1}}{A_{n}} = \frac{1}{3} \Rightarrow \frac{3 + d}{17 - d} = \frac{1}{3}$

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

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