Question

In n A.M.'s are introduced between 3 and 17 such that the ratio of the last mean to the first mean is 3 : 1, then the value of n is

Options

• 6

• 8

• 4

•  none of these.

Answer 1

6

Let 

\[A_1 , A_2 , A_3 , A_4 . . . . A_n\] be the n arithmetic means between 3 and 17.
Let d be the common difference of the A.P. 3,

\[A_1 , A_2 , A_3 , A_4 , . . . . A_n\] and 17.
Then, we have:

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

Now, 

\[A_1\] = 3 + d = 3 + \[\frac{14}{n + 1}\] = \[\frac{3n + 17}{n + 1}\]

And, 

\[A_n = 3 + nd = 3 + n\left( \frac{14}{n + 1} \right) = \frac{17n + 3}{n + 1}\]

\[\therefore \frac{A_n}{A_1} = \frac{3}{1}\]

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

\[ \Rightarrow \frac{17n + 3}{3n + 17} = \frac{3}{1}\]

\[ \Rightarrow 17n + 3 = 9n + 51\]

\[ \Rightarrow 8n = 48\]

\[ \Rightarrow n = 6\]