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.
Solution
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,
\[A_n = 3 + nd = 3 + n\left( \frac{14}{n + 1} \right) = \frac{17n + 3}{n + 1}\]
\[\therefore \frac{A_n}{A_1} = \frac{3}{1}\]
\[ \Rightarrow \frac{\left( \frac{17n + 3}{n + 1} \right)}{\left( \frac{3n + 17}{n + 1} \right)} = \frac{3}{1}\]
\[ \Rightarrow \frac{17n + 3}{3n + 17} = \frac{3}{1}\]
\[ \Rightarrow 17n + 3 = 9n + 51\]
\[ \Rightarrow 8n = 48\]
\[ \Rightarrow n = 6\]
