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Question
If n arithmetic means are inserted between 1 and 31 such that the ratio of the first mean and nth mean is 3 : 29, then the value of n is
Options
10
12
13
14
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Solution
14
The given series is 1, . . . . . . . . . . . , 31
There are n A.M.s between 1 and 31:
\[1, A_1 , A_2 , A_3 , . . . . . , A_n , 31\]
Common difference, d = \[\frac{31 - 1}{n + 1} = \frac{30}{n + 1}\]
Here, we have:
\[\frac{A_1}{A_n} = \frac{3}{29}\]
\[ \Rightarrow \frac{1 + d}{1 + nd} = \frac{3}{29}\]
\[ \Rightarrow \frac{1 + \frac{30}{n + 1}}{1 + n \times \frac{30}{n + 1}} = \frac{3}{29}\]
\[ \Rightarrow \frac{n + 1 + 30}{n + 1 + 30n} = \frac{3}{29}\]
\[ \Rightarrow \frac{n + 31}{31n + 1} = \frac{3}{29}\]
\[ \Rightarrow 29n + 899 = 93n + 3\]
\[ \Rightarrow 64n = 896\]
\[ \Rightarrow n = 14\]
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