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

विकल्प

• 10

• 12

• 13

• 14

Answer 1

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:

\[\frac{A_1}{A_n} = \frac{3}{29}\]

\[ \Rightarrow \frac{1 + d}{1 + nd} = \frac{3}{29}\]

\[ \Rightarrow \frac{1 + \frac{30}{n + 1}}{1 + n \times \frac{30}{n + 1}} = \frac{3}{29}\]

\[ \Rightarrow \frac{n + 1 + 30}{n + 1 + 30n} = \frac{3}{29}\]

\[ \Rightarrow \frac{n + 31}{31n + 1} = \frac{3}{29}\]

\[ \Rightarrow 29n + 899 = 93n + 3\]

\[ \Rightarrow 64n = 896\]

\[ \Rightarrow n = 14\]