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
(a) 10
(b) 12
(c) 13
(d) 14

Answer 1

(d) 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:
$$\begin{gathered} \frac{A_1}{A_n}=\frac{3}{29} \\ \Rightarrow \frac{1+d}{1+nd}=\frac{3}{29} \\ \Rightarrow \frac{1+{\displaystyle \frac{30}{n+1}}}{1+n\times {\displaystyle \frac{30}{n+1}}}=\frac{3}{29} \\ \Rightarrow \frac{n+1+30}{n+1+30n}=\frac{3}{29} \\ \Rightarrow \frac{n+31}{31n+1}=\frac{3}{29} \\ \Rightarrow 29n+899=93n+3 \\ \Rightarrow 64n=896 \\ \Rightarrow n=14 \end{gathered}$$