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

The correct option is D

14

The given series is 1 ...... 31

There are n A.M. between 1 and 31 : 1, $A_1,A_2,A_3,\dots \dots ,A_n,3]$

Common difference, $d=\frac{31-1}{n+1}=\frac{30}{n+1}$

Here, we have:

$\frac{A_1}{A_n}=\frac{3}{29}$

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

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

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

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

$\Rightarrow 29n+899=93n+3$

$\Rightarrow 64n=896$

$\Rightarrow n=14$