If n arithmetic means are inserted between 1 and 31 such that the ratio of the first mean and nth mean is 3 : 29,

Solution:

(b) 12

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

There are $\mathrm{n}$ A.M.s between 1 and 31: $1, A_{1}, A_{2}, A_{3}, \ldots . A_{n}, 31$

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

Here, we have:

$\frac{A_{1}}{A_{n}}=\frac{3}{29}$

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

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

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

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

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

$\Rightarrow 64 n=896$

$\Rightarrow n=14$