**Question:**

There is n arithmetic means between 9 and 27. If the ratio of the last mean to the first mean is 2 : 1, find the value of n.

**Solution:**

To find: The value of n

Given: (i) The numbers are 9 and 27

Formula used: (i) $d=\frac{b-a}{n+1}$, where, $d$ is the common difference

n is the number of arithmetic means

(ii) $A_{n}=a+n d$

We have 9 and 27,

Using Formula, $d=\frac{b-a}{n+1}$

$d=\frac{27-9}{n+1}$

$d=\frac{18}{n+1}$

Using Formula, $A_{n}=a+n d$

First mean i.e., $A_{1}=9+$ (1) $\left(\frac{18}{n+1}\right)$

$=9+\frac{18}{n+1}$

$=\frac{9 n+9+18}{n+1}$

$A_{1}=\frac{9 n+27}{n+1}$

Last mean i.e., $A_{n}=9+(n)\left(\frac{18}{n+1}\right)$

$=9+\frac{18 n}{n+1}$

$=\frac{9 n+9+18 n}{n+1}$

$A_{n}=\frac{27 n+9}{n+1}$… (ii)

The ratio of the last mean to the first mean is 2 : 1

$\Rightarrow \frac{A_{n}}{A_{1}}=\frac{2}{1}$

Substituting the value of $A_{1}$ and $A_{n}$ from eqn. (i) and (ii)

$\Rightarrow \frac{\frac{27 n+9}{n+1}}{\frac{9 n+27}{n+1}}=\frac{2}{1}$

$\Rightarrow \frac{27 n+9}{9 n+27}=\frac{2}{1}$

$\Rightarrow 27 n+9=18 n+54$

$\Rightarrow 9 n=45$

$\Rightarrow n=5$

Ans) The value of n is 5