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.
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