Question

There are $n$ A.M's between $3$ and $54$, such that the $8^{th}$ mean: $(n-2{)}^{th}$ mean $::3:5$. The value of $n$ is.

Answer 1

$\Rightarrow$ $3,A_1,A_2,A_3......A_n,54$ are in $AP$

$\Rightarrow$ $d=\frac{b-a}{n+1}=\frac{54-3}{n+1}$

$\therefore$ $d=\frac{51}{n+1}$

$\Rightarrow$ $A_8=a+8d$

$=3+8\left(\frac{51}{n+1}\right)$

$=\frac{3n+3+408}{n+1}$

$=\frac{3n+411}{n+1}$

$\Rightarrow$ $A_{n-2}=a+(n-2)d$

$=3+(n-2)\frac{51}{n+1}$

$=\frac{3n+3+51n-102}{n+1}$

$=\frac{54n-99}{n+1}$

$\Rightarrow$ Now, $\frac{A_8}{A_{n-2}}=\frac{\frac{3n+411}{n+1}}{\frac{54n-99}{n+1}}$

$\Rightarrow$ $\frac{3}{5}=\frac{3n+411}{54n-99}$

$\therefore$ $15n+2055=162n-297$

$\therefore$ $2352=147n$

$\therefore$ $n=16$