Question

The n arithmetic means between 20 and 80 are such that the first mean : last mean = 1 : 3. Find the value of n.

Answer 1

Let $A_1,A_2,\cdots A_n$ be n AMs between 20 and 80.

Then, 20, $A_1,A_2,\cdots ,A_n$, 80 are in AP.

Let their common difference be d. Then,

$T_{n+2}=80\Rightarrow 20+(n+2-1)d=80\Rightarrow d=\frac{60}{(n+1)}$ . . . (i)

First mean, $A_1=T_2=(20+d)=(20+\frac{60}{n+1})$ [using (i)]

$=\frac{(20n+80)}{(n+1)}$

Last mean, $A_n=T_{n+1}=[20+(n+1-1\left)d\right]$

$=\{20+\frac{n\times 60}{(n+1)}\}$ [using (i)]

$=\frac{(80n+20)}{(n+1)}$

$\therefore \frac{A_1}{A_n}=\frac{1}{3}\Rightarrow \frac{(20n+80)}{(n+1)}\times \frac{(n+1)}{(80n+20)}=\frac{1}{3}$

$\Rightarrow 3(20n+80)=(80n+20)\Rightarrow 20n=220\Rightarrow n=11$

Hence, n = 11