Question

If $n$ arithmetic mean's are inserted between $20$ and $80$, such that the ratio of first mean to the last mean is $1:3$, then the value of $n$ is

Answer 1

Given : $20,A_1,A_2,\dots ,A_n,80$ are in A.P.
Now,
$d=\frac{80-20}{n+1}=\frac{60}{n+1}$
From the given condition, we get
$\frac{A_1}{A_n}=\frac{1}{3}\Rightarrow \frac{20+d}{80-d}=\frac{1}{3}\Rightarrow 60+3d=80-d\Rightarrow d=5\Rightarrow \frac{60}{n+1}=5\therefore n=11$