Question

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

Answer 1

Let $A_1,A_2,A_3,...A_n$ be $n$ arithmetic means between $20$ and $80$

And let $d$ be the common difference between the terms of the A.P.

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

$=\frac{80-20}{n+1}=\frac{60}{n+1}$

Now, $A_1=20+d$

$=20+\frac{60}{n+1}=20\left(\frac{n+4}{n+1}\right)$

And $A_n=20+nd=20+\frac{60n}{n+1}=20\left(\frac{4n+1}{n+1}\right)$

Thus, $\frac{A_1}{A_n}=\frac{1}{3}$(given)

$\frac{20+d}{20+nd}=\frac{1}{3}$

$\frac{20\left(\frac{n+4}{n+1}\right)}{20\left(\frac{4n+1}{n+1}\right)}=\frac{1}{3}$

$\Rightarrow \frac{n+4}{4n+1}=\frac{1}{3}$

$\Rightarrow 4n+1=3n+12$

$\Rightarrow n=11$

$\therefore$ The value of $n$ is $11$