Question

If $n$arithmetic means 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

Step $1:$ Solve for the value of $d.$

Given,

The first term is $a=20$ and the last term is $l=80$

Number of A.M.'s $=n$

Total no. of terms in A.P $=$no. of means$+$ first term$+$ last term

$N=n+2$

Let common difference$=d,$ $l=a+(N-1)d$

$$\begin{gathered} 80=20+(n+2-1)d \\ 80-20=(n+1)d \\ 60=(n+1)d \end{gathered}$$

$d$ $=\frac{60}{n+1}...\left(1\right)$

First mean$=a+d=20+d$

Last mean $=a+nd=20+nd$

The ratio of first mean to the last mean is $1:3$(given),i.e.

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

$3(20+d)$$=20+nd$

$60+3d$ $=20+nd$

$40$ $=nd-3d$

$40$ $=(n-3)d$

$d$$=\frac{40}{n-3}...\left(2\right)$

Step $2:$ Solve for the value of $n.$

Equating $\left(1\right)$ and $\left(2\right)$

$\frac{60}{n+1}$ $=\frac{40}{n-3}$

$60(n-3)$ $=40(n+1)$

$60n-180$$=40n+40$

$20n$ $=220$

$n$ $=11$

Hence, the value of $n=11$,i.e. $11$ arithmetic means are inserted between $20$ and $80.$