Insert arithmetic means between 16 and 65 such that the 5th AM is 51.

To find: The number of arithmetic means

Given: (i) The numbers are 16 and 65

(ii) $5^{\text {th }}$ arithmetic mean is 51

Formula used: (i) $d=\frac{b-a}{n+1}$, where, $d$ is the common difference

n is the number of arithmetic means

(ii) $A_{n}=a+n d$

We have 16 and 65,

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

$d=\frac{65-16}{n+1}$

$d=\frac{49}{n+1}$

Using Formula, $A_{n}=a+n d$

Fifth arithmetic mean, $A_{5}=a+5 d$

$=16+5\left(\frac{49}{n+1}\right)$

$A_{5}=16+\left(\frac{245}{n+1}\right)$

$\mathrm{A}_{5}=51$ (Given)

Therefore, $A_{5}=16+\left(\frac{245}{n+1}\right)=51$

$\Rightarrow 16+\left(\frac{245}{n+1}\right)=51$

$\Rightarrow\left(\frac{245}{n+1}\right)=51-16$

$\Rightarrow\left(\frac{245}{n+1}\right)=35$

$\Rightarrow 245=35 n+35$

$\Rightarrow 210=35 n$

$\Rightarrow n=6$

The number of arithmetic means are 6.  

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

$d=\frac{65-16}{6+1}$

$d=\frac{49}{7}$

$d=7$

Using Formula, $A_{n}=a+n d$

First arithmetic mean, $\mathrm{A}_{1}=\mathrm{a}+\mathrm{d}$

$=16+7$

$=23$

Second arithmetic mean, $\mathrm{A}_{2}=\mathrm{a}+2 \mathrm{~d}$

$=16+2(7)$

$=16+14$

$=30$

Third arithmetic mean, $A_{3}=a+3 d$

$=16+3(7)$

$=16+21$

= 37

Fourth arithmetic mean, $\mathrm{A}_{4}=\mathrm{a}+4 \mathrm{~d}$

$=16+4(7)$

$=16+28$

$=44$

Fifth arithmetic mean, $\mathrm{A}_{5}=\mathrm{a}+5 \mathrm{~d}$

$=16+5(7)$

$=16+35$

$=51$

Sixth arithmetic mean, $A_{6}=a+6 d$

$=16+6(7)$

$=16+42$

$=58$

Ans) The six arithmetic means between 1 and 65 are $23,30,37,44,51$ and $58 .$