Suppose the given two numbers are $A$ and $B$. Let the $n$ arithmetic means be $a_1, a_2, a_3, a_4, a_5, \ldots \ldots \ldots, a_n$\nNow arithmetic progression is given as\n$A, a_1, a_2, \ldots \ldots \ldots . . a_n, B$. (For $n+2$ terms)\n$d($ common difference $)=\frac{B-A}{n+1}$\nas $B=A+(n+2-1) d$\nNow $p=$ first of $n$ arithmetic means $=a_1$\nSo $p=A+d=A+\frac{B-A}{n+1} \Rightarrow p=\frac{n A+B}{n+1}$\nFor harmonic progression, suppose $x_1, x_2, \ldots \ldots, x_n$ to be harmonic means, then harmonic progression for $n+2$ terms is given by\n$A, x_1, x_2, x_3, \ldots \ldots \ldots \ldots, x_n, B$\nwhich means $\frac{1}{A}, \frac{1}{x_1}, \frac{1}{x_2}, \ldots \ldots \ldots, \frac{1}{B}$ are in A.P.\nHere $\frac{1}{B}=\frac{1}{A}+(n+2-1) D$ or $D=\frac{A-B}{A B(n+1)}$\nand the $1^{\text {st }}$ term is,\n$\frac{1}{x_1}=\frac{1}{A}+D=\frac{1}{A}+\frac{A-B}{A B(n+1)}=\frac{n B+A}{(n+1) A B}$\nor $q=1^{\text {st }}$ harmonic mean $=\frac{(n+1) A B}{n B+A}$\nNow we have to prove that $q$ does not lies between $p$ and $\left(\frac{n+1}{n-1}\right)^2 p$\nWe know that $n+1>n-1$\nThus $\left(\frac{n+1}{n-1}\right)^2>1$ or $p<\left(\frac{n+1}{n-1}\right)^2 p$\nSo to prove the given, we have to show that $q$ is less than $p$. For this Let $\frac{p}{q}=\frac{(n A+B)(n B+A)}{(n+1)^2 A B}$\nThen $\frac{p}{q}-1=\frac{n\left(A^2+B^2\right)+A B\left(n^2+1\right)-(n+1)^2 A B}{(n+1)^2 A B}$\n\n\n \nHere R.H.S. is always greater than zero, which means L.H.S. is also greater than zero thus $\frac{p}{q}-1>0$ or $\frac{p}{q}>1$ or $p>q$\nHence proved that $q$ cannot lie between $p$ and $\left(\frac{n+1}{n-1}\right)^2 p$