Question

If $p$ be the first of $n$ arithmetic means between two numbers, and $q$ the first of $n$ harmonic means between the same two numbers,then $q$ cannot lie between $p$ and ${\left(\frac{n+1}{n-1}\right)}^{k}p$. Find $k$

Answer 1

$p=\frac{na+b}{n+1}\text{and}q=\frac{ab(n+1)}{a+nb}$

By eliminating $b,$ we get the equation

$n{a}^2-\left\{\right(n+1)p+(n+1\left)q\right\}a+npq=0$

For real roots of the above equation, we must have

$\left\{\right(n+1)p+(n+1)q{\}}^2-4n^{2}pq>0⟹(n+1{)}^2{p}^2-2pq(n^2+1)+(n-1{)}^2{q}^2>0$

$\Rightarrow \left\{\right(n+1{)}^{2}p-(n-1{)}^{2}q\}(p-q)>0⟹[{\left(\frac{n+1}{n-1}\right)}^{2}p-q](p-q)>0$

$\Rightarrow q<{\left(\frac{n+1}{n-1}\right)}^{2}p\text{and}q<p$

Hence, proved.