Question

Let $n\ge 1$ be a positive integer. Suppose, between two distinct positive real numbers $k$ and $l,n$ arithmetic means $a_1,a_2,...,a_n$ and $n$ harmonic means $h_1,h_2,....,h_n$ are inserted. For $i=1,2,....,n$ and $j=1,2,....,n$, let $p(i,j)=a_i{h}_j$ then

• A. $p(i,j)=kl$ for some $i$ and $j$
• B. If $n$ is even, there are an even number of ordered pairs $(i,j)$, such that $p(i,j)=kl$
• C. If $n$ is odd, there are an even number of ordered pairs $(i,j)$, such that $p(i,j)=kl$
• D. Whenever $n$ is odd there is an $i$ such that $p(i,i)=kl$

Answer 1

Answer: (A), (B), (D)

The correct options are
A If $n$ is even, there are an even number of ordered pairs $(i,j)$, such that $p(i,j)=kl$
B $p(i,j)=kl$ for some $i$ and $j$
D Whenever $n$ is odd there is an $i$ such that $p(i,i)=kl$

If between two distinct positive real numbers $a$ and $b,$ $n$ arithmetic means $a_1,a_2........a_n$ are inserted.

Then we can write as: $\Rightarrow k,a_1,a_2,a_3,.........a_n,l$, .....$\left(1\right)$

Where all the numbers collectively form an $A.P$.

For an $A.P.,$ the last term $l=a+(n-1)d,$ where $a$ is the first term, $n$ is total number of terms and $d$ is the common difference between terms.

For the given Arithmetic progression, the first term is $k,$ the total number of terms are $n+2$ and the last term is $l$.

So we can write, $l=k+(n+2-1)d_A$

$\to d_A=\frac{l-k}{n+1}$

Hence from the given $A.P.,$ we can write any general term as $\Rightarrow a_i=k+(i-1)d_A$

$\Rightarrow a_i=k+\frac{(i-1)(l-k)}{n+1}$

$\Rightarrow a_i=\frac{k(n+1)+(i-1)(l-k)}{n+1}$

Now if $n$ harmonic means are inserted between $k$ and $l$, we can write as: $\Rightarrow \frac{1}{k},\frac{1}{h_1},\frac{1}{h_2},....\frac{1}{h_n},\frac{1}{l}$ ...$\left(2\right)$

As We know that by inverting the term of a harmonic mean, the new set of terms becomes an Arithmetic progression. Hence the term in eq.$\left(2\right)$ are now in arithmetic progression.

For an $A.P.,$ the last term $l=a+(n-1)d,$ where $a$ is the first term, $n$ is total number of terms and $d$ is the common difference between terms.

Hence $\frac{1}{l}=\frac{1}{k}+(n+2-1)d_H$

$\Rightarrow d_H=\frac{k-l}{kl(n+1)}$

So we can write that $\frac{1}{h_j}=\frac{1}{k}+(j-1)d_H$

$\Rightarrow \frac{1}{h_j}=\frac{1}{k}+\frac{(j-1)(k-l)}{kl(n+1)}$

$\Rightarrow \frac{1}{h_j}=\frac{l(n+1)+(j-1)(k-l)}{kl(n+1)}$

$\Rightarrow h_j=$$\frac{kl(n+1)}{l(n+1)+(j-1)(k-l)}$

Now $P(i,j)=a_i,h_j$

$\Rightarrow P(i,j)=$$\frac{k(n+1)+(i-1)(l-k)}{l(n+1)+(j-1)(k-l)}$ $.kl$

So if $P(i,j)=kl$,

Then $k(n+1)+(i-1)(l-k)=l(n+1)+(j-1)(k-l)$

$\Rightarrow (l-k)(i+j-2)=(l-k)(n+1)$

$\Rightarrow i+j=n+3$ ....$\left(3\right)$

Hence we can say that $P(i,j)=kl$ for some $(i,j),$ related with the equation $i+j=n+3$

$\to$ Now if $n$ is even in eq.$\left(3\right)$, then $i+j=even+odd=odd$

As $i,j$ can be maximum at $(n+2)$, because there are only $(n+2)$ terms. $n$ is even, so is $n+2$, So there will be even number of ordered pairs of $(i,j),$ such that $P(i,j)$

$\to$ Now if $n$ is odd then $i+j=odd+odd=even=n+3$

As $i,j$ can be maximum at $(n+2)$, because there are only $(n+2)$ terms. $n$ is odd, so is $(n+2)$, So there will be odd number of ordered pairs of $(i,j),$ such that $P(i,j)$

`So when $n$ is odd, then $n+3$ is even, and in that case there will lie $i=\frac{n+3}{2},j=\frac{n+3}{2}$

So id $n$ is odd, then at some point where $i=j$. we can say that $P(i,i)=kl$.

Hence correct options are $A,B$ and $C$.