Question

Between two quantities ‘a’ and ‘b’ are inserted ‘n’ arithmetic means and also ‘n’ harmonic means. If ‘x’ and ‘y’ be the rth terms of A.P and H.P, respectively, then prove that $\dfrac{x}{a}+\dfrac{b}{y}$ is independent of ‘r’ and ‘n’.

Answer 1

Hint: Use the formula ${{A}_{n}}$ and ${{H}_{n}}$ which equal to $\left( \dfrac{nb+a}{n+1} \right)$ and $\left( \dfrac{\left( n+1 \right)+ab}{na+b} \right)$ for quantities a, b which represents the nth arithmetic and harmonic mean. For getting the value of x, y substitute n=r-1 and n+1=r. Then find the values $\dfrac{x}{a}$ and $\dfrac{b}{y}$ to get the sum and find the desired result.

Complete step by step answer:
We are given two quantities a and b and in between them ‘n’ arithmetic and ‘n’ harmonic means are inserted.

We will use the formula that if two quantities are given let’s suppose ‘c’ and ‘d’ then nth arithmetic means will be represented by ${{A}_{n}}$ so,

${{A}_{n}}=\dfrac{nd+c}{n+1}$

Now for quantities a, b ${{A}_{n}}$ will be,

${{A}_{n}}=\dfrac{nb+a}{n+1}......(i)$

We will use the formula that if two quantities are given let’s suppose ‘c’ and ‘d’ then the nth harmonic mean will be represented by \[{{H}_{n}}\] so,

\[{{H}_{n}}=\dfrac{\left( n+1 \right)cd}{nc+d}\]

Now for quantities a, b ${{H}_{n}}$ will be,

\[{{H}_{n}}=\dfrac{\left( n+1 \right)ab}{na+b}..........(ii)\]

We are also given that x, y are rth terms of A.P and H.P respectively.

Let the general term be represented by ${{T}_{r}}$

So, ${{T}_{r}}$ of AP is ${{T}_{r-1}}$ (mean) of A.P = x and ${{T}_{r}}$ of HP is \[{{H}_{r-1}}\]

(mean) of HP = y.

Now multiplying n = r – 1 or n + 1 = r in equation (i) and (ii), we get,

$x=\dfrac{\left( r-1 \right)b+a}{r}$ and $y=\dfrac{rab}{\left( r-1 \right)a+b}$

Now we will find out what is $\left( \dfrac{x}{a} \right),$

$\dfrac{x}{a}=\dfrac{\left( r-1 \right)b+a}{ra}.........(iii)$

And now let’s find out what is $\left( \dfrac{y}{b} \right)$ ,

$\dfrac{y}{b}=\dfrac{rab}{\left( r-1 \right)a+b}\times \dfrac{1}{b}=\dfrac{ra}{\left( r-1 \right)a+b}$

So,

$\dfrac{b}{y}=\dfrac{1}{\left( \dfrac{y}{b} \right)}=\dfrac{\left( r-1 \right)a+b}{ra}..........(iv)$
Now we will get sum of equation (iii) and (iv), we get
$\begin{align}

  & \dfrac{x}{a}+\dfrac{b}{y}=\dfrac{\left( r-1 \right)b+a}{ra}+\dfrac{b+\left( r-1 \right)a}{ra} \\
 & \dfrac{x}{a}+\dfrac{b}{y}=\dfrac{rb-b+a+b+ra-a}{ra} \\

\end{align}$

On simplifying we get,

$\dfrac{x}{a}+\dfrac{b}{y}=\dfrac{rb+ra}{ra}=\dfrac{r\left( a+b \right)}{ra}$

So we can write,

$\dfrac{x}{a}+\dfrac{b}{y}=\dfrac{a+b}{a}$

So, the sum of $\left( \dfrac{x}{a}+\dfrac{b}{y} \right)$ is independent of ‘r’ or ‘n’.


Note: Students should use the formula ${{A}_{n}}$ and ${{H}_{n}}$ which represent the nth term of A.M and H.M. They should be also careful while finding values of $\dfrac{x}{a}$ and $\dfrac{b}{y}$ to get the desired results.

Students generally make mistake in taking the formula ${{A}_{n}}=\dfrac{nd+c}{n+1}$ and \[{{H}_{n}}=\dfrac{\left( n+1 \right)cd}{nc+d}\].