Understanding Harmonic Progression (HP) Basics

A Harmonic Progression (HP) is a sequence where the reciprocals of the terms form an Arithmetic Progression (AP).

Let the HP be denoted by $a_1, a_2, a_3, \dots, a_k, \dots$.

Then, the sequence of reciprocals $\frac{1}{a_1}, \frac{1}{a_2}, \frac{1}{a_3}, \dots, \frac{1}{a_k}, \dots$ forms an AP.

Let this corresponding AP be $b_1, b_2, b_3, \dots, b_k, \dots$, where $b_k = \frac{1}{a_k}$.

The general formula for the k-th term of an AP is $b_k = b_1 + (k-1)d$, where $b_1$ is the first term and $d$ is the common difference.

Using the Given Information for HP

The problem states:

• The m-th term of the HP is $n$. This means $a_m = n$.
• The n-th term of the HP is $m$. This means $a_n = m$.
• It's given that $m \ne n$.

Translating this into the corresponding AP:

• The m-th term of the AP is $b_m = \frac{1}{a_m} = \frac{1}{n}$.
• The n-th term of the AP is $b_n = \frac{1}{a_n} = \frac{1}{m}$.

Calculating the AP's Common Difference ($d$) and First Term ($b_1$)

We can set up two equations using the AP formula $b_k = b_1 + (k-1)d$:

1. For the m-th term: $b_m = b_1 + (m-1)d = \frac{1}{n}$
2. For the n-th term: $b_n = b_1 + (n-1)d = \frac{1}{m}$

To find the common difference $d$, subtract the second equation from the first:

$$ (b_1 + (m-1)d) - (b_1 + (n-1)d) = \frac{1}{n} - \frac{1}{m} $$

Simplify the left side:

$$ (m-1 - (n-1))d = \frac{1}{n} - \frac{1}{m} $$

$$ (m-1-n+1)d = \frac{m-n}{mn} $$

$$ (m-n)d = \frac{m-n}{mn} $$

Since we know $m \ne n$, we can divide both sides by $(m-n)$:

$$ d = \frac{1}{mn} $$

Now, substitute the value of $d$ back into the first AP equation ($b_1 + (m-1)d = \frac{1}{n}$) to find $b_1$:

$$ b_1 + (m-1)\left(\frac{1}{mn}\right) = \frac{1}{n} $$

$$ b_1 = \frac{1}{n} - \frac{m-1}{mn} $$

Find a common denominator ($mn$):

$$ b_1 = \frac{m}{mn} - \frac{m-1}{mn} $$

$$ b_1 = \frac{m - (m-1)}{mn} $$

$$ b_1 = \frac{m - m + 1}{mn} $$

$$ b_1 = \frac{1}{mn} $$

So, the first term of the AP is $b_1 = \frac{1}{mn}$ and the common difference is $d = \frac{1}{mn}$.

Determining the (m+n)th Term of the HP

We want to find the (m+n)th term of the HP, which is $a_{m+n}$.

This is equal to the reciprocal of the (m+n)th term of the AP, $b_{m+n}$.

Let's calculate $b_{m+n}$ using the AP formula $b_k = b_1 + (k-1)d$:

$$ b_{m+n} = b_1 + ((m+n)-1)d $$

Substitute the values $b_1 = \frac{1}{mn}$ and $d = \frac{1}{mn}$:

$$ b_{m+n} = \frac{1}{mn} + (m+n-1)\left(\frac{1}{mn}\right) $$

$$ b_{m+n} = \frac{1}{mn} + \frac{m+n-1}{mn} $$

Combine the terms over the common denominator:

$$ b_{m+n} = \frac{1 + (m+n-1)}{mn} $$

$$ b_{m+n} = \frac{1 + m + n - 1}{mn} $$

$$ b_{m+n} = \frac{m+n}{mn} $$

Finally, find the (m+n)th term of the HP ($a_{m+n}$) by taking the reciprocal:

$$ a_{m+n} = \frac{1}{b_{m+n}} $$

$$ a_{m+n} = \frac{1}{\frac{m+n}{mn}} $$

$$ a_{m+n} = \frac{mn}{m+n} $$

This result matches the second option provided.