If The $p^{\text{th}}$ Term Of An A.P. Is $\frac{1}{q}$ And The $q^{\text{th}}$ Term Is $\frac{1}{p}$, Show That The Sum Of The First $pq$ Terms Is $\frac{1}{2}(pq + 1$\].

Feb 26, 2025 by ADMIN 172 views

If the $p^{\text{th}}$ term of an A.P. is $\frac{1}{q}$ and the $q^{\text{th}}$ term is $\frac{1}{p}$ , show that the sum of the first $pq$ terms is $\frac{1}{2}(pq + 1)$

In this article, we will explore the concept of an arithmetic progression (A.P.) and use it to derive a formula for the sum of the first $pq$ terms. We will start by defining the $p^{\text{th}}$ and $q^{\text{th}}$ terms of an A.P. and then use this information to find the sum of the first $pq$ terms.

An arithmetic progression is a sequence of numbers in which the difference between any two consecutive terms is constant. This constant difference is called the common difference. The general form of an A.P. is given by:

a, a + d, a + 2d, a + 3d, \ldots

where $a$ is the first term and $d$ is the common difference.

We are given that the $p^{\text{th}}$ term of an A.P. is $\frac{1}{q}$ and the $q^{\text{th}}$ term is $\frac{1}{p}$ . Using the general form of an A.P., we can write:

a + (p - 1)d = \frac{1}{q} \tag{1}

a + (q - 1)d = \frac{1}{p} \tag{2}

To find the sum of the first $pq$ terms, we can use the formula for the sum of an A.P.:

S_n = \frac{n}{2}(2a + (n - 1)d)

where $S_n$ is the sum of the first $n$ terms, $a$ is the first term, and $d$ is the common difference.

We want to find the sum of the first $pq$ terms, so we will substitute $n = pq$ into the formula:

S_{pq} = \frac{pq}{2}(2a + (pq - 1)d)

To find the value of $a + (pq - 1)d$ , we can use the given information to find the value of $a$ and $d$ .

We can solve equations (1) and (2) simultaneously to find the values of $a$ and $d$ . Subtracting equation (2) from equation (1), we get:

(p - 1)d - (q - 1)d = \frac{1}{q} - \frac{1}{p}

Simplifying, we get:

(p - q)d = \frac{p - q}{pq}

Dividing both sides by $p - q$ , we get:

d = \frac{1}{pq}

Substituting this value of $d$ into equation (1), we get:

a + (p - 1)\frac{1}{pq} = \frac{1}{q}

Simplifying, we get:

a = \frac{1}{q} - \frac{p - 1}{pq}

a = \frac{1}{q} - \frac{1}{q} + \frac{1}{pq}

a = \frac{1}{pq}

Now that we have found the values of $a$ and $d$ , we can substitute them into the formula for the sum of the first $pq$ terms:

S_{pq} = \frac{pq}{2}(2\frac{1}{pq} + (pq - 1)\frac{1}{pq})

Simplifying, we get:

S_{pq} = \frac{pq}{2}(\frac{2}{pq} + \frac{pq - 1}{pq})

S_{pq} = \frac{pq}{2}(\frac{2 + pq - 1}{pq})

S_{pq} = \frac{pq}{2}(\frac{pq + 1}{pq})

S_{pq} = \frac{1}{2}(pq + 1)

In this article, we have shown that the sum of the first $pq$ terms of an A.P. is $\frac{1}{2}(pq + 1)$ , given that the $p^{\text{th}}$ term is $\frac{1}{q}$ and the $q^{\text{th}}$ term is $\frac{1}{p}$ . We have used the general form of an A.P. and the formula for the sum of an A.P. to derive this result.
Q&A: If the $p^{\text{th}}$ term of an A.P. is $\frac{1}{q}$ and the $q^{\text{th}}$ term is $\frac{1}{p}$ , show that the sum of the first $pq$ terms is $\frac{1}{2}(pq + 1)$

A: An arithmetic progression is a sequence of numbers in which the difference between any two consecutive terms is constant. This constant difference is called the common difference.

A: The general form of an A.P. is given by:

a, a + d, a + 2d, a + 3d, \ldots

where $a$ is the first term and $d$ is the common difference.

A: The $p^{\text{th}}$ term of an A.P. is given by:

a + (p - 1)d

A: The $q^{\text{th}}$ term of an A.P. is given by:

a + (q - 1)d

A: We can use the formula for the sum of an A.P.:

S_n = \frac{n}{2}(2a + (n - 1)d)

where $S_n$ is the sum of the first $n$ terms, $a$ is the first term, and $d$ is the common difference.

A: We can solve equations (1) and (2) simultaneously to find the values of $a$ and $d$ . Subtracting equation (2) from equation (1), we get:

(p - 1)d - (q - 1)d = \frac{1}{q} - \frac{1}{p}

Simplifying, we get:

(p - q)d = \frac{p - q}{pq}

Dividing both sides by $p - q$ , we get:

d = \frac{1}{pq}

Substituting this value of $d$ into equation (1), we get:

a + (p - 1)\frac{1}{pq} = \frac{1}{q}

Simplifying, we get:

a = \frac{1}{q} - \frac{p - 1}{pq}

a = \frac{1}{q} - \frac{1}{q} + \frac{1}{pq}

a = \frac{1}{pq}

A: The sum of the first $pq$ terms of an A.P. is given by:

S_{pq} = \frac{pq}{2}(2\frac{1}{pq} + (pq - 1)\frac{1}{pq})

Simplifying, we get:

S_{pq} = \frac{pq}{2}(\frac{2}{pq} + \frac{pq - 1}{pq})

S_{pq} = \frac{pq}{2}(\frac{2 + pq - 1}{pq})

S_{pq} = \frac{pq}{2}(\frac{pq + 1}{pq})

S_{pq} = \frac{1}{2}(pq + 1)

A: This result shows that the sum of the first $pq$ terms of an A.P. is $\frac{1}{2}(pq + 1)$ , given that the $p^{\text{th}}$ term is $\frac{1}{q}$ and the $q^{\text{th}}$ term is $\frac{1}{p}$ . This result can be used to find the sum of the first $pq$ terms of an A.P. in various mathematical and real-world applications.