Proof of the formulas of arithmetic progression

General term of an A.P.

Let us express the progression in its algebraic form, where \(a_n\) denotes the nth term and \(d\) represents the common difference. The progression can be represented as:

\( a_1, a_1 + d, a_1 + 2d, a_1 + 3d, a_1 + 4d, \cdots \)

It should be evident that the general term of an arithmetic progression is defined as follows.

\( a_n = a_1 + (n - 1)d \)

Allow it some time; the pattern in the progression will become apparent upon closer examination again and again.

Sum of first n terms of an A.P.

This document aims to present two proofs regarding the subject matter, akin to derivations. However, before that, it is essential to recount the origin of this initiative.

We consider the problem given to Carl Friedrich Gauss to solve when he was just 10 years old. He was asked to find the sum of the positive integers from 1 to 100. He immediately replied that the sum is 5050. How did he do that?

He wrote :

\( S = 1 + 2 + 3 + \cdots + 99 + 100 \)

And then, reversed the numbers to write

\( S = 100 + 99 + 98 + \cdots + 2 + 1 \)

He observed that adding these two progression term-wise results in making every term equal to 101, such as \(1 + 100, 2 + 99, 3 + 98,\cdots\). Thus, by adding them term-wise, he obtained:

\( 2S = 101 + 101 + 101 + \cdots + 101 = 101 \times 100 \)

So, the sum is:

\( S = \frac{101 \cdot 100}{2} = 5050 \)

We are going to use this very method to find a general solution for the sum of the first n terms of an Arithmetic progression.

Let us say \(S_n\) is the sum of the first n terms of an Arithmetic progression,

\(S_n=a_1+(a_1+d)+(a_1+2d)+(a_1+3d)+\cdots +(a_1+(n-1)d)\)

or,

\( S_n=(a_1+(n-1)d)+(a_1+(n-2)d)+(a_1+(n-3)d)+ \cdots +a_1 \)

Adding these two term-wise wise he got,

\( 2S_n=(2a_1+(n-1)d)+(2a_1+(n-1)d)+(2a_1+(n-1)d)+ \cdots n \:times \)
\( \implies 2S_n=n\{2a_1+(n-1)d\} \)

Dividing both sides by 2 gives us our final equation,

\( S_n=\frac{n}{2}\{2a_1+(n-1)d\} \)

Second Proof of the sum of the first n terms of an A.P.

We will now illustrate this concept through the use of summation. For individuals who may not be acquainted with the principles of summation, we encourage you to bypass this section and return to the discussion on geometric progression. You may revisit this topic after gaining familiarity with the forthcoming sections.

\[ \begin{aligned} & \quad S_n = \sum_{i=1}^{n}a_i \\ \implies & \quad S_n=\sum_{i=1}^{n} a_1+(i-1)d = \sum_{i=1}^{n} a_1 + \sum_{i=1}^{n}(i-1)d \\ \implies & \quad S_n= na_1 + d \sum_{i=1}^{n-1}i = na_1 + \frac{n(n-1)}{2}d \end{aligned} \]
\( \Rightarrow S_n = \frac{n}{2} \{2a_1 + (n - 1)d\} \)
Validity of Arithmetic Progression
It is important to note that the terms of an arithmetic progression are not restricted solely to integers. Rather, both the terms and the common difference can encompass any real or complex number. This flexibility allows for a broader application of arithmetic progressions beyond traditional numerical constraints.