Arithmetic Sequence and Arithmetic Series*

An idea to solve a particular problem in mathematics may play a crucial role in solving more general problems in mathematics. The present article clearly illustrate this point. In this article, we will see how an idea to solve a particular problem related to a particular arithmetic sequence may be used to solve a general problem related to general arithmetic sequence.

 In mathematics, we often encounter the following type of sequences:

\[1, 2, 3, 4, 5, 6, \ldots\]

\[1, 3, 5, 7, 9, 11, \ldots\]

\[2, 4, 6, 8, 10, 12, \ldots\]

\[2, 5, 8, 11, 14, 17, \ldots\]

\[\pi, \pi + 2, \pi + 4, \pi + 6, \ldots\]

\[-5\pi, -4\pi, -3\pi, -2\pi, -\pi, 0, \pi, 2\pi, 3\pi, \ldots\]

\[1 + \sqrt{2}, 1 + 2 \sqrt{2}, 1 + 3\sqrt{2}, \ldots\]

We can see that in each of the above sequences, the difference between two consecutive terms is a constant. For example, in the first sequence, the difference is $1$, while in the second, third and fifth, it is $2$. The consecutive differences in fourth, sixth and seventh sequences are $3, \pi$ and $\sqrt{2}$, respectively.

Such type of sequences are  very special in mathematics, and hence they are given a special name: Arithmetic Sequence or Arithmetic Progression. We can define such a sequence more formally in the following way.

An arithmetic sequence is a sequence $\{a_1, a_2, a_3, \ldots, a_{k-1}, a_k, \ldots\}$ of real or complex numbers such that 

\[a_2 - a_1 = a_3 - a_2 = a_4 - a_3 = \cdots = a_k - a_{k-1} = \cdots.\]

The difference between consecutive terms is a constant which is called the common difference of the given arithmetic sequence. If we denote the common difference by $d$, then we have

\[a_2 - a_1 = d\]

which implies

\[a_2 = a_1 + d.\]

Furthermore, 

\[a_3 - a_2 = d\]

which implies

\[a_3 = a_2 + d = (a_1 + d) + d = a_1 + 2d.\]

More generally, by mathematical induction, we can prove that

\[a_n = a_1 + (n-1)d~\text{for}~n = 1, 2, 3, \ldots.\]

Because of the above relation, in mathematics, sometimes the arithmetic sequence is defined as a sequence $\{a_1, a_2, a_3, \ldots\}$ such that $a_n = a_1 + (n-1)d$ for $n = 1, 2, 3, \ldots$. Here the term $a_1$ is called the first term of the sequence and $d$ is called the common difference of the sequence, and $a_n$ is called the $n$-th term of the sequence. Thus forming an arithmetic sequence is quite easy: Fix a first term $a_1$ and the common difference $d$, and then write other terms using the relation $a_n = a_1 + (n-1)d$. 

One remark: The arithmetic sequence may be defined in group also. Readers familiar with the concept of "Group" in Abstract Algebra will understand this. We will not discuss this here.

One natural question arises: How to add (quickly) the finite number of consecutive terms of an arithmetic sequence? Can we find some formula for this? Before answering this question, let us first consider some particular examples.

Example-$1$. Consider the sequence of natural numbers:

\[1, 2, 3, 4, 5, 6, 7, , 9, 10, 11,\ldots\]

What is $1 + 2 + 3 + \cdots + 50$?

One way of doing this is just add the numbers one by one. But this will be a tedious task if the numbers of terms (summands) in the sum is very large, say $1000$ or $50000$ or $100000000$, etc. Let us try to find some other way of doing this. Write

\[S = 1 + 2 + 3 + \cdots + 48 + 49 + 50.\] 

The summation $S$ can also be written as

\[S = 50 + 49 + 48 + \cdots + 3 + 2 + 1.\]

We have just reversed the order of the summands.  Now observe that if we add the first term in first summation to the first term in the second summation, we get $1 + 50 = 51$. Similary, the sum of second terms in the first summation and the second summation is again $2 + 49 = 51$. In general, the sum of any term in the first summation with the corresponding term in the second summation is $51$. Thus, if we add both the summation, the resultant will be

\[2S = \underbrace{51 + 51 + 51+ \cdots + 51 + 51 + 51}_{50~\text{times}}.\]

This summation is easy to compute, because summands are equal. We can easily compute

\[2S = 50 \times 51.\]

See how easily we could obtain twice of $S$! But we wanted to compute $S$. But this is obvious. We can just divide the value of $2S$ by $2$ to get the value of $S$. Therefore

\[S = \frac{50 \times 51}{2} = 1275.\]

Example-$2$. Consider the sequence of natural numbers:

\[1, 2, 3, 4, 5, 6, 7, , 9, 10, 11,\ldots\]

What is $1 + 2 + 3 + \cdots + n$?

Here we are interested in finding the sum of first $n$ natural numbers. The previous example is a particular case of this general example, where $n = 50$. Can you apply the same trick to find the sum? Try it. 

Let us write

\[S_n = 1 + 2 + 3 + \cdots + (n-2) + (n-1) + n.\] 

The summation $S_n$ can also be written as

\[S_n = n + (n-1) + (n-2) + \cdots + 3 + 2 + 1.\]

Again, observe that the sum of any term in the first summation with the corresponding term in the second summation is $n + 1$. Thus, if we add both the summation, the resultant will be

\[2S_n = \underbrace{(n + 1) + (n + 1) + (n + 1) + \cdots + (n + 1)  + (n + 1)  + (n + 1) }_{n~\text{times}}.\]

This summation is easy to compute, because summands are equal. We can easily compute

\[2S_n = n(n+1).\]

We can just divide the value of $2S_n$ by $2$ to get the value of $S_n$. Therefore

\[S_n = \frac{n(n +1)}{2}.\]

This is the nice formula for the sum of first $n$ natural numbers. See how the idea for a particular case perfectly worked for the more general case. Put $n == 50$ to get the same answer in the first example.

Let us consider the second sequence written in the beginning of this article: $\{1, 3, 5, 7, 9, \ldots\}$. You can see that the common difference of this arithmetic sequence is $2$. Since the first term $a_1$ is $1$, we can use the formula for $n$-th term and write $a_n = 1 + 2(n-1) = 2n - 1$. Thus the first $n$ odd natural numbers are $\{1, 3, 5, \ldots, 2n-5, 2n -3, 2n-1\}$.

Example-$3$. What is the sum of first $n$ odd numbers. That is,what is 

\[1 + 3 + 5 + \cdots + (2n-1)?\]

We can repeat the same procedure. Let us write

\[T_n = 1 + 3 + 5 + \cdots + (2n-5) + (2n-3) + (2n-1).\] 

The summation $T_n$ can also be written as

\[T_n = (2n-1) + (2n-3) + (2n-5) + \cdots + 5 + 3 + 1.\]

Again, observe that the sum of any term in the first summation with the corresponding term in the second summation is $2n$. Thus, if we add both the summation, the resultant will be

\[2T_n = \underbrace{2n  + 2n + 2n + \cdots + 2n + 2n + 2n}_{n~\text{times}}.\]

This summation is easy to compute, because summands are equal. We can easily compute

\[2T_n = 2n^2.\]

We can just divide the value of $2T_n$ by $2$ to get the value of $T_n$. Therefore

\[T_n = n^2.\]

This is the nice formula for the sum of first $n$ odd natural numbers. Many of you must have been aware of these two formulas.

Now you can try the following question.

Question $1$. What is the sum of first $n$ even natural numbers.

The above idea can be implemented for the sum of finite number of terms of any arithmetic sequence. Let $\{a_1, a_2, \ldots, a_n\}$ be first $n$ terms of an arithmetic sequence $\{a_1, a_2, \ldots, a_n, \ldots\}$. We would like to determine the summation

\[S = a_1 + a_2 + \ldots + a_{n-1} + a_n.\]

If $d$ is the common difference of this sequence, then we can write

\[S = a_1 + (a_1 + d) + \ldots + (a_1 + (n-2)d) + (a_1 + (n-1)d).\]

Rewrite the above equation as

\[S = (a_1 + (n-1)d) + (a_1 + (n-2)d) + \ldots + a_2 + a_1. \]

Now, additing both the equations, we get

\[2S = n[2a_1 + (n-1)d)].\]

This yields

\[S = \frac{n}{2}[2a_1 + (n-1)d].\]

This is the formula for the sum of first $n$ terms of an arithmetic sequence. We can write this formula in slightly different form also. Write the above formula as

\[S = \frac{n}{2}[a_1 + (a_1 + (n-1)d)].\]

Since $a_n = a_1 + (n-1)d$, we can write

\[S = \frac{n}{2}(a_1 + a_n).\]

Thus the sum of the first $n$-terms of an arithmetic sequence is half of $n$ times sum of first term and the $n$-th term. The formulas in Example $2$ and Example $3$ can easily be deduced from this formula. In Example $2$, we have $a_1 = 1$, $d = 1$ and $a_n = n$. In Example $3$, we have $a_1 = 1$, $d = 2$ and $a_n = 2n - 1$. We can substitute these values in any one of the formula for the sum of first $n$ terms of an arithmetic sequence to get the same result.

The sum of terms of an arithmetic sequence is called arithmetic series. We have just derived the the formula for the sum of first $n$-terms of an arithmetic series.

*This article is the English translation of the author's original article समान्तर अनुक्रम और समान्तर श्रेणी .