Sum of the First $n$ Terms of an Arithmetic Progression

Consider a finite arithmetic progression $a_1, a_2, a_3, \ldots, a_{n-2}, a_{n-1}, a_n$ . Let $S_n$ denote the sum of the terms of this progression:

S_n = a_1 + a_2 + a_3 + \ldots + a_{n-2} + a_{n-1} + a_n.

Gauss’s Method

First, consider a problem whose solution will suggest how to derive the formula. Consider the arithmetic progression $1, 2, 3, \ldots, 98, 99, 100$ and find the sum of its terms.

Write the sum in two ways:

\begin{aligned} S_{100} &= 1 + 2 + 3 + \ldots + 98 + 99 + 100 \\ S_{100} &= 100 + 99 + 98 + \ldots + 3 + 2 + 1 \end{aligned}

Adding these equalities:

2S_{100} = \underbrace{101 + 101 + 101 + \ldots + 101 + 101 + 101}_{100 \text{ terms}}

Therefore $2S_{100} = 101 \times 100$ ; $S_{100} = 5050$ .

ℹNote — Carl Friedrich Gauss

It is said that the great German mathematician Carl Friedrich Gauss (1777—1855) came up with this solution at the age of 5.

The Sum Formula

Using the same approach to find $S_n$ , write the sum in two ways:

S_n = a_1 + (a_1 + d) + (a_1 + 2d) + \ldots + (a_1 + (n-1)d),

S_n = a_n + (a_n - d) + (a_n - 2d) + \ldots + (a_n - (n-1)d).

Adding these equalities:

2S_n = (a_1 + a_n) + (a_1 + a_n) + \ldots + (a_1 + a_n) = n(a_1 + a_n).

⚡Theorem — Sum of the First n Terms of an Arithmetic Progression

S_n = \frac{a_1 + a_n}{2} \cdot n

Substituting the expression $a_n = a_1 + d(n-1)$ gives the equivalent formula:

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

The latter formula is convenient when the first term and common difference are given.

Examples

✎Example — Sum of All Three-Digit Multiples of 6

Problem. Find the sum of all three-digit numbers that are multiples of 6.

Solution. These numbers form an arithmetic progression with first term $a_1 = 102$ and common difference $d = 6$ . Then $a_n = 102 + 6(n-1) = 6n + 96$ . Since $a_n < 1000$ , the number of terms is the largest natural solution of $6n + 96 < 1000$ :

6n < 904; \quad n < 150\tfrac{2}{3}.

So $n = 150$ . Now compute the sum:

S_{150} = \frac{2 \cdot 102 + 6 \cdot (150 - 1)}{2} \cdot 150 = 82\,350.

Answer: 82,350.

✎Example — Finding a Term from the Sum

Problem. The sum of the first seventy-five terms of an arithmetic progression is 450. Find the thirty-eighth term.

Solution. Let the first term and common difference be $a_1$ and $d$ respectively. Then:

S_{75} = \frac{2a_1 + 74d}{2} \cdot 75 = 75(a_1 + 37d) = 450.

Since $a_{38} = a_1 + 37d$ , the desired term equals:

a_{38} = 450 \div 75 = 6.

Answer: 6.

✎Example — Quadratic Sum Formula Implies AP

Problem. Prove that a sequence whose sum of the first $n$ terms can be computed by the formula $S_n = an^2 + bn$ , where $a$ and $b$ are constants, is an arithmetic progression.

Solution. We have:

a_{n+1} = S_{n+1} - S_n = a(n+1)^2 + b(n+1) - an^2 - bn = 2an + a + b.

The equality $a_{n+1} = 2an + a + b$ shows that the sequence $a_2, a_3, \ldots, a_n, \ldots$ is an arithmetic progression with common difference $2a$ . If we show that $a_2 - a_1 = 2a$ , the proof is complete.

We have: $a_1 = S_1 = a + b$ ; $a_2 = 2a + a + b = 3a + b$ . Then $a_2 - a_1 = 2a$ .

Thus $a_1, a_2, a_3, \ldots, a_n, \ldots$ is an arithmetic progression.

Exercises

✏Exercise — Sum of Arithmetic Progression Problems

32.1. What is the sum of the first seven terms of an arithmetic progression $(a_n)$ if $a_1 = 9$ and $a_7 = 15$ ?

32.3. Find the sum of the first twelve terms of an arithmetic progression with first term $a_1 = -6$ and common difference $d = 4$ .

32.7. The arithmetic progression $(a_n)$ is defined by $a_n = -4n + 1$ . Find the sum of its first thirty-two terms.

32.15. What is the sum of the first $n$ :

Natural numbers;
Odd natural numbers?

Sum of the First nnn Terms of an Arithmetic Progression

Gauss’s Method

The Sum Formula

Examples

Exercises

Sum of the First $n$ Terms of an Arithmetic Progression