Sum of the First n Terms of Arithmetic Sequences

Finding the sum of the first $n$ terms of an arithmetic sequence is an important problem in the study of sequences. Mastering the summation formula allows you to quickly compute the sum of many terms, avoiding the tedious work of adding them one by one.

The Story of Gauss

Before presenting the formula, let us share a famous mathematical story.

At the end of the 18th century, when the German mathematician Gauss was still a primary school student, his teacher asked the whole class to compute the sum $1+2+3+\cdots+100$, hoping to keep the students busy for a while. But Gauss gave the answer very quickly: $5050$.

His method was to pair the first and last of the 100 numbers:

$$\begin{aligned} &(1 + 100) + (2 + 99) + (3 + 98) + \cdots + (50 + 51) \\ &= 101 + 101 + 101 + \cdots + 101 \quad (\text{50 pairs in total}) \\ &= 101 \times 50 = 5050 \end{aligned}$$

This clever method inspired the derivation of the arithmetic sequence summation formula.

Derivation of the Formula

Method 1: Reverse Order Addition

Let $S_n$ be the sum of the first $n$ terms of the arithmetic sequence $\{a_n\}$:

$S_n = a_1 + a_2 + a_3 + \cdots + a_n$

Write this expression in reverse order:

$S_n = a_n + a_{n-1} + a_{n-2} + \cdots + a_1$

Adding the two equations:

$$\begin{aligned} 2S_n &= (a_1 + a_n) + (a_2 + a_{n-1}) + (a_3 + a_{n-2}) + \cdots + (a_n + a_1) \end{aligned}$$

Since $a_1 + a_n = a_2 + a_{n-1} = a_3 + a_{n-2} = \cdots$ (a property of arithmetic sequences), there are $n$ pairs, and each pair sums to $a_1 + a_n$. Therefore:

$2S_n = n(a_1 + a_n)$

Method 2: Substituting the General Term Formula

Substitute $a_n = a_1 + (n-1)d$ into the formula above:

$$\begin{aligned} S_n &= \frac{n(a_1 + a_n)}{2} \\ &= \frac{n[a_1 + a_1 + (n-1)d]}{2} \\ &= \frac{n[2a_1 + (n-1)d]}{2} \end{aligned}$$

The Geometric Meaning of the Formula

From a geometric viewpoint, the sum of the first $n$ terms of an arithmetic sequence can be regarded as the area of a trapezoid:

• Upper base: $a_1$ (first term)
• Lower base: $a_n$ (last term)
• Height: $n$ (number of terms)

Trapezoid area formula: $S = \frac{(\text{upper base} + \text{lower base}) \times \text{height}}{2} = \frac{(a_1 + a_n) \times n}{2}$

This exactly corresponds to Form 1 of the arithmetic sequence summation formula!

Worked Examples

Example 1: Using Form 1

Find the sum of all terms of the arithmetic sequence $2, 5, 8, 11, \ldots, 98$.

Solution:

First determine the number of terms. Given $a_1 = 2$, $d = 3$, $a_n = 98$.

From $a_n = a_1 + (n-1)d$: $98 = 2 + (n-1) \times 3$ $96 = 3(n-1)$ $n = 33$

Using Form 1: $S_{33} = \frac{33 \times (2 + 98)}{2} = \frac{33 \times 100}{2} = 1650$

Example 2: Using Form 2

In the arithmetic sequence $\{a_n\}$, $a_1 = 5$ and $d = 3$. Find the sum of the first 20 terms $S_{20}$.

Solution:

Using Form 2: $S_{20} = 20 \times 5 + \frac{20 \times 19}{2} \times 3 = 100 + 570 = 670$

Example 3: Finding the Number of Terms from the Sum

In the arithmetic sequence $\{a_n\}$, $a_1 = 1$ and $d = 2$. If the sum of the first $n$ terms is $S_n = 100$, find $n$.

Solution:

Using Form 2: $S_n = n \times 1 + \frac{n(n-1)}{2} \times 2 = n + n(n-1) = n^2$

Therefore: $n^2 = 100$ $n = 10 \quad (\text{discarding the negative value})$

Practice Problems

Exercise 1

Find the sum of the first 50 terms of the arithmetic sequence $1, 3, 5, 7, \ldots$.

Exercise 2

In the arithmetic sequence $\{a_n\}$, $a_1 = -10$ and $d = 3$. After how many terms is the sum minimal?

Exercise 3

The sum of the first $n$ terms of the arithmetic sequence $\{a_n\}$ is $S_n = n^2 + 2n$. Find the general term formula $a_n$.

Exercise 4

Adapted from a postgraduate entrance examination problem

The sum of the first $n$ terms of the arithmetic sequence $\{a_n\}$ is $S_n$. Given $S_3 = 9$ and $S_6 = 36$, find $S_9$.

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