Arithmetic Series

A series is what you get when you add the terms of a sequence. Adding $1 + 2 + 3 + \dots + 100$ one number at a time takes a while, but there’s a famous shortcut, and it works for every arithmetic series.

Key ideas

Series and Sₙ

A series is the sum of the terms of a sequence. $S_n$ means the sum of the first $n$ terms:

$$S_n = t_1 + t_2 + \dots + t_n$$

An arithmetic series is the sum of an arithmetic sequence, like $3 + 7 + 11 + 15$.

The pairing trick

To add $1 + 2 + \dots + 100$, write the sum forwards and backwards and add them:

$$\begin{aligned} S &= 1 + 2 + 3 + \dots + 100 \\ S &= 100 + 99 + 98 + \dots + 1 \\ 2S &= 101 + 101 + 101 + \dots + 101 \end{aligned}$$

There are $100$ pairs, each adding to $101$, so $2S = 100 \times 101$ and $S = 5050$.

The formulas

The same trick works for any arithmetic series: every pair adds to $t_1 + t_n$, and there are $n$ pairs, so $2S_n = n(t_1 + t_n)$.

$$S_n = \frac{n}{2}(a + t_n) \qquad\text{or}\qquad S_n = \frac{n}{2}\big(2a + (n - 1)d\big)$$

• Use the first when you know the last term.
• Use the second when you know the common difference.

If you know the last term but not $n$, find $n$ first with $t_n = a + (n - 1)d$.

Worked examples

Example 1: The classic

Find $1 + 2 + 3 + \dots + 100$ using the formula.

Solution. $n = 100$, $a = 1$, $t_{100} = 100$:

$$S_{100} = \frac{100}{2}(1 + 100) = 50(101) = 5050$$

Example 2: Using d

Find the sum of the first $20$ terms of $3 + 7 + 11 + \dots$

Solution. $a = 3$, $d = 4$, $n = 20$:

$$S_{20} = \frac{20}{2}\big(2(3) + 19(4)\big) = 10(6 + 76) = 820$$

Example 3: Finding n first

Find the sum $5 + 8 + 11 + \dots + 302$.

Solution. First find how many terms there are:

$$5 + (n - 1)(3) = 302 \quad\Rightarrow\quad 3(n - 1) = 297 \quad\Rightarrow\quad n = 100$$

Then use the first and last terms:

$$S_{100} = \frac{100}{2}(5 + 302) = 50(307) = 15\,350$$

Example 4: Seats in a theatre

A theatre has $18$ seats in the first row, and each row has $2$ more seats than the row in front. How many seats are in the first $20$ rows?

Solution. $a = 18$, $d = 2$, $n = 20$:

$$S_{20} = \frac{20}{2}\big(2(18) + 19(2)\big) = 10(36 + 38) = 740$$

There are $740$ seats.

Common mistakes

Using the term formula instead of the sum formula. $t_n$ is one term; $S_n$ is the total of $n$ terms.

Miscounting the terms. In $5 + 8 + \dots + 302$, there are $100$ terms, not $302$ or $297 \div 3 = 99$. Solve for $n$ carefully.

Halving only part of the formula. $\dfrac{n}{2}(a + t_n)$ means half of $n$, times the whole bracket.

Using $nd$ instead of $(n - 1)d$. The formula has $(n - 1)d$, matching the general term.

Practice

1. (Warm-up) Find the sum of the first $10$ terms of $2 + 4 + 6 + \dots$

Solution

$$S_{10} = \frac{10}{2}\big(2(2) + 9(2)\big) = 5(22) = 110$$

2. (Warm-up) Find $1 + 2 + 3 + \dots + 50$.

Solution

$$\frac{50}{2}(1 + 50) = 25(51) = 1275$$

3. (Warm-up) An arithmetic series has $15$ terms, first term $4$, and last term $60$. Find its sum.

Solution

$$S_{15} = \frac{15}{2}(4 + 60) = \frac{15}{2}(64) = 480$$

4. (Core) Find the sum of the first $25$ terms of $10 + 7 + 4 + \dots$

Solution

$a = 10$, $d = -3$:

$$S_{25} = \frac{25}{2}\big(20 + 24(-3)\big) = \frac{25}{2}(-52) = -650$$

5. (Core) Find the sum $6 + 13 + 20 + \dots + 216$.

Solution

$6 + 7(n - 1) = 216$ gives $n - 1 = 30$, so $n = 31$.

$$S_{31} = \frac{31}{2}(6 + 216) = \frac{31}{2}(222) = 3441$$

6. (Core) Find the sum of the first $40$ odd numbers, $1 + 3 + 5 + \dots$ What do you notice?

Solution

$$S_{40} = \frac{40}{2}\big(2(1) + 39(2)\big) = 20(80) = 1600$$

$1600 = 40^2$. In fact, the sum of the first $n$ odd numbers is always $n^2$.

7. (Core) You save $25 in the first week and increase your savings by $5 each week. How much have you saved after $26$ weeks?

Solution

$a = 25$, $d = 5$, $n = 26$:

$$S_{26} = \frac{26}{2}\big(2(25) + 25(5)\big) = 13(50 + 125) = 13(175) = 2275$$

You’ve saved $2275.

8. (Challenge) How many terms of $3 + 7 + 11 + \dots$ add up to $465$?

Solution

$$\begin{aligned} \frac{n}{2}\big(6 + 4(n - 1)\big) &= 465 \\ \frac{n}{2}(4n + 2) &= 465 \\ n(2n + 1) &= 465 \\ 2n^2 + n - 465 &= 0 \end{aligned}$$

The quadratic formula gives $n = 15$ or $n = -15.5$. Only a positive whole number works, so $15$ terms. Check: $\tfrac{15}{2}(3 + 59) = 465$. ✓

9. (Challenge) Use the forwards-and-backwards trick to prove that $S_n = \dfrac{n}{2}(a + t_n)$ for any arithmetic series.

Solution

$$\begin{aligned} S_n &= a + (a + d) + (a + 2d) + \dots + t_n \\ S_n &= t_n + (t_n - d) + (t_n - 2d) + \dots + a \end{aligned}$$

Add the two lines column by column. In each column the $d$‘s cancel, leaving $a + t_n$. There are $n$ columns:

$$2S_n = n(a + t_n) \quad\Rightarrow\quad S_n = \frac{n}{2}(a + t_n)$$