Arithmetic Sequences and Series

Arithmetic Sequences and Series — Moosa Academy

Once the step between terms is constant, two formulas do all the work: one jumps straight to any term you want, and one adds up a whole stretch of them without touching a calculator more than a few times. A sequence is the list; a series is what you get when you add that list together.

Theorem The two core formulas

The $n$ th term is $a_n = a_1 + (n-1)d$ , and the sum of the first $n$ terms is $S_n = \dfrac{n}{2}(a_1 + a_n)$ .

Here $a_1$ is the first term, $d$ the common difference, and $n$ the position.

Term formula:

a_n = a_1 + (n-1)d

Sum, knowing both ends:

S_n = \dfrac{n}{2}(a_1 + a_n)

Sum, knowing $d$ instead:

S_n = \dfrac{n}{2}\bigl[2a_1 + (n-1)d\bigr]

Note the $(n-1)$ , not $n$ . Reaching the $12$ th term takes only eleven steps, because the first term is already there before any step is taken.

Concept Sigma notation

A series is often written compactly with the summation sign:

\displaystyle\sum_{k=1}^{n} f(k)

The letter $k$ starts at the value written below and climbs to the value written above, and $f(k)$ is the rule generating each term. Everything the notation describes is still an ordinary series — the same two formulas apply.

Example Finding a distant term

a) Find the 12th term of $9,\ 16,\ 23,\ 30,\ \ldots$

Common difference:

16 - 9 = 7

Substitute:

a_{12} = 9 + (12-1)(7)

Simplify:

a_{12} = 9 + 77

$a_{12} = 86$

b) Write a rule for $5,\ -13,\ -31,\ \ldots$

Common difference:

-13 - 5 = -18

Substitute:

a_n = 5 + (n-1)(-18)

Expand:

a_n = 5 - 18n + 18

$a_n = -18n + 23$

c) Given $a_5 = 19$ and $d = 6$ , find $a_1$

Substitute into the term formula:

19 = a_1 + (5-1)(6)

Simplify:

19 = a_1 + 24

$a_1 = -5$

Part (c) shows the formula works backwards just as happily. Any three of $a_1$ , $d$ , $n$ , $a_n$ determine the fourth.

Example Arithmetic means

Find four arithmetic means between $-8$ and $22$ .

Four means plus the two end terms gives

n = 6

The end terms are

a_1 = -8

and

a_6 = 22

Substitute:

22 = -8 + (6-1)d

Simplify:

30 = 5d

, so

d = 6

Build the sequence:

-8,\ -2,\ 4,\ 10,\ 16,\ 22

The four means are $-2,\ 4,\ 10,\ 16$

The trap is counting $n = 4$ . The means sit between the given numbers, so both ends must be counted too.

Example Summing a series

a) Find $12 + 19 + 26 + \cdots + 180$

Common difference:

19 - 12 = 7

Find how many terms:

180 = 12 + (n-1)(7)

Simplify:

168 = 7(n-1)

, so

n - 1 = 24

and

n = 25

Apply the sum formula:

S_{25} = \dfrac{25}{2}(12 + 180)

Simplify:

S_{25} = \dfrac{25}{2}(192) = 25 \times 96

$S_{25} = 2400$

b) Evaluate $\displaystyle\sum_{k=1}^{15}(3k + 2)$

Number of terms:

n = 15

First term at

k = 1

3(1) + 2 = 5

Last term at

k = 15

3(15) + 2 = 47

Apply the sum formula:

S_{15} = \dfrac{15}{2}(5 + 47)

Simplify:

S_{15} = \dfrac{15}{2}(52) = 15 \times 26

$S_{15} = 390$

In both parts the work is the same: find $n$ , find the two end terms, then average them and multiply by how many there are. That is all the sum formula really says.

Summary

A sequence is a list of terms; a series is that list added together.
The $n$ th term is $a_n = a_1 + (n-1)d$ — note the $(n-1)$ , not $n$ .
Knowing both ends, the sum is $S_n = \dfrac{n}{2}(a_1 + a_n)$ .
Knowing $d$ instead, use $S_n = \dfrac{n}{2}\bigl[2a_1 + (n-1)d\bigr]$ .
Any three of $a_1$ , $d$ , $n$ , $a_n$ give you the fourth.
Inserting $m$ arithmetic means between two numbers makes $n = m + 2$ terms in total.
Sigma notation is only shorthand — the same formulas evaluate it.