1.3 Arithmetic Sequence (A.S.)

1. The Definition

Let’s start with an example! If the first term of a sequence is $u_1 = 5$, and a fixed value, say $d=3$, is always added in order to find the next term, the following sequence is generated:

$5, 8, 11, 14, 17, \dots$

Such a sequence is called arithmetic. That is, in an arithmetic sequence the difference between any two consecutive terms is constant.

EXAMPLE 1

  • If $u_1 = 1, d=2$ the sequence is $\mathbf{1, 3, 5, 7, 9, \dots}$
  • If $u_1 = 2, d=2$ the sequence is $\mathbf{2, 4, 8, 10, 12, \dots}$
  • If $u_1 = -10, d=5$ the sequence is $\mathbf{-10, -5, 0, 5, 10, \dots}$
  • If $u_1 = 10, d=-3$ the sequence is $\mathbf{10, 7, 4, 1, -2, \dots}$

Notice that the common difference $d$ may also be negative!

2. QUESTION A: What is the general formula for $u_n$?

If we know $u_1$ and $d$, then the general term is given by:

$u_n = u_1 + (n-1)d$

Indeed, let us think: In order to find $u_5$, we start from $u_1$ and then add 4 times the difference $d$.

Hence, $u_5 = u_1 + 4d$
Similarly, $u_{10} = u_1 + 9d$
$u_{50} = u_1 + 49d$
In general, $u_n = u_1 + (n-1)d$

EXAMPLE 2

In an arithmetic sequence let $u_1 = 3$ and $d = 5$. Find (a) the first four terms and (b) the 100th term.

Solution:

(a) $\mathbf{3, 8, 13, 18}$
(b) Now the general formula is needed:
$u_{100} = u_1 + 99d = 3 + 99(5) = \mathbf{498}$

EXAMPLE 3

In an arithmetic sequence let $u_1 = 100$ and $u_{16} = 145$. Find $u_7$.

Solution:

Since $u_1$ is known, $d$ is needed. Exploit the information for $u_{16}$ first:
$u_{16} = u_1 + 15d$
$145 = 100 + 15d$
$45 = 15d$
$d = 3$
Therefore, $u_7 = u_1 + 6d = 100 + 6(3) = \mathbf{118}$

⚠️ REMEMBER

Usually, the first task in an A.S. is to find the basic elements, $u_1$ and $d$, and then everything else!

EXAMPLE 4

In an arithmetic sequence let $u_{10} = 42$ and $u_{19} = 87$. Find $u_{100}$.

Solution:

The formula for $u_{10}$ and $u_{19}$ takes the form:
$u_{10} = u_1 + 9d \quad \Rightarrow \quad u_1 + 9d = 42 \quad \text{(a)}$
$u_{19} = u_1 + 18d \quad \Rightarrow \quad u_1 + 18d = 87 \quad \text{(b)}$
Subtract (b) - (a):
$18d - 9d = 87 - 42$
$9d = 45$
$d = 5$
Then, (a) gives $u_1 = 42 - 9d = 42 - 9(5) = -3$
Since $u_1 = -3$ and $d = 5$, any term can be found!
$u_{100} = u_1 + 99d = -3 + 99(5) = \mathbf{492}$

3. QUESTION B: What is the sum $S_n$ of the first $n$ terms?

Formula (1): Use if $u_1$ and the last term $u_n$ are known.
$S_n = \frac{n(u_1 + u_n)}{2}$
Formula (2): Use if $u_1$ and $d$ (the basic elements) are known.
$S_n = \frac{n}{2}[2u_1 + (n-1)d]$

EXAMPLE 5

For the A.S. $3, 5, 7, 9, 11, \dots$ find $S_3$ and $S_{101}$.

Solution:

Here $u_1 = 3$ and $d = 2$. For $S_3$ the result is direct:
$S_3 = 3 + 5 + 7 = \mathbf{15}$
For $S_{101}$ formula (2) is used:
$S_{101} = \frac{101}{2}[2u_1 + 100d] = \frac{101}{2}[2(3) + 100(2)]$
$S_{101} = \frac{101}{2}[206] = 101 \times 103 = \mathbf{10403}$

EXAMPLE 6

Find $10 + 20 + 30 + \dots + 200$

Solution:

This is an arithmetic sequence with $u_1 = 10$ and $d = 10$. The number of terms is clearly $20$ and $u_{20} = 200$.
$S_{20} = \frac{20(u_1 + u_{20})}{2} = 10(10 + 200) = \mathbf{2100}$

EXAMPLE 7

Show that $1 + 2 + 3 + \dots + n = \frac{n(n+1)}{2}$

Solution:

This is the simplest arithmetic series with $u_1 = 1$ and $d = 1$. Asking for $S_n$:
$S_n = \frac{n(u_1 + u_n)}{2} = \mathbf{\frac{n(1 + n)}{2}}$
For example: $1+2+3+\dots+100 = \frac{100 \times 101}{2} = 5050$

EXAMPLE 8

The 3rd term of an A.S. is zero while the sum of the first 15 terms is -300. Find the first term and the sum of the first ten terms.

Solution:

Organize the data:
GIVEN: $u_3 = 0, \quad S_{15} = -300$
ASK FOR: $u_1, \quad S_{10}$
The formulas for $u_3$ and $S_{15}$ give:
$u_3 = u_1 + 2d \quad \Rightarrow \quad u_1 + 2d = 0$
$S_{15} = \frac{15}{2}(2u_1 + 14d) \quad \Rightarrow \quad 15u_1 + 105d = -300$
Solve the system:
$u_1 + 2d = 0$
$15u_1 + 105d = -300$
This obtains $u_1 = \mathbf{8}$ and $d = -4$.
Finally:
$S_{10} = \frac{10}{2}(2u_1 + 9d) = 5(16 - 36) = \mathbf{-100}$

4. Notice for Consecutive Terms

Let $a, x, b$ be consecutive terms of an arithmetic sequence (whether these are the first three terms or some other three consecutive terms).

The common difference is equal to:

$x - a = b - x$

Hence, $2x = a + b$, that is: $x = \frac{a + b}{2}$

($x$ is the mean of $a$ and $b$)

EXAMPLE 9

Let $x+1, 3x, 6x-5$ be consecutive terms of an A.S. Find $x$.

Solution:

It holds: $(3x) - (x+1) = (6x-5) - (3x)$
$\Rightarrow 2x - 1 = 3x - 5$
$\Rightarrow \mathbf{x = 4}$
(Indeed, the three terms are 5, 12, 19)

EXAMPLE 10

Let $a, 10, b, a+b$ be consecutive terms of an A.S. Find $a$ and $b$.

Solution:

Clearly $10 - a = b - 10 = (a+b) - b$
that is: $10 - a = b - 10 = a$
Hence:
$10 - a = a \Rightarrow 2a = 10 \Rightarrow \mathbf{a = 5}$
$b - 10 = a \Rightarrow b - 10 = 5 \Rightarrow \mathbf{b = 15}$

EXAMPLE 11

Let $100, a, b, c, 200$ be consecutive terms of an A.S. Find the values of $a, b$ and $c$.

Solution:

Notice that $100, b, 200$ are also in an arithmetic sequence.
Thus $b$ is the mean of $100$ and $200$, that is $\mathbf{b = 150}$.
Now:
$a$ is the mean of $100$ and $150$, that is $\mathbf{a = 125}$
$c$ is the mean of $150$ and $200$, that is $\mathbf{c = 175}$