1.3 Arithmetic Sequence (A.S.)

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!

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, $u_7 = u_1 + 6d = 100 + 6(3) = \mathbf{118}$

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!
$u_{100} = u_1 + 99d = -3 + 99(5) = \mathbf{492}$

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: $$\begin{aligned} S_{101} &= \dfrac{101}{2}[2u_1 + 100d] = \frac{101}{2}[2(3) + 100(2)] \\ &= \dfrac{101}{2}[206] = 101 \times 103 = \mathbf{10403} \end{aligned}$$

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 $u_{20} = 200$.

$$S_{20} = \dfrac{20(u_1 + u_{20})}{2} = 10(10 + 200) = \mathbf{2100}$$

EXAMPLE 7

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

Solution:

This is the simplest arithmetic series with $u_1 = 1$ and $d = 1$. Asking for $S_n$: $$S_n = \dfrac{n(u_1 + u_n)}{2} = \mathbf{\dfrac{n(1 + n)}{2}}$$

For example, $$1+2+3+\dots+100 = \dfrac{100 \times 101}{2} = 5050$$

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: $$\begin{aligned} u_3 &= u_1 + 2d \implies u_1 + 2d = 0 \\ S_{15} &= \frac{15}{2}(2u_1 + 14d) \implies 15u_1 + 105d = -300 \end{aligned} $$

This obtains $u_1 = \mathbf{8}$ and $d = -4$. Finally: $$S_{10} = \dfrac{10}{2}(2u_1 + 9d) = 5(16 - 36) = \mathbf{-100}$$

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 $\mathbf{c = 175}$