Arithmetic Sequences

Arithmetic Sequences are ordered sets of numbers where consecutive terms are separated by a constant difference. They can be expressed algebraically with the explicit formula:

$\textcolor{olive}{t_n} =\textcolor{red}{a} + \textcolor{green}{d}(\textcolor{blue}{n}-1)$

$\textcolor{blue}{n}$ represents the sequence position
$\textcolor{teal}{tₙ}$ represents the term value in position $n$
$\textcolor{brown}{t_{n-m}}$ represents the term value $m$ previous of the $n$th position
$\textcolor{red}{a}$ represents the first term in any sequence
$\textcolor{green}{d}$ represents the common difference

Arithmetic Sequences are similar to linear functions since they involve constant differences. An arithmetic sequence can be expressed in function notation as:

$\textcolor{teal}{f(n)} = \textcolor{red}{a} + \textcolor{green}{d}(\textcolor{blue}{n}-1)$

When graphed, the points of an arithmetic sequence should be visualized discretely rather than continuouly like a regular linear graph. This is because the terms in a sequence are distinct and separate from one another:

Graph of a Continuous Linear Sequence. It resembles a linear function.

Graph of a Discrete Linear Sequence. It resembles a linear scatter plot.

The continuous linear graph has a domain of all real numbers, which consists of positive and negative values which can extend to both infinities. The discrete linear graph conversely has a domain of all natural numbers, which only consists of detached positive integers starting from 1 which can extend to positive infinity depending on the sequence.

The formula for finding the difference between two consecutive terms is:

$\textcolor{green}{d} = \textcolor{teal}{t_n} - \textcolor{brown}{t_{n-1}}$

The formula for determining the recursive arthmetic function is:

$\textcolor{teal}{t_n}= t_{n-1}+\textcolor{green}{d}, t_1=\textcolor{red}{a}$

NOTE: The first term must be listed after the recursive formula since there is no previous term(s) it relies on to determine its value.

Example

For the sequence $-9, -6, -3, 0...,$ identify the following:

The explicit formula
The recursive formula
The $30$th term

i. In order to determine the explicit formula, we must first identify the difference:

$\textcolor{green}{d} = \textcolor{olive}{\text{next}} - \textcolor{brown}{\text{previous}}$

We can first determine the difference between the second and first terms:

$\textcolor{green}{d_1} = \textcolor{teal}{-6} - (\textcolor{brown}{-9})$

$\textcolor{green}{d_1 = 3}$

We can then determine the difference between the third and second terms:

$\textcolor{green}{d_2} = \textcolor{olive}{-3} - (\textcolor{brown}{-6})$

$\textcolor{green}{d_2 = 3}$

We can then plug the difference and the starting term into the arithmetic sequence formula to get the explicit formula:

$\textcolor{teal}{t_n} = \textcolor{red}{a} + \textcolor{green}{d}(\textcolor{blue}{n}-1)$

$\textcolor{teal}{t_n} = (\textcolor{red}{-9}) + \textcolor{green}{3}(\textcolor{blue}{n}-1)$

$\textcolor{teal}{t_n} = (\textcolor{red}{-9}) + 3\textcolor{blue}{n} -3$

$\textcolor{teal}{t_n} = 3\textcolor{blue}{n} - 12$

ii. We can find the recursive formula using the equation shown at the start of the lesson, in addition to plugging in the initial value ($-9$) and difference ($3$).

$\textcolor{teal}{t_n}=\textcolor{brown}{t_{n-1}}+\textcolor{green}{d}, t₁ =\textcolor{red}{a}$

$\textcolor{teal}{t_n}=\textcolor{brown}{t_{n-1}}+\textcolor{green}{3}, t₁ =\textcolor{red}{-9}$

iii. We can find the $30$th term by plugging the sequence number into the explicit formula, then solving for the rest:

$\textcolor{olive}{t_n} = 3(\textcolor{blue}{30}) - 12$

$\textcolor{olive}{t_n} = 90 - 12$

$\textcolor{olive}{t_n = 78}$

Determine the general term, $t_n$, for artithmetic sequence where $t_{10} = 300$ and $t_{15} = 325$.

Show Answer

First, we can find the respective equations by plugging the values into the arithmetic sequence formula:

$\textcolor{teal}{t_n} = \textcolor{red}{a} + \textcolor{green}{d}(\textcolor{blue}{n}-1)$

We can determine the first equation by plugging in the appropriate values and simplifying:

$\text{E}1: \textcolor{teal}{t_{10}} = \textcolor{red}{a} + \textcolor{green}{d}(\textcolor{blue}{10}-1)$

$\text{E}1: \textcolor{teal}{300} = \textcolor{red}{a} + 9\textcolor{green}{d}$

We can determine the second equation by plugging in the appropriate values and simplifying:

$\text{E}2: \textcolor{teal}{t_{15}} = \textcolor{red}{a} + \textcolor{green}{d}(\textcolor{blue}{15}-1)$

$\text{E}2: \textcolor{teal}{325} = \textcolor{red}{a} + 14\textcolor{green}{d}$

We can then subtract one equation from the other in order to determine the difference. In this case, we will subtract Equation 1 from Equation 2:

$\textcolor{olive}{325} - \textcolor{olive}{300} = \textcolor{red}{a} - \textcolor{red}{a} + 14\textcolor{green}{d} - 9\textcolor{green}{d}$

$\textcolor{olive}{25} = 5\textcolor{green}{d}$

$\cfrac{\textcolor{olive}{25}}{5} = \cfrac{\cancel{5}\textcolor{green}{d}}{\cancel{5}}$

$\textcolor{green}{d = 5}$

We can plug the difference into one of the equations in order to get the initial value, $a$. In this case, we will use the first equation:

$\textcolor{teal}{300} = \textcolor{red}{a} + 9\textcolor{green}{d}$

$\textcolor{teal}{300} = \textcolor{red}{a} + 9(\textcolor{green}{5})$

$\textcolor{teal}{300} = \textcolor{red}{a} + 45$

$\textcolor{red}{a = 255}$

We can now use all values to determine the general term (also known as the explicit formula):

$\textcolor{teal}{t_n} = \textcolor{red}{a} + \textcolor{green}{d}(\textcolor{blue}{n}-1)$

$\textcolor{teal}{t_n} = \textcolor{red}{255} + \textcolor{green}{5}(\textcolor{blue}{n}-1)$

$\textcolor{teal}{t_n} = 5n + \textcolor{red}{255}-5$

$\textcolor{teal}{t_n} = 5n + 250$

Therefore, we can determine that the general term is $\boldsymbol{t_n = 5n + 250}$.

Example

Determine the # of terms for the sequence $3,15,27,...,495$

First, we can find the difference in order to determine the explicit formula:

$\textcolor{green}{d} = \textcolor{teal}{t_n} - \textcolor{brown}{t_{n-1}}$

We can first determine the difference between the second and first terms:

$\textcolor{green}{d_1} = \textcolor{teal}{15} - \textcolor{brown}{3}$

$\textcolor{green}{d_1 = 12}$

We can then determine the difference between the third and second terms:

$\textcolor{green}{d_2} = \textcolor{teal}{27} - \textcolor{brown}{15}$

$\textcolor{green}{d_2 = 12}$

Next, we can plug all pertinent values into the explicit formula. We can then divide both sides by the same value to get the # of terms:

$\textcolor{teal}{t_n} = \textcolor{red}{a} + \textcolor{green}{d}(\textcolor{blue}{n}-1)$

$\textcolor{teal}{495} = \textcolor{red}{3} + \textcolor{green}{12}(\textcolor{blue}{n}-1)$

$\textcolor{teal}{495} = 12\textcolor{blue}{n} + \textcolor{red}{3} - \textcolor{red}{12}$

$12\textcolor{blue}{n} = \textcolor{red}{504}$

$ \cfrac{\cancel{12}\textcolor{blue}{n}}{\cancel{12}} = \cfrac{\textcolor{red}{504}}{12}$

$\textcolor{blue}{n = 42}$

Therefore, we can determine that this sequence has $\boldsymbol{42}$ terms.

Marisa deposits $$1200$ in a savings account that pays simple interest at $2$%. She makes no other deposits. (Simple interest means that interest increases by a constant rate each year).

Write the sequence of the year-end balances over $4$ years.
Determine the general term formula in function notation.
At the end of some year, the balance in the account is $$13200$. For how long has the original deposit earned interest?

Show Answer

i. In order to determine the sequence, we must first identify how much the bank account increases year to year. We can do this by determining how much Marisa makes after a year:

$A_1 = \textcolor{red}{P}(1+\textcolor{green}{r})$

$A_1 = \textcolor{red}{12000}(1+\textcolor{green}{0.02})$

$A_1 = \textcolor{red}{12000}(1.02)$

$A_1 = 12240$

We can then determine the difference between the principal amount and the amount after a year to determine how much interest Marisa makes each year:

$\textcolor{green}{d} = 12240 - 12000$

$\textcolor{green}{d = $240}$

Since Marisa makes an additional $$240$ every year, we can multiply this amount by the # of years to determine our sequence:

$A = 12000 + 240n$

We can first determine Marisa's total amount after the second year:

$A_2 = 12000 + 240(2)$

$A_2 = 12000 + 480$

$A_2 = $12480$

We can then determine Marisa's total amount after the third year:

$A_3 = 12000 + 240(3)$

$A_3 = 12000 + 720$

$A_3 = $12720$

We can finally determine Marisa's total amount after the fourth year:

$A_4 = 12000 + 240(4)$

$A_4 = 12000 + 960$

$A_4 = $12960$

Therefore, we can determine our $4$-year sequence as $\boldsymbol{12000, 12240, 12480, 12720, 12960}$.

ii. We can determine the function notation by plugging in the pertinent values. This should be simple as function notation for arithmetic functions is identical to the explicit function:

$f(n) = \textcolor{red}{a} + \textcolor{green}{d}(\textcolor{blue}{n}-1)$

$f(n) = \textcolor{red}{12000} + \textcolor{green}{240}(\textcolor{blue}{n}-1)$

$f(n) = \textcolor{red}{12000} + 240n - 240$

$f(n) = 240n + 11760$

Therefore, we can determine the function notation for this formula as $\boldsymbol{f(n) = 240n + 11760}$.

iii. We can identify the length of time the deposit has gained interest by plugging the balance into the function notation formula:

$13200 = 240n + 11760$

$240n = 1440$

$\cfrac{\cancel{240}n}{\cancel{240}} = \cfrac{1440}{240}$

$n = 6$

Therefore, we can determine that the deposit has been gaining interest for $\boldsymbol{5} \; [\textbf{years}]$. Since the first term represents the starting investment, the sixth term represents the end of the fifth year.

Arithmetic Sequences

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