Precalculus by Richard Wright

Jesus said, “Father, forgive them, for they do not know what they are doing.” And they divided up his clothes by casting lots. Luke 23:34 NIV

Summary: In this section, you will:

- Write the explicit rule for an arithmetic sequence.
- Write the recursive rule for an arithmetic sequence.
- Evaluate the sum for an arithmetic series.

SDA NAD Content Standards (2018): PC.7.2

You are saving money to buy your first car. You save $100 every month with a goal of $3000. That means your total money is $100 the first month, $200 the second month, $300 the third month, etc. These totals are an arithmetic sequence.

An arithmetic sequence with a pattern of a common difference, *d*. We can derive the explicit formula by looking at a simple example.

$$ 1, 3, 5, 7, 9, … $$

Write the numbers using the pattern which in this example is adding 2.

$$ 1, 1+2, 1+2+2, 1+2+2+2, 1+2+2+2+2 $$

$$ 1, 1+\color{blue}{1}(2), 1+\color{blue}{2}(2), 1+\color{blue}{3}(2), 1+\color{blue}{4}(2) $$

The blue terms are 1 less than the term number, so the rule becomes

$$ a_n = 1 + (n - 1)2 $$

Notice that the first term was 1 and the common difference was 2, replace those numbers in the rule to get the general formula for the *n*^{th} term.

$$ a_n = a_1 + (n - 1)d $$

$$ a_n = a_1 + (n - 1)d $$

where *a*_{1} is the first term and *d* is the common difference

This simplifies to

$$ a_n = dn + c $$

where \(c = a_1 - d\).

This is **linear**, so any time an explicit rule for a sequence is linear, it is arithmetic.

$$ a_1 = a_1, a_{n} = a_{n-1} + d $$

Write the rule for the *n*^{th} term for 4, 7, 10, 13, 16, ….

The first term is 4, so \(a_1 = 4\). Subtract terms to find the common difference, *d*.

$$ 7 - 4 = 3; 10 - 7 = 3; 13 - 10 = 3; 16 - 13 = 3 $$

It appears that the common difference is 3, so \(d = 3\).

$$ a_n = a_1 + (n - 1)d $$

$$ a_n = 4 + (n - 1)3 $$

Simplify.

$$ a_n = 4 + 3n - 3 $$

$$ a_n = 3n + 1 $$

Write the rule for the *n*^{th} term for 16, 12, 8, 4, 0, ….

\(a_n = -4n + 20\)

The 7^{th} term of an arithmetic sequence is 26, and the 15^{th} term is 50. Write the rule for the *n*^{th} term.

This gives two points \(\color{blue}{a_7 = 26}\) and \(\color{red}{a_{15} = 50}\). Substitute each point into the formula to obtain two equations to solve for *a*_{1} and *d*.

$$ a_n = a_1 + (n - 1)d $$

$$ \begin{align} \color{blue}{26} &= \color{blue}{a_1 + (7 - 1)d} \\ \color{red}{50} &= \color{red}{a_1 + (15 - 1)d} \end{align} $$

$$ \begin{align} \color{blue}{26} &= \color{blue}{a_1 + 6d} \\ \color{red}{50} &= \color{red}{a_1 + 14d} \end{align} $$

Solve the system of equations with something like elimination (or substitution or matrices or Cramer's Rule).

$$ \require{enclose} \begin{align} (-1)\color{blue}{26} &= \color{blue}{a_1 + 6d} \\ \color{red}{50} &= \color{red}{a_1 + 14d} \\ \enclose{top}{24} &= \enclose{top}{\quad \quad 8d} \end{align} $$

$$ d = 3 $$

$$ 26 = a_1 + 6d $$

$$ 26 = a_1 + 6(3) $$

$$ a_1 = 8 $$

Now write the rule for the *n*^{th} term.

$$ a_n = a_1 + (n - 1)d $$

$$ a_n = 8 + (n - 1)3 $$

$$ a_n = 8 + 3n -3 $$

$$ a_n = 3n + 5 $$

The 5^{th} term of an arithmetic sequence is 24, and the 9^{th} term is 44. Write the rule for the *n*^{th} term.

\(a_n = 5n - 1\)

Write the recursive rule for 4, 1, −2, −5, ….

The common difference appears to be −3, and the first term is 4. \(d = -3, a_1 = 4\)

$$ a_1 = a_1, a_{n} = a_{n-1} + d $$

$$ a_1 = 4, a_{n} = a_{n-1} - 3 $$

Write the recursive rule for 5, 7, 9, 11, ….

\(a_1 = 5, a_{n} = a_{n-1} + 2\)

What is the sum of all the odd numbers less than 20? Start listing the numbers.

$$ 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 $$

Bend the last half of the list under the first and add.

1 | + | 3 | + | 5 | + | 7 | + | 9 |

+ 19 | + | 17 | + | 15 | + | 13 | + | 11 |

20 | + | 20 | + | 20 | + | 20 | + | 20 |

5(20) = 100

Notice the first and last term added to 20. Also, notice that there were 10 terms, but the 20 was multiplied by 5 or \(\frac{10}{2}\). Several more tests could be done to get a pattern. It produces the formula for the sum of an arithmetic series.

$$ S_n = \frac{n}{2}\left(a_1 + a_n\right) $$

$$ S_n = \frac{n}{2}\left(a_1 + a_n\right) $$

where \(a_n = a_1 + (n-1)d\) and the lower limit is 1.

Find the sum of the integers 1 to 35.

This is an arithmetic series with *d* = 1 and *a _{n}* = 1. The last term is \(a_{35} = 35\), so

$$ S_n = \frac{n}{2}\left(a_1 + a_n\right) $$

$$ S_{35} = \frac{35}{2}\left(1 + 35\right) $$

$$ S_{35} = \frac{35}{2} (36) $$

$$ S_{35} = 630 $$

Find the 40^{th} partial sum of –10 + –7 + –4 + –1 + ….

$$ S_n = \frac{n}{2} \left(a_1 + a_n\right) $$

From the series, *d* = 3 and \(a_1 = -10\). *a _{n}* is not known, so use the explicit rule formula for the arithmetic sequence.

$$ a_n = a_1 + (n-1)d $$

$$ a_{n} = -10 + (n-1)(3) $$

$$ a_n = 3n - 13 $$

Because the problem asks for the 40th partial sum, *n* = 40.

$$ a_{40} = 3(40) - 13 = 107 $$

Now that all the variables are known, fill in the sum formula.

$$ S_n = \frac{n}{2} \left(a_1 + a_n\right) $$

$$ S_{40} = \frac{40}{2}\left(-10 + 107\right) $$

$$ S_{40} = 1940 $$

Find the sum of the first 20 terms of 8 + 7 + 6 + 5 + ….

–30

Evaluate \(\displaystyle \sum_{i=1}^{80} (4i - 20)\).

The explicit rule in this problem is 4*i* – 20 which is linear, so the problem is arithmetic. *n* = 80 because that is the upper limit. To find *a*_{1} and *a _{n}* plug the 1 and 80 into the \(4i - 20\).

$$ a_1 = 4(1) - 20 = -16 $$

$$ a_{80} = 4(80) - 20 = 300 $$

$$ S_{n} = \frac{n}{2} \left(a_1 + a_n\right) $$

$$ S_{80} = \frac{80}{2} \left(a_1 + a_{80}\right) $$

$$ S_{80} = \frac{80}{2} (-16 + 300) $$

$$ S_{80} = 11360 $$

Evaluate \(\displaystyle \sum_{i=1}^{100} (-2i + 16)\)

–8500

$$ a_n = a_1 + (n - 1)d $$

where *a*_{1} is the first term and *d* is the common difference

This simplifies to

$$ a_n = dn + c $$

where \(c = a_1 - d\).

This is **linear**, so any time an explicit rule for a sequence is linear, it is arithmetic.

$$ a_1 = a_1, a_{n} = a_{n-1} + d $$

$$ S_n = \frac{n}{2}\left(a_1 + a_n\right) $$

where \(a_n = a_1 + (n-1)d\) and the lower limit is 1.

Helpful videos about this lesson.

- 12, 14, 16, 18, 20, …
- 15, 20, 25, 30, 35, …
- –4, –1, 2, 5, 8, …
- The 6
^{th}term of an arithmetic sequence is 63, and the 10^{th}term is 107. - The 12
^{th}term of an arithmetic sequence is –80, and the 20^{th}term is –136. - 4, 10, 16, 22, 28, …
- 25, 12, –1, –14, –27, …
- –50, –29, –8, 13, 34, …
- \(16 + 15 + 14 + 13 + \cdots + -4\)
- Find the 13
^{th}partial sum: \(53 + 57 + 61 + 65 + 69 + \cdots\) - Find the 100
^{th}partial sum: \(-34 + -36 + -38 + -40 + -42 + \cdots\) - \(\displaystyle \sum_{i=1}^{15} (8i - 50)\)
- \(\displaystyle \sum_{n=1}^{20} (-4n + 3)\)
- \(\displaystyle \sum_{k=1}^{50} (12k - 1)\)
- You are saving money to buy your first car. You save $100 every month with a goal of $3000. (a) Write a rule for the
*n*^{th}term for the amount of money you have saved. (b) How many months until you have saved your $3000? (c) And what kind of car do you want? - (10-02) Evaluate \(\displaystyle \sum_{i=4}^{8} (2i + 10)\).
- (10-02) Evaluate \(\displaystyle \sum_{n=1}^{25} 2n^2\).
- (10-01) Write the first five terms of \(a_n = n^2 - n\).
- (9-06) Use Cramer's Rule to solve \(\left\{\begin{align} 2x - 3y &= -4 \\ 4x + 5y &= 14 \end{align}\right.\).
- (9-04) Use an inverse matrix to solve \(\left\{\begin{align} 3x - 3y &= 0 \\ 2x - y &= 2 \end{align}\right.\).

Write the rule for the *n*^{th} term.

Write the recursive rule for the sequence.

Find the sum of the series.

Problem Solving

Mixed Review

- \(a_n = 2n + 10\)
- \(a_n = 5n + 10\)
- \(a_n = 3n - 7\)
- \(a_n = 11n - 3\)
- \(a_n = -7n + 4\)
- \(a_1 = 4, a_{n} = a_{n-1} + 6\)
- \(a_1 = 25, a_{n} = a_{n-1} - 13\)
- \(a_1 = -50, a_{n} = a_{n-1} +21\)
- 126
- 1001
- –13300
- 210
- –780
- 15250
*a*= 100_{n}*n*; 30 months- 110
- 11050
- 0, 2, 6, 12, 20
- (1, 2)
- (2, 2)