Precalculus by Richard Wright

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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

10-03 Arithmetic Sequences and Series

Mr. Wright teaches the lesson.

Summary: In this section, you will:

SDA NAD Content Standards (2018): PC.7.2

Used Cars
Used cars. (Pixabay/MichaelGaida)

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.

Arithmetic Sequences

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 + 1(2), 1 + 2(2), 1 + 3(2), 1 + 4(2)

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

an = 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 nth term.

an = a1 + (n − 1)d

Explicit Rule for Arithmetic Sequences

an = a1 + (n − 1)d

where a1 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.

Recursive Rule for Arithmetic Sequence

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

Write an Explicit Rule for Arithmetic Sequence

Write the rule for the nth term for 4, 7, 10, 13, 16, ….

Solution

The first term is 4, so a1 = 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.

an = a1 + (n − 1)d

an = 4 + (n − 1)3

Simplify.

an = 4 + 3n − 3

an = 3n + 1

Write the rule for the nth term for 16, 12, 8, 4, 0, ….

Answer

an = −4n + 20

Find Explicit Rule Based on Two Terms

The 7th term of an arithmetic sequence is 26, and the 15th term is 50. Write the rule for the nth term.

Solution

This gives two points a7 = 26 and a15 = 50. Substitute each point into the formula to obtain two equations to solve for a1 and d.

an = a1 + (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 = a1 + 6d

26 = a1 + 6(3)

a1 = 8

Now write the rule for the nth term.

an = a1 + (n − 1)d

an = 8 + (n − 1)3

an = 8 + 3n − 3

an = 3n + 5

The 5th term of an arithmetic sequence is 24, and the 9th term is 44. Write the rule for the nth term.

Answer

an = 5n − 1

Write the Recursive Rule for an Arithmetic Sequence

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

Solution

The common difference appears to be −3, and the first term is 4. d = −3, a1 = 4

a1 = a1, an = an−1 + d

a1 = 4, an = an−1 − 3

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

Answer

a1 = 5, an = an−1 + 2

Arithmetic Series

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.

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

Sum of an Arithmetic Series

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

where an = a1 + (n − 1)d and the lower limit is 1.

Find the Sum of an Arithmetic Series

Find the sum of the integers 1 to 35.

Solution

This is an arithmetic series with d = 1 and an = 1. The last term is a35 = 35, so n = 35.

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

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

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

S35 = 630

Find the Sum of an Arithmetic Series

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

Solution

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

From the series, d = 3 and a1 = −10. an is not known, so use the explicit rule formula for the arithmetic sequence.

an = a1 + (n − 1)d

an = −10 + (n − 1)(3)

an = 3n − 13

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

a40 = 3(40) − 13 = 107

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

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

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

S40 = 1940

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

Answer

–30

Find the Sum of an Arithmetic Series

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

Solution

The explicit rule in this problem is 4i – 20 which is linear, so the problem is arithmetic. n = 80 because that is the upper limit. To find a1 and an plug the 1 and 80 into the 4i − 20.

a1 = 4(1) − 20 = −16

a80 = 4(80) − 20 = 300

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

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

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

S80 = 11360

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

Answer

–8500

Lesson Summary

Explicit Rule for Arithmetic Sequences

an = a1 + (n − 1)d

where a1 is the first term and d is the common difference

This simplifies to

an = dn + c

where c = a1d.

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

Recursive Rule for Arithmetic Sequence

a1 = a1, an = an − 1 + d


Sum of an Arithmetic Series

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

where an = a1 + (n − 1)d and the lower limit is 1.

Helpful videos about this lesson.

Practice Exercises

  1. Write the rule for the nth term.
  2. 12, 14, 16, 18, 20, …
  3. 15, 20, 25, 30, 35, …
  4. –4, –1, 2, 5, 8, …
  5. The 6th term of an arithmetic sequence is 63, and the 10th term is 107.
  6. The 12th term of an arithmetic sequence is –80, and the 20th term is –136.
  7. Write the recursive rule for the sequence.
  8. 4, 10, 16, 22, 28, …
  9. 25, 12, –1, –14, –27, …
  10. –50, –29, –8, 13, 34, …
  11. Find the sum of the series.
  12. 16 + 15 + 14 + 13 + ⋯ + −4
  13. Find the 13th partial sum: 53 + 57 + 61 + 65 + 69 + ⋯
  14. Find the 100th partial sum: −34 + −36 + −38 + −40 + −42 + ⋯
  15. \(\displaystyle \sum_{i=1}^{15} (8i - 50)\)
  16. \(\displaystyle \sum_{n=1}^{20} (-4n + 3)\)
  17. \(\displaystyle \sum_{k=1}^{50} (12k - 1)\)
  18. Problem Solving
  19. 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 nth 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?
  20. Mixed Review
  21. (10-02) Evaluate \(\displaystyle \sum_{i=4}^{8} (2i + 10)\).
  22. (10-02) Evaluate \(\displaystyle \sum_{n=1}^{25} 2n^2\).
  23. (10-01) Write the first five terms of an = n2n.
  24. (9-06) Use Cramer's Rule to solve \(\left\{\begin{align} 2x - 3y &= -4 \\ 4x + 5y &= 14 \end{align}\right.\).
  25. (9-04) Use an inverse matrix to solve \(\left\{\begin{align} 3x - 3y &= 0 \\ 2x - y &= 2 \end{align}\right.\).

Answers

  1. an = 2n + 10
  2. an = 5n + 10
  3. an = 3n − 7
  4. an = 11n − 3
  5. an = −7n + 4
  6. a1 = 4, an = an−1 + 6
  7. a1 = 25, an = an−1 − 13
  8. a1 = −50, an = an−1 + 21
  9. 126
  10. 1001
  11. –13300
  12. 210
  13. –780
  14. 15250
  15. an = 100n; 30 months
  16. 110
  17. 11050
  18. 0, 2, 6, 12, 20
  19. (1, 2)
  20. (2, 2)