Precalculus by Richard Wright
Are you not my student and
has this helped you?
This book is available
to download as an epub.
Precalculus by Richard Wright
Are you not my student and
has this helped you?
This book is available
to download as an epub.
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:
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 + 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
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.
$$ a_1 = a_1, a_{n} = a_{n-1} + d $$
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
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 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
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 an = a1 + (n − 1)d and the lower limit is 1.
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 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
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
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 = a1 − d.
This is linear, so any time an explicit rule for a sequence is linear, it is arithmetic.
a1 = a1, an = an − 1 + d
$$ S_n = \frac{n}{2}\left(a_1 + a_n\right) $$
where an = a1 + (n − 1)d and the lower limit is 1.
Helpful videos about this lesson.