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Introduction to Sequences - At A Glance:
Sequences, especially arithmetic and geometric ones, are good for word problems.
Sequence story problems come in two main flavors. If these flavors were ice cream, they'd be vanilla and rocky road. We may have to find
- the value of a particular term a_{n}. This is the standard vanilla problem.
- the value of n at which the terms do something in particular. This is the more complicated, rocky road problem.
In general, it's good strategy to write out the first few terms of the sequence in question so we can see the pattern of the terms. Maybe we can do it with ice cream cone in hand.
Sample Problem
An old story goes that a peasant won a reward from the king, and asked for rice: one grain to be placed on the first square of a chessboard, two grains on the second square, four on the third square, and so on. Each square was to contain double the number of grains on the previous square.
- How many grains of rice would be on the 32nd square?
- Which square would contain exactly 512 grains of rice?
Answer.
If we look at the first few terms, we can see the pattern:
The n^{th} square contains 2^{n – 1} grains of rice. We have
a_{n} = 2^{n – 1}
where n is the square and a_{n} is how many grains of rice are on that square. Now we're prepared to answer the questions.
- The question "How many grains of rice would be on the 32nd square?" is asking us to find the value of the 32nd term. No problem:
a_{32} = 2^{31} = 2,147,483,648.
- The question "Which square would contain exactly 512 grains of rice?" is asking us what value of n makes a_{n} = 512. We use the formula we have for a_{n} and solve for n:
512 = a_{n}
= 2^{n – 1}
log _{2} 512 = n – 1
9 = n – 1
n = 10
This means the 10th square would contain exactly 512 grains of rice.
Be Careful: One type of sequence problem asks us to find a value of a_{n}. Another asks us to find a value of n. We should be sure to provide the correct information in our answer.
Sometimes when we're asked to find a value of n, we might solve an equation or inequality and get a value of n that isn't a whole number. This is where the road can get bumpy, so we could take a lick of our rocky road cone, and then we'd try out the whole numbers to either side and see which gives a better answer.
Example 1
Liana runs 0.25 miles on January 1st. She runs 0.5 miles farther each day. - How far does Liana run on January 8th?
- What is the first day on which Liana runs 5 miles?
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This story problem describes an arithmetic sequence where n is the day in January and a_{n} is how many miles Liana runs that day. The first term is how many miles Liana runs on the first day: a_{1} = 0.25 The difference is the number of miles by which she increases her run each day: d = 0.5 Since this is an arithmetic sequence, the n^{th} term is a_{n} = a_{1} + (n – 1)d. This is the distance Liana runs on the n^{th} day in January. 1. On January 8th, the 8th day of January, Liana runs a_{8} miles. We use the formula to find a_{8}. a_{8} = a_{1} + (7)d = 0.25 + (7)(0.5) = 3.75 On the 8th day of January, Liana runs 3.75 miles. 2. We want to know the first day on which Liana runs 5 miles. This means we want to know the smallest value of n for which a_{n} ≥ 5. Using the formula for an arithmetic sequence, we want to know for which n we first have a_{1} + (n – 1)d ≥ 5 We put in the values of a_{1} and d and solve the inequality: 0.25 + (n – 1)(.5) ≥ 5 (n – 1)(.5) ≥ 4.75 n – 1 ≥ 9.5 n ≥ 10.5 A sequence doesn't have a 10.5th term, so that can't be the right answer. Let's find a_{10} and a_{11} and see which is a better choice. a_{10} = 0.25 + (9)(.5) = 4.75 a_{11} = 0.25 + (10)(.5) = 5.25 On January 10th Liana runs 4.75 miles, so she didn't run 5 miles that day. On January 11th she ran more than 5 miles. January 11th is the first day she runs 5 miles. The best advice we can give for story problems is to write out the first few terms of the sequence so you can see the pattern. Like avoiding brain freeze from eating ice cream too fast, the only way to get better is to practice. | |
Exercise 1
A giant cookie sits on a plate. Cookie Monster eats half the cookie with one bite. With another bite he eats half of the remaining cookie.
Cookie Monster keeps taking bites, eating half the remaining cookie with each bite.
(a) After 5 bites, what fraction of the cookie has Cookie Monster eaten?
(b) How many bites does it take before less than one one-thousandth of the cookie remains?
Answer
After one bite, half the cookie remains. After two bites, one-fourth of the cookie remains. With each bite, the remaining portion of cookie gets halved.
Let's put this information in a table:
After the n^{th} bite, the fraction of cookie remaining is .
(a) After 5 bites, the fraction of cookie remaining is
This means Cookie Monster has eaten of the cookie.
(b) We want to find the first value of n for which
Since , we need to solve the inequality
Multiplying both sides by 2^{n} and by 1000, we have
1000 < 2^{n}
log _{2} 1000 < n
9.966 < n
In order for n to be a whole number greater than 9.966, we must have n = 10. To make sure, we can check the values of a_{9} and a_{10}.
It takes 10 bites before less than one one-thousandth of the cookie remains.
Exercise 2
A dress is listed at $150. At the end of each week the price is reduced by 10%.
(a) What is the price of the dress after 8 weeks?(b) After how many weeks will the dress be less than $50?
(hint: If 10% of the price is taken off, then 90% of the price remains.)
Answer
After one week, the price will be reduced by 10%. This means the dress will cost 90% of its original price, so it will cost
0.9(150) = 135.
After two weeks the price will be reduced by 10% again, so the dress will cost
0.9(135) = (0.9)(0.9)(150).
Continuing in this fashion, after n weeks the price of the dress will be
a_{n} = (.9)^{n}(150).
(a) After 8 weeks, the price of the dress will be
a_{8} = (0.9)^{8}(150) = 64.57.
The dress will cost $64.57.
(b) This question is asking for what value of n we have a_{n}<50.
At this point it's tempting to take logs of both sides.
If we do that, we get
n < 10.4.
That doesn't make sense, because we're looking for a statement of the form
n > ...
The weirdness is because is negative, and negative signs do funny things to inequalities. To avoid the negatives, let's rearrange the fractions before taking the log.
Since n must be a whole number, we can conclude that after n = 11 weeks the dress will be less than $50.
Exercise 3
A dress is listed at $75. At the end of each week the price is reduced by $2.
(a) What is the price of the dress after 8 weeks?(b) After how many weeks will the dress be less than $50?
Answer
This problem describes an arithmetic sequence. The dress initially costs a_{1} = 75. Each week the price is reduced by $2, so d = -2. After n weeks the price of the dress will be
a_{n} = 75 + (n – 1)(-2).
(a) After 8 weeks the price of the dress is
a_{8} = 75 + (7)(-2) = 61.
The dress will be $61.
(b) We want to find the first n for which a_{n}<50.
75 + (n – 1)(-2) < 50
75 – 2n + 2 < 50
27 < 2n
13.5 < n
After 14 weeks the dress will be less than $50. We could confirm this answer by calculating a_{13} = 51 and a_{14} = 49.
Exercise 4
During her first week of biking, Jenny bikes 3km per day. Each week, she bikes 3km farther per day than she did the previous week.
(a) How many weeks does it take Jenny to get up to 15km per day?(b) How far does Jenny bike on her 31st day?
Answer
This problem describes an arithmetic sequence. During her 1st week Jenny bikes a_{1} = 3 km per day. Each week she adds d = 3 km to her distance. During her n^{th} week she bikes
a_{n} = 3 + (n – 1)(3)km per day.
(a) This question is asking for the smallest n with a_{n} = 15.
15 = 3 + (n – 1)3
12 = (n – 1)3
4 = n – 1
5 = n
During her 5th week of biking, Jenny will go 15 km per day.
(b) This is a slightly sneaky question, since it's phrased in terms of days instead of weeks. Her 31st day will be during her 5th week of biking, so Jenny will go 15 km that day.
Exercise 5
A particular drug decays in the system so that 1 day after taking a dose, 30% of the drug remains in the bloodstream. Paco takes one pill.
(a) What percent of the drug remains in Paco's bloodstream 5 days after taking one dose?(b) How many days before less than 1% of the drug remains in Paco's bloodstream?
Answer
This problem describes a geometric sequence. One day after taking a dose, 30% of the drug remains. Two days after, 30% of that 30%, or (.3)^{2}, remains. After n days, (.3)^{n} of the drug remains.
(a) After 5 days, the portion of the drug in Paco's bloodstream is
a_{5} = (.3)^{5} = .00243
or 0.2 %.
(b) The question is asking for the smallest n with a_{n}<.01. As with question 2(b), we rearrange numbers before taking logs.
After 4 days, less than 1% of the drug will remain in Paco's bloodstream.