Sequences Exercises

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 2ⁿ and by 1000, we have

1000 < 2ⁿ

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 a₉ and a₁₀.

It takes 10 bites before less than one one-thousandth of the cookie 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

a₈ = (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.

Answer

This problem describes an arithmetic sequence. The dress initially costs a₁ = 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

a₈ = 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 a₁₃ = 51 and a₁₄ = 49.

Answer

This problem describes an arithmetic sequence. During her 1st week Jenny bikes a₁ = 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.

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

a₅ = (.3)⁵ = .00243

or 0.243 %.

(b) The question is asking for the smallest n with a_n < 0.01. 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.