We here at Shmoop R Us would be remiss if we didn't provide you with some real world information and advice regarding all of these sequences and series. This is especially true considering all the different applications they have. Besides, we know how boring the straightforward problems can be sometimes. This can be especially true when they decide to present themselves in the form of a monotone teacher who drones on in dreary classrooms day after day after day after day. So bombs away…

And if you'd like applications, and there are many more, don't be afraid to take a look our Best of the Web section.

A local radio station is holding a contest to help people get to know the newly elected mayor. The morning DJ is asking listeners to call in and name the new mayor's favorite movie. If the third caller does not name the movie correctly, the $100 prize money is increased by $50 for the next day. This $50 addition continues every morning until a caller answers correctly. If no one answers for 11 days, how much will the lucky caller win?

Solution: Let's be honest, this isn't a monster of a problem to tackle by hand. However, it might be quicker to use an arithmetic sequence. After all, an arithmetic sequence is created whenever we have the same amount added or subtracted over and over again.

Remember the explicit form for an arithmetic sequence looks like

{*a _{n}*} =

In our situation, the first prize was $100 so that is *a*_{1}. The common difference is $50, which we should put in for *d*.

{*a _{n}*} = 100 + 50(

Since we would like to find the prize money after 11 days, we can just throw 11 in for *n* and calculate.

*a*_{11 }= 100 + 50(11 – 1)*a*_{11 }= 100 + 50(10)*a*_{11 }= 100 + 500*a*_{11 }= 600

So the lucky winner earned a healthy sum of $600 for calling in and randomly guessing *Dumb and Dumber*. What can we say? The mayor has great taste.

The rival radio station is holding a contest to help people get to know the newly elected mayor as well. They're competitors, what do you expect? *Their* afternoon DJ is asking listeners to call in and name the newly mayor's favorite song. If the fifth caller does not name the song correctly, the $5 prize money is doubled for the next day. This doubling continues every morning until a caller answers correctly. If no one answers for 11 days, which station has to pay out more cash to the winner?

Solution: A caller on day one would win $5. A caller on day two would win $10. Day three: $20. All of these are barely worth the phone call. But, 11 days…

We could go on and on doubling the prize money by hand until the buttons on our calculators fall off. Or, we could create a rule for geometric sequence created by this potentially lucrative situation.

This time we're going to need the general form for a geometric sequence:

{*a _{n}*} =

The first $5 prize can be substituted for *a*_{1} while the common ratio, *r*, is 2. This means we have an equation that looks like…

{*a _{n}*} = 5(2)

Since we are after the prize money on the 11^{th} day, 11 once again goes in for *n*.

*a*_{11 }= 5(2)^{11-1}*a*_{11 }= 5(2)^{10}*a*_{11 }= 5(1,024)*a*_{11 }= 5,120

Ouch! $5,120 later the marketing intern at the rival radio station has resigned for personal reasons. How was anyone supposed to know the mayor was a closet Justin Bieber fan?

You've just graduated from a fine institution and as a graduation present received a cool $1,000 cash from a really eccentric, very rich great aunt. Score. The kicker is she made her money via her impressive investment portfolio. Auntie Em has taken it upon herself to invest this $1,000. She has guaranteed you 8% interest over the next 20 years at which time they money will be yours.

How much will the money be worth when you finally get your hands on it?

Solution: Lets start with what we can figure out easily. Our interest here is going to compound annually so at the end of the first year the gift will be worth…

$1,000 + $1,000 (8%) =

$1,000 + $1,000 (0.08) =

$1,000 + $80 =

$1,080

At the end of the second year the gift will be worth…

$1,080 + $1,080 (8%) =

$1,080 + $1,080 (0.08) =

$1,080 + $86.4 =

$1,116.4

The third year…

$1,116.4 + $1,116.4 (8%) =

$1,116.4 + $1,116.4 (0.08) =

$1,116.4 + $89.31 =

$1,205.71

Things are looking up. Unfortunately, while the gift is meant to teach us patience we have none at this point. We certainly don't have enough to do this process 20 times.

Luckily, we have a bit of a pattern here. We have gone from a gift worth $1,000 to $1,080, $1,116.4, and then $1,206.71. Dividing any of these values by the previous years values yield a ratio of 1.08. This means that $1,206.71 is really just $1,000(1.08)(1.08)(1.08). We simply need to apply this increase 20 times.

Writing this is a geometric sequence we have:

*a _{n}* =

a

Notice that we will have to go to *n *= 21 for our solution because *n* = 1 actually gives us our initial value. This isn't a big deal and simply means the years are just shifted by one in our equation.

*a*_{21} = 1000(1.08)^{21-1}*a*_{21} = 1000(1.08)^{21-1}*a*_{21} = 1000(1.08)^{20}*a*_{21} = $4,660.96

Patience is indeed a virtue…if by virtue we mean worth exactly $3,660.96.

A ball is bounced from a height of 10 feet. Each time it bounces it travels half of its original height. What total distance does the ball travel?

Solution: Discounting the drop, the ball travels 10 total feet on the first bounce. It then travels 5 total feet on the second and 2.5 on the third. Be careful. Each of these bounces is accounting for the distance the ball travels going up and then down again. So the bounces look like this:

*b*_{1} = 10*b*_{2} = 5*b*_{3} = 2.5*b*_{4} = 1.25

…

Maybe it's just us, but this looks like it could be a geometric series with a first term of 10 and a common ratio of ½. That means we have a series that looks like…

All we have to do now is find the sum and then add in the original drop. Since the sum of an infinite geometric series is given by we can write the following:

Of course, 20 feet + 10 feet = 30 feet. This means the ball travels a total of 30 feet in the air. This is really Shmooptacular considering the fact that the ball will never *really* stop bouncing. After all, we can always cut a number in half.

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