- Topics At a Glance
- Sequences
- Defining Sequences and Evaluating Terms
- Patterns
- Sequences Can Start at
*n*= 0 - Special Types of Sequences
- Arithmetic Sequences
- Geometric Sequences
- Comparing Arithmetic and Geometric Sequences
- Visualizing Sequences
- 2-D Graphs
- Convergence and Divergence of Sequences
- Other Useful Sequence Words
- Heads and Tails
- Word Problems
**In the Real World**- I Like Abstract Stuff; Why Should I Care?
**How to Solve a Math Problem**

There are three steps to solving a math problem.

- Figure out what the problem is asking.

- Solve the problem.

- Check the answer.

Some of the world's most expensive television sets cost $140,000 each.

If you deposit $400 in the bank and earn 10% interest every year, how many years will it be before you're able to buy this television set?

Answer.

1. Figure out what the problem is asking.

The problem is describing a geometric sequence. If you earn 10% interest per year, each year you'll have 1.1 times as much money as you did the previous year.

After one year you'll have

400(1.1),

after two years you'll have

400(1.1)(1.1),

and after *n* years you'll have

*a _{n}* = 400(1.1)

We want to know what value of *n* makes *a _{n}* ≥ 140,000.

2. Solve the problem.

Solve the inequality.

140,000 < *a _{n}*

140,000 < 400(1.1)^{n}

350 < (1.1)^{n}

log _{1.1}350 < *n*

61.46 < *n*

It would take 62 years before you had enough money to buy the television.

3. Check the answer.

We'll check the answer by finding *a*_{61} and *a*_{62}.

After 61 years, the amount of money you have is

*a*_{61} = 400(1.1)^{61} = 133971.92.

That's not quite enough. After 62 years you have

*a*_{62} = 400(1.1)^{62} = 147369.11.

That is enough.