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# Visual Patterns

## Describe non-arithmetic sequences by finding a rule.

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Finding the Next Term in a Sequence

You drop a rubber ball from a height of 48 inches. Each time it bounces, it reaches lower and lower heights. The following sequence shows its height with each successive bounce. How high will the ball bounce on its fifth bounce?

48, 36, 27, 20.25,...

### Sequence of Numbers

When looking at a sequence of numbers, consider the following possibilities.

• There could be a common difference (the same value is added or subtracted) to progress from each term to the next.

Example: 5,8,11,14,\begin{align*}5, 8, 11, 14, \ldots\end{align*} (add 3)

• There could be a common ratio (factor by which each term is multiplied) to progress from one term to the next.

Example: 9,3,1\begin{align*}9, 3, 1\end{align*}, 13\begin{align*}\frac{1}{3} \ldots\end{align*} (\begin{align*}\left ( \right .\end{align*} multiply by 13)\begin{align*}\left . \frac{1}{3} \right )\end{align*}

• If the terms are fractions, perhaps there is a pattern in the numerator and a different pattern in the denominators.

Example: 19,38,57,76,\begin{align*}\frac{1}{9}, \frac{3}{8}, \frac{5}{7}, \frac{7}{6}, \ldots\end{align*} (numerator (+2), denominator (-1))

• If the terms are growing rapidly, perhaps the difference between the term values is increasing by some constant factor.

Example: 2,5,9,14,\begin{align*}2, 5, 9, 14, \ldots\end{align*} (add 3, add 4, add 5, ...)

• The terms may represent a particular type of number such as prime numbers, perfect squares, cubes, etc.

Example: 2,3,5,7,\begin{align*}2, 3, 5, 7, \ldots\end{align*} (prime numbers)

• Consider whether each term is the result of performing an operation on the two prior terms.

Example: 2,5,7,12,19,\begin{align*}2, 5, 7, 12, 19, \ldots\end{align*} (add the previous two terms)

• Consider the possibility that the value is connected to the term number:

Example: 0,2,6,12,\begin{align*}0, 2, 6, 12, \ldots\end{align*}

In this example (0×1)=0,(1×2)=2,(2×3)=6,(3×4)=12,\begin{align*}(0 \times 1) = 0, (1 \times 2) = 2, (2 \times 3) = 6, (3 \times 4) = 12, \ldots\end{align*}

This list is not intended to be a comprehensive list of all possible patterns that may be present in a sequence but they are a good place to start when looking for a pattern.

Let's find the next two terms in the following sequences.

1. 160,80,40,20,,\begin{align*}160, 80, 40, 20, \underline{\;\;\;\;\;\;\;}, \underline{\;\;\;\;\;\;\;}\end{align*}

Each term is the result of multiplying the previous term by 12\begin{align*}\frac{1}{2}\end{align*}. Therefore, the next terms are:

12(20)=10\begin{align*}\frac{1}{2}(20)=10\end{align*} and 12(10)=5\begin{align*}\frac{1}{2}(10)=5\end{align*}

1. 0,3,7,12,18,,\begin{align*}0, 3, 7, 12, 18, \underline{\;\;\;\;\;\;\;}, \underline{\;\;\;\;\;\;\;}\end{align*}

The difference between the first two terms (30)\begin{align*}(3-0)\end{align*} is 3, the difference between the second and third terms (73)\begin{align*}(7-3)\end{align*} is 4, the difference between the third and fourth terms (127)\begin{align*}(12-7)\end{align*} is 5 and the difference between the fourth and fifth terms (1812)\begin{align*}(18-12)\end{align*} is 6. Each time we add one more to get the next term. The next difference will be 7, so 18+7=25\begin{align*}18+7=25\end{align*} for the sixth term. To get the seventh term, we add 8, so 25+8=33\begin{align*}25+8=33\end{align*}.

1. 9,5,4,1,3,,\begin{align*}9, 5, 4, 1, 3, \underline{\;\;\;\;\;\;\;}, \underline{\;\;\;\;\;\;\;}\end{align*}

This sequence requires that we look at the previous two terms. To get the third term, the second term was subtracted from the first: 95=4\begin{align*}9-5=4\end{align*}. To get the fourth term, the third term is subtracted from the second: 54=1\begin{align*}5-4=1\end{align*}. Similarly: 41=3\begin{align*}4-1=3\end{align*}. Now, to get the next terms, continue the pattern:

13=2\begin{align*}1-3=-2\end{align*} and 3(2)=5\begin{align*}3-(-2)=5\end{align*}

### Examples

#### Example 1

Earlier, you were asked to find how high the ball will bounce on its fifth bounce.

Each successive term in the sequence is the result of multiplying the previous term by \begin{align*}\frac{3}{4}\end{align*}. Therefore, the next term, the fifth, is:

\begin{align*}\frac{3}{4}(20.25)=15.1875\end{align*}.

Therefore, the ball reaches a height of 15.1875 inches on its fifth bounce.

Find the next two terms in each of the following sequences.

#### Example 2

\begin{align*}-5, -1, 3, 7, \underline{\;\;\;\;\;\;\;}, \underline{\;\;\;\;\;\;\;}\end{align*}

Each term is the previous term plus 4. Therefore, the next two terms are 11 and 15.

#### Example 3

\begin{align*}\frac{1}{3}, \frac{2}{3}, \frac{7}{9}, \frac{5}{6} \underline{\;\;\;\;\;\;\;}, \underline{\;\;\;\;\;\;\;}\end{align*}

The pattern here is somewhat hidden because some of the fractions have been reduced. If we “unreduced” the second and fourth terms we get the sequence: \begin{align*}\frac{1}{3}\end{align*}, \begin{align*}\frac{4}{6}\end{align*}, \begin{align*}\frac{7}{9}\end{align*},\begin{align*}\frac{10}{12}, \underline{\;\;\;\;\;\;\;}, \underline{\;\;\;\;\;\;\;}\end{align*}. Now the pattern can be observed to be that the numerator and denominator each increase by 3. So the next two terms are \begin{align*}\frac{13}{15}\end{align*} and \begin{align*}\frac{16}{18}\end{align*}. Reducing the last term gives us the final answer of \begin{align*}\frac{13}{15}\end{align*} and \begin{align*}\frac{8}{9}\end{align*}.

#### Example 4

\begin{align*}1, 4, 9, 16, \underline{\;\;\;\;\;\;\;}, \underline{\;\;\;\;\;\;\;}\end{align*}

This sequence is the set of perfect squares or the term number squared. Therefore the \begin{align*}5^{th}\end{align*} and \begin{align*}6^{th}\end{align*} terms will be \begin{align*}5^2=25\end{align*} and \begin{align*}6^2=36\end{align*}.

### Review

Find the next three terms in each sequence.

1. \begin{align*}15, 21, 27, 33, \underline{\;\;\;\;\;\;\;}, \underline{\;\;\;\;\;\;\;}, \underline{\;\;\;\;\;\;\;}\end{align*}
2. \begin{align*}-4, 12, -36, 108, \underline{\;\;\;\;\;\;\;}, \underline{\;\;\;\;\;\;\;}, \underline{\;\;\;\;\;\;\;}\end{align*}
3. \begin{align*}51, 47, 43, 39, \underline{\;\;\;\;\;\;\;}, \underline{\;\;\;\;\;\;\;}, \underline{\;\;\;\;\;\;\;}\end{align*}
4. \begin{align*}100, 10, 1, 0.1, \underline{\;\;\;\;\;\;\;}, \underline{\;\;\;\;\;\;\;}, \underline{\;\;\;\;\;\;\;}\end{align*}
5. \begin{align*}1, 2, 4, 8, \underline{\;\;\;\;\;\;\;}, \underline{\;\;\;\;\;\;\;}, \underline{\;\;\;\;\;\;\;}\end{align*}
6. \begin{align*}\frac{7}{2}, \frac{5}{3}, \frac{3}{4}, \frac{1}{5}, \underline{\;\;\;\;\;\;\;}, \underline{\;\;\;\;\;\;\;}, \underline{\;\;\;\;\;\;\;}\end{align*}

Find the missing terms in the sequences.

1. \begin{align*}1, 4, \underline{\;\;\;\;\;\;\;}, 16, 25, \underline{\;\;\;\;\;\;\;}\end{align*}
2. \begin{align*}\frac{2}{3}, \frac{3}{4}, \underline{\;\;\;\;\;\;\;}, \frac{5}{6}, \underline{\;\;\;\;\;\;\;}\end{align*}
3. \begin{align*}0, 2, \underline{\;\;\;\;\;\;\;}, 9, 14,\underline{\;\;\;\;\;\;\;}\end{align*}
4. \begin{align*}1, \underline{\;\;\;\;\;\;\;}, 27, 64, 125, \underline{\;\;\;\;\;\;\;}\end{align*}
5. \begin{align*}5, \underline{\;\;\;\;\;\;\;}, 11, 17, 28, \underline{\;\;\;\;\;\;\;}, 73\end{align*}
6. \begin{align*}3, 8, \underline{\;\;\;\;\;\;\;}, 24, \underline{\;\;\;\;\;\;\;}, 48\end{align*}
7. \begin{align*}1, 1, 2, \underline{\;\;\;\;\;\;\;}, 5, \underline{\;\;\;\;\;\;\;}, 13\end{align*}
8. Do any of the problems above have a constant difference? If so, which ones and what is the constant?
9. Do any of the problems above have a common ratio? If so, which ones and what is the ratio?

To see the Review answers, open this PDF file and look for section 11.1.

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