# Geometric Sequence

TOPIC: GEOMETRIC SEQUENCE/PROGRESSION

Sub-topics:

• Geometric Means
• Mean Proportional

LEARNING OBJECTIVES:

• To recognize geometric sequence/progression
• To solve for the geometric means and mean proportional
• To find the nth term of a given geometric sequence
• To apply the concepts of geometric sequence to solve real-life problems

MATH CONCEPTS

• GEOMETRIC SEQUENCE/PROGRESSION – It is a type of sequence where each term has a common ratio. After the first term, the succeeding terms are generated by multiplying to a constant number.
• COMMON RATIO- The common ratio, usually denoted as ‘r’, is calculated by dividing any term by the term preceding it.
• FINDING THE NTH TERM OF A GEOMETRIC SEQUENCE $a_{n}=a_{1}r^{n-1}$ where $a_{n}$ is the value of the nth term, $a_{1}$ is the first term, r is the common ratio, and n is the position of the term.
• GEOMETRIC MEANS – These are the terms between any two given terms of a geometric sequence.
• MEAN PROPORTIONAL – A term between two terms of a geometric sequence. To get this value, find the product of the two terms and get the square root of the answer.

IMPORTANCE

• Geometric sequence can enhance the ability of the brain to look for common patterns among numbers.
• Like any other type of sequence, mastering this topic can be a good foundation in understanding functions.

DISCUSSION

GEOMETRIC SEQUENCE

Consider the sequence 3, 12, 48, 192, … Obviously, the value of the terms are increasing and the terms are not increasing randomly but in a specific order. Notice that after the first term, 3, the succeeding terms are generated by multiplying it by 4.

The aforementioned number pattern is a good example of geometric sequence. Geometric sequence has a general form $a,\, ar,\,ar^{2},\,ar^{3},...,\,ar^{n-1}$, where a  is the first term, r  is the common ratio, and n refers to the position of the nth term.

Thus, the sequence 3, 12, 48, 192, 768, 3072, … can be expressed as:

$3,\,(3\times 4),\,(3\times 4)^{2},\,(3\times 4^{3}),\,(3\times 4^{4}),\,(3\times 4^{5}),\,...,\,(3\times 4^{n-1})$

COMMON RATIO

If an arithmetic sequence has the concept of common difference, a geometric sequence has a common ratio. Remember that a geometric sequence always has a common ratio. If the common ratio is not present, then the given sequence is not a geometric one. Common ratio can be obtained by simply dividing the current term to the previous term. In general,

$r=\frac{a_{2}}{a_{1}}=\frac{a_{3}}{a_{2}}=\frac{a_{4}}{a_{3}}=\frac{a_{5}}{a_{4}}=...=\frac{a_{n}}{a_{n-1}}$

For example, let’s have the sequence 7, 14, 28, 56, 112, 224, 448, ….

$r=\frac{14}{7}=\frac{28}{14}=\frac{56}{28}=\frac{112}{224}=\frac{448}{224}=2$

ILLUSTRATIVE EXAMPLES

1) Determine if the following given is an example of geometric sequence.

1. 6, 12, 24, 48, 96, …
2. $\frac{1}{2500},\,\frac{1}{500},\,\frac{1}{100},\,\frac{1}{20},...$
3. $\frac{3}{25},\,\frac{5}{27},\,\frac{7}{29},\,\frac{9}{31}, ...$
4. $2x,\,4x^{3},\,8x^{5},\,16x^{7},\,32x^{9},\,...$

SOLUTIONS:

DISCUSSION

FINDING THE NTH TERM OF A GEOMETRIC SEQUENCE

One of the important skills that we should learn about is finding the nth term of a geometric sequence.  The formula is $a_{n}=a_{1}r^{n-1}$ where $a_{n}$ is the value of the nth term, $a_{1}$ is the first term, r is the common ratio, and n is the position of the term. Remember that appropriate identification of each element is needed.

ILLUSTRATIVE EXAMPLES

Solve for the specified term of each geometric sequence.

1. Write the first 6 terms of the geometric sequence whose $a_{1}=1$ and $r=1/2$
2. $a_{1}=6,\,r=\frac{1}{3}$. Find $a_{12}$.
3. 3, 12, 48,…, find the 9th term.
4. 2, -6, 18, -54,…, find the 10th term.

SOLUTIONS:

1) Using the given condition, we just need to list down the first 6 terms. Simply multiply the first term to the common ratio which is ½ then repeat the same process until the 6th term is obtained.

1, 1/2, 1/4, 1/8, 1/16, 1/32

2) Use the formula: $a_{n}=a_{1}r^{n-1},\,a_{1}=6,\,r=\frac{1}{3},\,n=12$

$a_{12}=(6)(1/3)^{12-1}=(6)(1/3)^{11}$

$a_{12}=3.38702\times 10^{-5}\,or \, 0.0000338702$

3) Use the formula: $a_{n}=a_{1}r^{n-1},\,a_{1}=3,\,r=4,\,n=9$

$a_{9}=(2)(4)^{9-1}=(2)(4)^{8}=(3)65536$

$a_{9}=196,608$

4) Use the formula: $a_{n}=a_{1}r^{n-1},\,a_{1}=2,\,r=-3,\,n=10$

$a_{10}=(2)(-3)^{10-1}=(2)(-3)^{9}=(2)(-19683)$

$a_{10}=-39,366$

DISCUSSION

GEOMETRIC MEANS AND MEAN PROPORTIONAL

Geometric means are the terms between any two given terms of a geometric sequence while mean proportional is a term between two terms of a geometric sequence. To solve for the value of mean proportional, simply find the product of the two terms and get the square root of the answer.

ILLUSTRATIVE EXAMPLES

A) Find the mean proportional for each pair of terms.

1. 6 and 24
2. 3 and 49
3. 1.5 and 7.5

SOLUTIONS:

$MP = \sqrt{ab}$, where a is the first term and b is the last term.

1. $MP = \sqrt{(6)(24)}=\sqrt{144}=\pm 12$
2. $MP = \sqrt{(3)(49)}=\sqrt{147}=\pm 7\sqrt{3}$
3. $MP = \sqrt{(1.5)(7.5)}=\sqrt{11.25}=\pm \frac{3\sqrt{5}}{2}$

B) Insert the indicated number of geometric means between the two given terms.

1. Four terms between 3 and 48.
2. Two terms between -12 and 96.

Solutions:

1) We need to insert four geometric terms between 3 and 48. Listing it down,

3, __ , __, __, __, 48

$a_{1},\, a_{2},\,a_{3},\,a_{4},\, a_{5},\,a_{6}$

Let $a_{6}=48,\,a_{1}=3,\,r=?,\,n=6$

Use the formula: $a_{n}=a_{1}r^{n-1}$

$a_{6}=a_{1}r^{6-1}=3r^{6-1}$

$48= 3r^{5}$   (divide both sides by 3)

$16= r^{5}$   (extract the fifth root of 16)

$r\approx 1.7411$

To complete the terms in the sequence, 3 x 1.7411 = 5.2233, 5.2233 x 1.7411 = 9.0943, 9.0943 x 1.7411 = 15.8341, 15.8341 x 1.7411 = 27.5687

The complete terms are 3, 5.22, 9.09, 15.83, 27. 57, 48.

2) We need to insert two geometric terms between -12 and 96. Listing it down,

-12, __, __, 96

$a_{1},\, a_{2},\,a_{3},\,a_{4}$

Let $a_{4}=96,\,a_{1}=-12,\,r=?,\,n=4$

Use the formula: $a_{n}=a_{1}r^{n-1}$

$a_{4}=a_{1}r^{4-1} =(-12) r^{4-1}$

$96=(-12) r^{3}$  (divide both sides by -12)

$-8=r^{3}$   extract the cube root of -8)

To complete the terms in the sequence, -12 x -2 = 24, 24 x -2 = -48.

The complete terms are -12, 24, -48, 96

1) CELLULAR TELEPHONES In 1990 the average monthly bill for cellular telephone service in the United States was $80.90. From 1990 through 1997, the average monthly bill decreased by about 8.6% per year. Source: Statistical Abstract of the United States 1. Write a rule for the average monthly cellular telephone bill $a_{n}$ (in dollars) in terms of the year. Let n = 1 represent 1990. 2. What was the average monthly cellular telephone bill in 1993? 3. When did the average monthly cellular telephone bill fall to$50?

Solutions:

1) Because the average monthly bill decreased by the same percent each year, the average monthly bills from year to year form a geometric sequence. Use $a_{1}=80.9$ and r = 1 – 0.086 = 0.914.

A rule for the average monthly bill is: $a_{n}=80.9(0.914)^{n-1}$

2) In 1993, n = 4. So, the average monthly bill was

$a_{4}=80.9(0.914)^{4-1} =80.9(0.914)^{3}$

$a_{4}\approx 61.77$

3) You want to find n such that $a_{n}=80.9(0.914)^{n-1}$

$80.9(0.914)^{n-1} = 50$

$(0.914)^{n-1} \approx 0.618$

$n-1=\frac{\log 0.618}{\log 0.914}$

$n \approx 6$

The average monthly cellular telephone bill reached \$50 in 1995 (when n = 6).

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