Algebra 1 · F.BF.A.1A · F.BF.A.2

Geometric sequences and the explicit formula

A ball bounces lower each time, always by the same fraction. How high is the eighth bounce, without working out the seven before it?

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Geometric sequences and the explicit formula

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When you reach for it

A phone tree passes news along fast. In round 1 the organizer calls 3 people. In round 2 each of those 3 calls 3 more, so 9 people are called. In round 3 each of those 9 calls 3 more, so 27 are called. How many are called in round 8?

You could keep multiplying by 3 until you reach round 8. The numbers 3, 9, 27, 81, \ldots form a geometric sequence, and its explicit formula gives any term straight from its position n, with no need to build the terms before it.

Reach for it whenever a quantity is multiplied by the same number at every step: a ball that rebounds to the same fraction of its last height, a yeast culture whose mass doubles every hour, a message that each person passes to the same number of new people. It is the multiplying twin of the arithmetic sequence, where the same number is added at every step.

The procedure

The common ratio

A geometric sequence is a sequence in which each term after the first is the term before it multiplied by one fixed number. That number is the common ratio, written r. Since a_{n+1} = a_n \cdot r, dividing both sides by a_n gives

r = \frac{a_{n+1}}{a_n}

so the ratio is any term divided by the one before it, and for a geometric sequence every such quotient is the same number. It can be a whole number, a fraction, a decimal or negative:

Sequence	First quotient	r
2, 14, 98, 686	\displaystyle \frac{14}{2}	7
80, 40, 20, 10	\displaystyle \frac{40}{80}	\displaystyle \frac{1}{2}
1000, 1100, 1210, 1331	\displaystyle \frac{1100}{1000}	1.1
2, -6, 18, -54	\displaystyle \frac{-6}{2}	-3

In each row the other quotients give the same r; for the last, \displaystyle \frac{18}{-6} = \frac{-54}{18} = -3 as well.

A negative ratio flips the sign at every step, so the terms alternate: positive, negative, positive. A ratio between 0 and 1 makes each term a fraction of the one before, so the terms shrink. A ratio above 1 makes them grow.

Two values are left out. The first term must not be 0: if a_1 = 0, every term is 0 \cdot r = 0, and the quotient \displaystyle \frac{0}{0} has no value, so no ratio can be named. The ratio must not be 0 either: then the list is a_1, 0, 0, 0, \ldots, and from the third term on the quotient is \displaystyle \frac{0}{0} again. With a_1 \ne 0 and r \ne 0, every term is a product of nonzero numbers, so no term is ever 0, and dividing by a term is always allowed.

The explicit formula

Write the terms out with the multiplications showing:

a_1, \quad a_1 r, \quad a_1 r \cdot r = a_1 r^2, \quad a_1 r^3, \quad \ldots

Getting from term 1 to term n takes n - 1 steps, and each step multiplies by r. Multiplying by r a total of n - 1 times is multiplying by r^{\,n-1}, so

a_n = a_1 \cdot r^{\,n-1}

The exponent counts steps, not positions, which is why it is n - 1 and not n. At n = 1 the formula gives a_1 \cdot r^0 = a_1, because r^0 = 1 for any nonzero r (the zero exponent), so it covers the first term too.

The arithmetic formula a_n = a_1 + (n - 1)d counts the same n - 1 steps. There each step adds d, so the steps together add (n - 1)d; here each step multiplies by r, so together they multiply by r^{\,n-1}.

For the phone tree, a_1 = 3 and r = 3, so a_n = 3 \cdot 3^{\,n-1}. Since 3 = 3^1, adding the exponents gives 3^{1 + (n-1)} = 3^n. Round 8 has 3^8 = 6561 calls.

The terms as a function of n

Read a_n = a_1 \cdot r^{\,n-1} as a function of n: put in a position, get out a term. The variable n sits in the exponent, which makes this an exponential function of n, defined only at n = 1, 2, 3, \ldots. Moving one place along always multiplies the output by r, so equal steps in n give equal ratios. An arithmetic sequence is the linear case: equal steps in n give equal differences.

That is what the graphs show. Plot each term as the point (n, a_n):

Figure 1Adding 3 each step puts the dots on a line. Doubling each step puts them on a curve that bends upward and overtakes the line.

The arithmetic dots lie on a straight line, because each step goes up by the same amount. The geometric dots lie on a curve that bends upward, because each step goes up by more than the last: 1, 2, 4, 8 in turn. The dots are not joined into a line or a curve, since there is no term in position 2.5; the dashed paths only show the shape the dots sit on. The geometric sequence starts behind and overtakes at n = 5, and its lead keeps growing after that.

Exponential functions with any real input, not only whole positions, are the subject of graphing exponential functions, where r is renamed the growth or decay factor.

Arithmetic, geometric or neither

Given a list of terms, work out every difference a_{n+1} - a_n and every ratio \displaystyle \frac{a_{n+1}}{a_n}. If all the differences are equal, the list is arithmetic. If all the ratios are equal, it is geometric. If neither, it is neither. Every consecutive pair has to be checked: any two numbers have a difference and a ratio, so the first pair alone proves nothing. A constant list such as 7, 7, 7, 7 is both, with d = 0 and r = 1, and it is the only kind of sequence that is both (Exercise 18).

Worked examples

Example 1

Finding the common ratio

Find the common ratio and the next term of each sequence.

(a) 7, 28, 112, 448 (b) 64, -48, 36, -27 (c) 2.5, 3, 3.6, 4.32

Solution

(a) \displaystyle \frac{28}{7} = 4, \displaystyle \frac{112}{28} = 4 and \displaystyle \frac{448}{112} = 4, so r = 4. The next term is 448 \cdot 4 = 1792.

(b) \displaystyle \frac{-48}{64} = -\frac{3}{4}, \displaystyle \frac{36}{-48} = -\frac{3}{4} and \displaystyle \frac{-27}{36} = -\frac{3}{4}, so \displaystyle r = -\frac{3}{4}. The next term is \displaystyle -27 \cdot \left(-\frac{3}{4}\right) = \frac{81}{4}. The signs alternate because r is negative, and the sizes shrink because each is \displaystyle \frac{3}{4} of the one before.

(c) \displaystyle \frac{3}{2.5} = 1.2, \displaystyle \frac{3.6}{3} = 1.2 and \displaystyle \frac{4.32}{3.6} = 1.2, so r = 1.2. The next term is 4.32 \cdot 1.2 = 5.184.

Example 2

Rebound heights

A ball is dropped from a height of 5 meters. Each rebound reaches 60\% of the height of the one before, so the first rebound reaches 3 meters. Write a formula for the height of the nth rebound, find the height of the fifth, and find the first rebound lower than 10 centimeters.

Solution

Taking 60\% of a height means multiplying it by 0.6, the same number each time, so the heights form a geometric sequence with a_1 = 3 and r = 0.6. (Check: 5 \cdot 0.6 = 3.)

a_n = 3(0.6)^{n-1}

The fifth rebound is four steps after the first:

a_5 = 3(0.6)^4 = 3(0.1296) = 0.3888

so about 0.39 meters. For the last question, 10 centimeters is 0.1 meters. There is no move yet that brings n down out of the exponent, so try terms, each 0.6 times the last:

n	a_n (meters)
5	0.3888
6	0.23328
7	0.139968
8	0.0839808

The seventh rebound is still above 0.1 and the eighth is below it, so the eighth rebound is the first lower than 10 centimeters.

Figure 2Each rebound is 0.6 times the one before, so the heights shrink by a smaller amount each time but stay above 0.

No term is ever 0, since each is a product of nonzero numbers. A real ball does stop; the model describes the bounces only while each really is 60\% of the last.

Example 3

A growing yeast culture

In a lab, a yeast culture has a mass of 0.1 grams. Every hour its mass doubles. Write a formula for the mass after n hours, find it after 10 hours, and find the first hour in which the formula gives more than 1000 grams, which is 1 kilogram.

Solution

After one hour the mass is 0.2 grams, and each hour multiplies it by 2, so a_1 = 0.2 and r = 2:

a_n = 0.2 \cdot 2^{\,n-1}

Writing 0.2 = 0.1 \cdot 2 and using the product of powers, 0.1 \cdot 2 \cdot 2^{\,n-1} = 0.1 \cdot 2^n. This second form reads directly: the starting area, doubled n times. After 10 weeks,

a_{10} = 0.1 \cdot 2^{10} = 0.1 \cdot 1024 = 102.4

grams, about a tenth of a kilogram. The question is when 0.1 \cdot 2^n first passes 1000, that is, when 2^n first passes 10\,000. Powers of 2 give 2^{13} = 8192 and 2^{14} = 16\,384. Hour 13 gives 819.2 grams and hour 14 gives 1638.4, so the formula first passes a kilogram in the fourteenth hour. A real culture runs short of food and room well before that and its growth slows: the model holds only while the yeast has what it needs to keep doubling.

Example 4

The formula from two consecutive terms

In a geometric sequence, a_4 = 54 and a_5 = 81. Find the explicit formula and a_7.

Solution

Two consecutive terms give the ratio directly: \displaystyle r = \frac{81}{54} = \frac{3}{2}. To find a_1, use the formula at a known term. The fourth term is three steps after the first, so

54 = a_1 \left(\frac{3}{2}\right)^3 = a_1 \cdot \frac{27}{8}, \qquad a_1 = 54 \cdot \frac{8}{27} = 16

and the formula is \displaystyle a_n = 16\left(\frac{3}{2}\right)^{n-1}. The sequence begins 16, 24, 36, 54, 81, which matches. The seventh term is two steps after the fifth, so \displaystyle a_7 = 81 \cdot \left(\frac{3}{2}\right)^2 = \frac{729}{4} = 182.25. The formula agrees: \displaystyle 16 \cdot \frac{729}{64} = \frac{729}{4}.

Example 5

Two terms with a gap: two answers

In a geometric sequence, a_2 = 12 and a_4 = 108. Find every possible explicit formula.

Solution

From a_2 to a_4 is two steps, so a_4 = a_2 \cdot r^2:

108 = 12r^2, \qquad r^2 = 9, \qquad r = 3 \text{ or } r = -3

Both are valid, because 3^2 and (-3)^2 are both 9. For each, the first term is one step back: \displaystyle a_1 = \frac{a_2}{r}.

r = 3: a_1 = 4, so a_n = 4 \cdot 3^{\,n-1}, giving 4, 12, 36, 108.
r = -3: a_1 = -4, so a_n = -4(-3)^{n-1}, giving -4, 12, -36, 108.

The missing third term is 36 or -36; the two given terms do not decide which. Had the question said a_4 = -108, the step would give r^2 = -9, which no real number satisfies, so no geometric sequence would fit. With an odd gap, such as a_1 and a_4, the step gives r^3, and a number has only one real cube root, so the answer is unique.

Example 6

Arithmetic, geometric or neither

Decide whether each list is arithmetic, geometric or neither.

(a) 4, 10, 25, 62.5 (b) 40, 31, 22, 13 (c) 1, 4, 9, 16 (d) 5, 10, 20, 30

Solution

(a) Ratios \displaystyle \frac{10}{4} = \frac{25}{10} = \frac{62.5}{25} = 2.5. Geometric, r = 2.5. (The differences 6, 15, 37.5 are not equal.)

(b) Differences 31 - 40 = 22 - 31 = 13 - 22 = -9. Arithmetic, d = -9. (The ratios \displaystyle \frac{31}{40} and \displaystyle \frac{22}{31} are not equal.)

(d) Ratios 2, 2, 1.5; differences 5, 10, 10. Neither: it starts like a geometric sequence with r = 2, but the last pair breaks the pattern, which is why every pair is checked.

Boundaries

The first term and the ratio must be nonzero. Otherwise a term is 0 and the ratio is the undefined \displaystyle \frac{0}{0}.

n is a positive whole number. The formula can be evaluated at n = 2.5, but there is no term there. The graph is dots, not a curve.

r = 1 gives a constant sequence. It is geometric and arithmetic at once, and the formula still holds: a_n = a_1.

Two terms with an even gap may give two sequences or none. The step gives r^2 (or r^4, and so on), which has two real solutions when it is positive and none when it is negative. An odd gap gives exactly one.

A few terms do not prove a sequence geometric. The formula describes a sequence known to multiply by the same number every step; a short list can fit other rules too, as closed form versus recursive form shows.

A model is geometric only while the factor stays fixed. The ball and the yeast culture follow the formula for as long as each step really multiplies by the same number.

Where this goes next

The same sequence can be described one step at a time, as a_n = r \cdot a_{n-1} with a starting value; that is the recursive formula, and closed form versus recursive form weighs the two descriptions. The exponential pattern seen here grows into exponential growth and decay, with inputs that are any real number, and the test of differences against ratios is what choosing a function family applies to real data. Adding up the terms of a geometric sequence, rather than finding one, is the subject of geometric series.

Exercises

The common ratio

Find the common ratio and the next term.

6, 30, 150, 750
\displaystyle 96, 24, 6, \frac{3}{2}
-5, 10, -20, 40
0.8, 1.2, 1.8, 2.7

The explicit formula

Write the explicit formula, then find the term asked for.

2, 12, 72, \ldots; find a_5.
729, 243, 81, \ldots; find a_7.
1, -4, 16, \ldots; find a_6.
a_1 = 200 and r = 0.9; find a_4.

From two terms

Find every explicit formula that fits, or explain why none does.

a_3 = 50 and a_4 = 10. Also find a_6.
a_2 = 10 and a_4 = 2.5. Also find a_3.
a_1 = 5 and a_4 = 40.
a_3 = 4 and a_5 = -36.

Arithmetic, geometric or neither

Decide which, and give d or r where there is one.

81, 54, 36, 24
7, 2, -3, -8
2, 5, 11, 23, 47

A setting

In the phone tree at the start, 3^n people are called in round n. In which round are more than 1000 people called for the first time?

Reasoning

Prove that any three consecutive terms a, b, c of a geometric sequence satisfy b^2 = ac. Then find every possible middle term b of a geometric sequence that runs 4, b, 81.
Prove that a sequence of nonzero terms that is both arithmetic and geometric is constant.

Answers

r = 5; next term 3750. Each quotient \displaystyle \frac{30}{6}, \displaystyle \frac{150}{30}, \displaystyle \frac{750}{150} is 5, and 750 \cdot 5 = 3750.
\displaystyle r = \frac{1}{4}; next term \displaystyle \frac{3}{8}. Each quotient is \displaystyle \frac{1}{4}, and \displaystyle \frac{3}{2} \cdot \frac{1}{4} = \frac{3}{8}.
r = -2; next term -80. Each quotient is -2, so the signs alternate, and 40 \cdot (-2) = -80.
r = 1.5; next term 4.05. \displaystyle \frac{1.2}{0.8} = \frac{1.8}{1.2} = \frac{2.7}{1.8} = 1.5, and 2.7 \cdot 1.5 = 4.05.
a_n = 2 \cdot 6^{\,n-1}; a_5 = 2 \cdot 6^4 = 2 \cdot 1296 = 2592.
\displaystyle a_n = 729\left(\frac{1}{3}\right)^{n-1}; \displaystyle a_7 = 729 \cdot \frac{1}{729} = 1.
a_n = (-4)^{n-1}; a_6 = (-4)^5 = -1024. The first term is 1, so it does not show in the formula.
a_n = 200(0.9)^{n-1}; a_4 = 200(0.9)^3 = 200(0.729) = 145.8.
\displaystyle a_n = 1250\left(\frac{1}{5}\right)^{n-1}; \displaystyle a_6 = \frac{2}{5}. Here \displaystyle r = \frac{10}{50} = \frac{1}{5}, and a_1 is two steps before a_3, so a_1 = 50 \cdot 5^2 = 1250. Then \displaystyle a_6 = a_4 \cdot \left(\frac{1}{5}\right)^2 = \frac{10}{25} = \frac{2}{5}.
\displaystyle a_n = 20\left(\frac{1}{2}\right)^{n-1} with a_3 = 5, or \displaystyle a_n = -20\left(-\frac{1}{2}\right)^{n-1} with a_3 = -5. Two steps give 10r^2 = 2.5, so \displaystyle r^2 = \frac{1}{4} and \displaystyle r = \pm\frac{1}{2}; then \displaystyle a_1 = \frac{a_2}{r} and a_3 = a_2 r.
a_n = 5 \cdot 2^{\,n-1}, the only one. Three steps give 5r^3 = 40, so r^3 = 8, and 2 is the only real number whose cube is 8.
None. Two steps give 4r^2 = -36, so r^2 = -9, and no real number squares to a negative.
Geometric, \displaystyle r = \frac{2}{3}. \displaystyle \frac{54}{81} = \frac{36}{54} = \frac{24}{36} = \frac{2}{3}; the differences -27, -18, -12 are not equal.
Arithmetic, d = -5. Each difference is -5; the ratios \displaystyle \frac{2}{7} and \displaystyle -\frac{3}{2} are not equal.
Neither. The differences are 3, 6, 12, 24 and the ratios are \displaystyle \frac{5}{2}, \frac{11}{5}, \frac{23}{11}, \frac{47}{23}. The differences double, but the terms themselves neither add nor multiply by a fixed number.
Round 7. Trying powers: 3^6 = 729 is not more than 1000, and 3^7 = 2187 is.
b = 18 or b = -18. With common ratio r the three terms are a, ar, ar^2, so b^2 = a^2r^2 = a \cdot ar^2 = ac. For 4, b, 81 this gives b^2 = 4 \cdot 81 = 324, so b = \pm 18. Both work: 4, 18, 81 has \displaystyle r = \frac{9}{2}, and 4, -18, 81 has \displaystyle r = -\frac{9}{2}.
Take three consecutive terms a, a + d, a + 2d. Equal ratios give \displaystyle \frac{a + d}{a} = \frac{a + 2d}{a + d}, so (a + d)^2 = a(a + 2d), which expands to a^2 + 2ad + d^2 = a^2 + 2ad. Then d^2 = 0, so d = 0 and every term equals the first.