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Algebra · Chapter 16

Geometric Sequences

Multiply by the same ratio again and again: find any term without listing them, add a whole sequence in one line, and sum infinitely many terms to a finite number.

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16.0a0%

Terms of a Geometric Sequence

In this section you will be able to spot a common ratio, jump straight to any term with $g_n = g_1 r^{n-1}$, and link two terms without finding $r$.

Two sequences share a ratio: one starts at 27, the other at 99, and term 15 of the first equals term 11 of the second. Find term 9 of the first. You never need the ratio itself.

In an arithmetic sequence you add the same number each time. In a geometric sequence you multiply by the same number each time: the common ratio rr.

To reach term nn from term 1, you multiply by rr exactly n−1n - 1 times. So gn=g1rn−1g_n = g_1 r^{n-1}. The same idea links any two terms: gn=gmrn−mg_n = g_m r^{n-m}.

Look: the middle of three equally spaced terms is their geometric mean: g82=g5⋅g11g_8^2 = g_5 \cdot g_{11}, because both sides equal g12r14g_1^2 r^{14}.

Exploreplay first, formulas later

6 terms, g1 = 3, r = 2: g_6 = 96, S_6 = 189, no finite S∞

first term g₁3
ratio r2
terms n6
Terms: 3, 6, 12, 24, 48, 96
Running total after each term
g6 = 3 × (2)5 = 96
gn=g1⋅rn−1g_n = g_1 \cdot r^{n-1}
S6 = 189
Sn=g1⋅1−rn1−rS_n = g_1 \cdot \frac{1 - r^n}{1 - r}
|r| ≥ 1: the total runs away, no limit

r = 2: g_6 = 96, S_6 = 189, the sum never settles

Name it

Definition 16.0.1 · Geometric Sequence

A geometric sequence is a sequence of numbers with the same ratio between consecutive terms.

1,2,4,8,16,32,…,10241, 2, 4, 8, 16, 32, \ldots, 1024

is a geometric sequence because there is always a ratio of 22 between consecutive terms.

Definition 16.0.2 · Geometric Sequence Notation

In general, the terms of a geometric sequence can be represented as

g1,g2,g3,g4,…,gng_1, g_2, g_3, g_4, \ldots, g_n

where

  • rr is the common ratio between consecutive terms
  • nn is the number of terms

Theorem 16.0.4 · nth Term in a Geometric Sequence

gn=g1⋅rn−1g_n = g_1 \cdot r^{n-1}
gn=gm⋅rn−mg_n = g_m \cdot r^{n-m}

Watch itpredict, then reveal

Example 16.1

AIME

Two geometric sequences a1,a2,a3,…a_1, a_2, a_3, \ldots and b1,b2,b3,…b_1, b_2, b_3, \ldots have the same common ratio, with a1=27a_1 = 27, b1=99b_1 = 99, and a15=b11a_{15} = b_{11}. Find a9a_9.

  1. 1

    Call the common ratio rr. Write a15a_{15} and b11b_{11} with the nnth-term formula.

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Try it

Problem 16.1.1

MATHCOUNTS

The fifth term of a geometric sequence of positive numbers is 1111 and the eleventh term is 55. What is the eighth term of the sequence?

Problem 16.1.2

MATHCOUNTS

The third and fourth terms of an arithmetic sequence are the first and second terms of a geometric sequence. If the first two terms of the arithmetic sequence are 5,25, 2, then what is the fourth term of the geometric sequence?

Problem 16.1.3

MATHCOUNTS

The numbers aa, bb, cc, and dd form a geometric sequence, in that order. If bb is three more than aa, and cc is nine more than bb, what is the value of dd?

Contest cornerreal contest problems

Problem 16.2.1(additional)

AIME

Call a 3-digit number geometric if it has 3 distinct digits which, when read from left to right, form a geometric sequence. Find the difference between the largest and smallest geometric numbers.

Problem 16.2.5(additional)

AMC 10

Positive integers aa, bb, and 20092009, with a<b<2009a < b < 2009, form a geometric sequence with an integer ratio. What is aa?

Problem 16.2.6(additional)

AIME

In an increasing sequence of four positive integers, the first three terms form an arithmetic progression, the last three terms form a geometric progression, and the first and fourth terms differ by 3030. Find the sum of the four terms.

Checkpoint for section 16.0a

Get 2 of 3 right to master this section. One try each, no hints.
1

The second term of a geometric sequence is 66 and the fifth term is 162162. What is the first term?

2

In a geometric sequence of positive numbers, g3=4g_3 = 4 and g7=64g_7 = 64. What is g5g_5?

3

What is the common ratio of the geometric sequence 81,−54,36,−24,…81, -54, 36, -24, \ldots?

16.0b0%

Geometric Series

In this section you will be able to add up a geometric sequence in one line, and find the sum of infinitely many terms when $|r| < 1$.

An infinite series adds up to 2005. Square every term and the new series adds up to ten times as much. What was the common ratio?

To add g1+g1r+⋯+g1rn−1g_1 + g_1 r + \cdots + g_1 r^{n-1}, multiply the sum by rr and subtract. Almost every term cancels, and you get Sn=g1rn−1r−1S_n = g_1 \frac{r^n - 1}{r - 1}.

When ∣r∣<1|r| < 1, the term rnr^n shrinks toward 0. The running total then settles on g11−r\frac{g_1}{1 - r}. When ∣r∣≥1|r| \ge 1, it never settles.

Look: squaring every term of a geometric series gives another geometric series, with first term g12g_1^2 and ratio r2r^2.

Exploreplay first, formulas later

8 terms, g1 = 1, r = 1/2: g_8 = 1/128, S_8 = 255/128, S∞ = 2

first term g₁1
ratio r1/2
terms n8
Terms: 1, 1/2, 1/4, 1/8, 1/16, 1/32, 1/64, …
Running total after each term
limit
g8 = 1 × (1/2)7 = 1/128
gn=g1⋅rn−1g_n = g_1 \cdot r^{n-1}
S8 = 255/128
Sn=g1⋅1−rn1−rS_n = g_1 \cdot \frac{1 - r^n}{1 - r}
limit 2; gap left = 1/128
S∞=g11−r=2S_\infty = \frac{g_1}{1-r} = 2

r = 1/2: g_8 = 1/128, S_8 = 255/128, limit 2, gap 1/128

Name it

Theorem 16.0.5 · Sum of All Terms in a Finite Geometric Sequence

For r≠1r \ne 1,

Sn=g1⋅rn−1r−1S_n = g_1 \cdot \frac{r^n - 1}{r - 1}

Theorem 16.0.6 · Sum of All Terms in an Infinite Geometric Sequence

For −1<r<1-1 < r < 1,

S∞=g11−rS_\infty = \frac{g_1}{1 - r}

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Example 16.2

AIME

For −1<r<1-1 < r < 1, let S(r)S(r) denote the sum of the geometric series

12+12r+12r2+12r3+⋯ .12 + 12r + 12r^2 + 12r^3 + \cdots.

Let aa between −1-1 and 11 satisfy S(a)S(−a)=2016S(a)S(-a) = 2016. Find S(a)+S(−a)S(a) + S(-a).

Geometric growth (its total is revealed at the end of the example)

first term g₁12
ratio r1/2
terms n8
Terms: 12, 6, 3, 3/2, 3/4, 3/8, 3/16, …
Running total after each term
limit
g8 = 12 × (1/2)7 = 3/32
gn=g1⋅rn−1g_n = g_1 \cdot r^{n-1}
S8 = 765/32
Sn=g1⋅1−rn1−rS_n = g_1 \cdot \frac{1 - r^n}{1 - r}
limit 24; gap left = 3/32
S∞=g11−r=24S_\infty = \frac{g_1}{1-r} = 24

r = 1/2: g_8 = 3/32, S_8 = 765/32, limit 24, gap 3/32

  1. 1

    Use the infinite-sum formula. What is S(r)S(r) as a fraction?

    Predict first:
    Click a choice or type its letter (A–E).
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Example 16.3

AIME

An infinite geometric series has sum 2005. A new series, obtained by squaring each term of the original series, has 10 times the sum of the original series. The common ratio of the original series is mn\frac{m}{n} where mm and nn are relatively prime integers. Find m+nm + n.

Geometric growth (its total is revealed at the end of the example)

r = ?
Running total after each term
limit
Running total of the squares
limit
S∞ = 2005
sum of the squares = 20050

Sum 2005; sum of the squares 20050

  1. 1

    Call the first term aa and the ratio rr. What equation does "sum 2005" give?

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Problem 16.2.4(additional)

AMC 10

The sum of an infinite geometric series is a positive number SS, and the second term in the series is 11. What is the smallest possible value of SS?

Problem 16.2.8(additional)

Let pp and qq be real numbers with ∣p∣<1|p| < 1 and ∣q∣<1|q| < 1 such that

p+pq+pq2+pq3+⋯=2andq+qp+qp2+qp3+⋯=3.p + pq + pq^2 + pq^3 + \cdots = 2 \quad\text{and}\quad q + qp + qp^2 + qp^3 + \cdots = 3.

Find 100pq100pq.

Contest cornerreal contest problems

Problem 16.2.2(additional)

AIME

The sum of the first 20112011 terms of a geometric sequence is 200200. The sum of the first 40224022 terms is 380380. Find the sum of the first 60336033 terms.

Problem 16.2.3(additional)

AIME

Suppose nn is a positive integer and dd is a single digit in base 10. Find nn if

n810=0.d25d25d25…\frac{n}{810} = 0.d25d25d25\ldots

4 terms, g1 = 9/25, r = 1/100: g_4 = ≈ 3.600e-7, S_4 = ≈ 0.3636, S∞ = 4/11

first term g₁9/25
ratio r1/100
terms n4
Terms: 9/25, 9/2500, 9/250000, ≈ 3.600e-7
Running total after each term
limit
g4 = 9/25 × (1/100)3 = ≈ 3.600e-7
gn=g1⋅rn−1g_n = g_1 \cdot r^{n-1}
S4 = ≈ 0.3636
Sn=g1⋅1−rn1−rS_n = g_1 \cdot \frac{1 - r^n}{1 - r}
limit 4/11; gap left = ≈ 3.636e-9
S∞=g11−r=411S_\infty = \frac{g_1}{1-r} = \tfrac{4}{11}

r = 1/100: g_4 = ≈ 3.600e-7, S_4 = ≈ 0.3636, limit 4/11, gap ≈ 3.636e-9

Problem 16.2.10(additional)

AIMEChallenge

Two distinct, real, infinite geometric series each have a sum of 11 and have the same second term. The third term of one of the series is 1/81/8, and the second term of both series can be written in the form m−np\frac{\sqrt{m} - n}{p}, where mm, nn, and pp are positive integers and mm is not divisible by the square of any prime. Find 100m+10n+p100m + 10n + p.

Checkpoint for section 16.0b

Get 2 of 3 right to master this section. One try each, no hints.
1

Find 1+3+9+27+81+243+7291 + 3 + 9 + 27 + 81 + 243 + 729.

2

Find the sum 18−6+2−23+⋯18 - 6 + 2 - \frac23 + \cdots.

3

Write 0.36‾=0.363636…0.\overline{36} = 0.363636\ldots as a fraction in lowest terms.

16.1

Practice Problems

In this section you will be able to mix arithmetic and geometric sequences in one problem.

Some problems hand you an arithmetic sequence and a geometric sequence that share terms. Write each condition as an equation, then let whole numbers narrow things down.

The other practice problems of 16.1 are the Try-it problems in 16.0a. Want every problem from this chapter on one page? Open the problem set.

Contest cornerreal contest problems

Problem 16.1.4

MATHCOUNTS

The positive integers AA, BB, and CC form an arithmetic sequence while the integers BB, CC, and DD form a geometric sequence. If CB=53\frac{C}{B} = \frac53, what is the smallest possible value of A+B+C+DA + B + C + D?

16.2

Additional Problems

In this section you will be able to use everything from this chapter on harder contest problems.

Two last challenges: a basketball game scored in sequences, and an AIME sequence that switches between geometric and arithmetic steps. Write out the first few terms and look for the pattern.

The other Additional Problems sit in sections 16.0a and 16.0b, next to the ideas they use.

Contest cornerreal contest problems

Problem 16.2.7(additional)

AMC 12

A high school basketball game between the Raiders and Wildcats was tied at the end of the first quarter. The number of points scored by the Raiders in each of the four quarters formed an increasing geometric sequence, and the number of points scored by the Wildcats in each of the four quarters formed an increasing arithmetic sequence. At the end of the fourth quarter, the Raiders had won by one point. Neither team scored more than 100100 points. What was the total number of points scored by the two teams in the first half?

Problem 16.2.9(additional)

AIMEChallenge

A sequence of positive integers with a1=1a_1 = 1 and a9+a10=646a_9 + a_{10} = 646 is formed so that the first three terms are in geometric progression, the second, third, and fourth terms are in arithmetic progression, and, in general, for all n≥1n \ge 1, the terms a2n−1,a2n,a2n+1a_{2n-1}, a_{2n}, a_{2n+1} are in geometric progression, and the terms a2n,a2n+1a_{2n}, a_{2n+1}, and a2n+2a_{2n+2} are in arithmetic progression. Let ana_n be the greatest term in this sequence that is less than 10001000. Find n+ann + a_n.