M20 Arithmetic and Geometric Sequences.pdf

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Arithmetic and

Geometric Sequences

At the end of this lecture, a student must be able to:

1

Differentiate a sequence from a set

2

Recognize arithmetic sequences and its properties

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Sequences

Definition

A sequenceis an ordered set of numbers. The numbers in the sequence are calledterms.

Examples:

1 2,4,6,8,10,12, . . . 2 2,3,5,7,11,13, . . . 3 20,−10,5,−5

2, 5 4,−

5 8, . . . 4 4,2,8,6,12,10, . . .

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Sequences

Definition

Examples:

1 2,4,6,8,10,12, . . . 2 2,3,5,7,11,13, . . . 3 20,−10,5,−5

2, 5 4,−

5 8, . . . 4 4,2,8,6,12,10, . . .

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Sequences

Definition

Examples:

1 2,4,6,8,10,12, . . . 2 2,3,5,7,11,13, . . . 3 20,−10,5,−5

2, 5 4,−

5 8, . . . 4 4,2,8,6,12,10, . . .

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The nth term of a sequence is denoted by an.

Examples:

1 Given the sequence 3,1,4,1,5,9,2,9,5,

a1 = 3, a2 = 1, a3 = 4,etc.

2 The terms in the sequence witha_n = 2n+ 3 for n≥1 are

5,7,9,11,13, ...

3 The terms in the sequence witha

n = 3(

√

3)n _for _n_≥_{1 are}

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Examples:

1 Given the sequence 3,1,4,1,5,9,2,9,5

,

a1 = 3, a2 = 1, a3 = 4,etc.

5,7,9,11,13, ...

n = 3(

√

3)n _for _n_≥_{1 are}

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Examples:

a1 = 3, a2 = 1, a3 = 4,etc.

5,7,9,11,13, ...

n = 3(

√

3)n _for _n_≥_{1 are}

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Examples:

a1 = 3, a2 = 1, a3 = 4,etc.

5,7,9,11,13, ...

n = 3(

√

3)n _for _n_≥_{1 are}

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Examples:

a1 = 3, a2 = 1, a3 = 4,etc.

5,7,9,11,13, ...

n = 3(

√

3)n _for _n_≥_{1 are}

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Examples:

a1 = 3, a2 = 1, a3 = 4,etc.

5,7,9,11,13, ...

n = 3(

√

3)n _for _n_≥_{1 are}

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Examples:

a1 = 3, a2 = 1, a3 = 4,etc.

5,7,9,11,13, ...

n = 3(

√

3)n _for _n_≥_{1 are}

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Arithmetic Sequences

Definition

Anarithmetic sequence orarithmetic progression is a sequence such that there existsd∈_R such that an+1−an=d for all n≥1.

The constantd is called the common difference of the sequence.

Examples: The following are arithmetic sequences:

1 10,12,14,16,18, . . ., where d= 2 2 10,6,2,−2,−6, . . ., where d=−4 3 log

23,log26,log212, . . ., where d= 1 since

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Arithmetic Sequences

Definition

1 10,12,14,16,18, . . .,

where d= 2

2 10,6,2,−2,−6, . . ., where d=−4 3 log

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Arithmetic Sequences

Definition

1 10,12,14,16,18, . . ., where d= 2

2 10,6,2,−2,−6, . . ., where d=−4 3 log

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Definition

1 10,12,14,16,18, . . ., where d= 2 2 10,6,2,−2,−6, . . ., where d=−4

3 log

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Definition

23,log26,log212, . . .,

where d= 1 since

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Arithmetic Sequences

Definition

23,log26,log212, . . ., where d=

1 since

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Definition

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Arithmetic Sequences

Definition

log₂6−log₂3

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Definition

log₂6−log₂3 = log₂ 6₃

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Definition

log₂6−log₂3 = log₂ 6₃ =

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Arithmetic Sequences

Definition

log₂6−log₂3 = log₂ 6₃ = log₂2 = 1

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Definition

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Example: Find the value of k so that the sequence k, 2k−2,

k+ 8 form an arithmetic progression.

Solution:

a2−a1 = a3−a2

(2k−2)−k = (k+ 8)−(2k−2)

k−2 = −k+ 10 2k = 12

k = 6

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Solution:

a2−a1 = a3−a2

(2k−2)−k = (k+ 8)−(2k−2)

k−2 = −k+ 10 2k = 12

k = 6

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Solution:

a2−a1 = a3−a2

(2k−2)−k = (k+ 8)−(2k−2)

k−2 = −k+ 10 2k = 12

k = 6

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Solution:

a2−a1 = a3−a2

(2k−2)−k = (k+ 8)−(2k−2)

k−2 = −k+ 10

2k = 12

k = 6

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Solution:

a2−a1 = a3−a2

(2k−2)−k = (k+ 8)−(2k−2)

k−2 = −k+ 10 2k = 12

k = 6

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Solution:

a2−a1 = a3−a2

(2k−2)−k = (k+ 8)−(2k−2)

k−2 = −k+ 10 2k = 12

k = 6

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Solution:

a2−a1 = a3−a2

(2k−2)−k = (k+ 8)−(2k−2)

k−2 = −k+ 10 2k = 12

k = 6

The sequence has terms 6,

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Solution:

a2−a1 = a3−a2

(2k−2)−k = (k+ 8)−(2k−2)

k−2 = −k+ 10 2k = 12

k = 6

The sequence has terms 6,10,

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Solution:

a2−a1 = a3−a2

(2k−2)−k = (k+ 8)−(2k−2)

k−2 = −k+ 10 2k = 12

k = 6

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The

n

th Term of an Arithmetic Sequence

Given an arithmetic sequence with first term a1 and common

differenced,

a2 = a1+d

a3 = a2+d= (a1+d) +d =a1+ 2d

a4 = a3+d= (a1+ 2d) +d=a1+ 3d

a5 = a4+d= (a4+ 3d) +d=a1+ 4d

.. .

The nth term of an arithmetic sequence is given by

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The

n

th Term of an Arithmetic Sequence

differenced,

a2 =

a1+d

a3 = a2+d= (a1+d) +d =a1+ 2d

a4 = a3+d= (a1+ 2d) +d=a1+ 3d

a5 = a4+d= (a4+ 3d) +d=a1+ 4d

.. .

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The

n

th Term of an Arithmetic Sequence

differenced,

a2 = a1+d

a3 = a2+d= (a1+d) +d =a1+ 2d

a4 = a3+d= (a1+ 2d) +d=a1+ 3d

a5 = a4+d= (a4+ 3d) +d=a1+ 4d

.. .

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The

n

differenced,

a2 = a1+d

a3 =

a2+d= (a1+d) +d =a1+ 2d

a4 = a3+d= (a1+ 2d) +d=a1+ 3d

a5 = a4+d= (a4+ 3d) +d=a1+ 4d

.. .

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The

n

th Term of an Arithmetic Sequence

differenced,

a2 = a1+d

a3 = a2+d=

(a1+d) +d =a1+ 2d

a4 = a3+d= (a1+ 2d) +d=a1+ 3d

a5 = a4+d= (a4+ 3d) +d=a1+ 4d

.. .

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The

n

differenced,

a2 = a1+d

a3 = a2+d= (a1+d) +d=

a1+ 2d

a4 = a3+d= (a1+ 2d) +d=a1+ 3d

a5 = a4+d= (a4+ 3d) +d=a1+ 4d

.. .

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The

n

th Term of an Arithmetic Sequence

differenced,

a2 = a1+d

a3 = a2+d= (a1+d) +d=a1+ 2d

a4 = a3+d= (a1+ 2d) +d=a1+ 3d

a5 = a4+d= (a4+ 3d) +d=a1+ 4d

.. .

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The

n

differenced,

a2 = a1+d

a3 = a2+d= (a1+d) +d=a1+ 2d

a4 =

a3+d= (a1+ 2d) +d=a1+ 3d

a5 = a4+d= (a4+ 3d) +d=a1+ 4d

.. .

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The

n

th Term of an Arithmetic Sequence

differenced,

a2 = a1+d

a3 = a2+d= (a1+d) +d=a1+ 2d

a4 = a3+d=

(a1+ 2d) +d=a1+ 3d

a5 = a4+d= (a4+ 3d) +d=a1+ 4d

.. .

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The

n

differenced,

a2 = a1+d

a3 = a2+d= (a1+d) +d=a1+ 2d a4 = a3+d= (a1+ 2d) +d=

a1+ 3d

a5 = a4+d= (a4+ 3d) +d=a1+ 4d

.. .

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The

n

th Term of an Arithmetic Sequence

differenced,

a2 = a1+d

a3 = a2+d= (a1+d) +d=a1+ 2d

a4 = a3+d= (a1+ 2d) +d=a1+ 3d

a5 = a4+d= (a4+ 3d) +d=a1+ 4d

.. .

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The

n

differenced,

a2 = a1+d

a3 = a2+d= (a1+d) +d=a1+ 2d

a4 = a3+d= (a1+ 2d) +d=a1+ 3d

a5 = a4+d= (a4+ 3d) +d=a1+ 4d

.. .

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The

n

th Term of an Arithmetic Sequence

differenced,

a2 = a1+d

a3 = a2+d= (a1+d) +d=a1+ 2d

a4 = a3+d= (a1+ 2d) +d=a1+ 3d

a5 = a4+d= (a4+ 3d) +d=a1+ 4d

.. .

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The

n

differenced,

a2 = a1+d

a3 = a2+d= (a1+d) +d=a1+ 2d

a4 = a3+d= (a1+ 2d) +d=a1+ 3d

a5 = a4+d= (a4+ 3d) +d=a1+ 4d

.. .

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Example: What is the 101st term in the sequence 2,5,8,11, . . .?

Solution:

Sincea1 = 2 andd= 3,

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Solution:

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Solution:

a101 =

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Solution:

a101 = 2 +

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Solution:

a101 = 2 + (101−1)(3)

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Solution:

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Example: How many numbers between 100 and 500 are divisible by 6?

Solution:

Smallest multiple of 6 greater than 100: 102. Largest multiple of 6 less than 500: 498.

Count number of terms in arithmetic sequence 102, 108, . . ., 498, where

a1 = 102, d= 6, and an= 498.

Solving for n,

498 = 102 + (n−1)6 498 = 96 + 6n

6n = 402

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Solution:

Smallest multiple of 6 greater than 100:

102. Largest multiple of 6 less than 500: 498.

a1 = 102, d= 6, and an= 498.

Solving for n,

498 = 102 + (n−1)6 498 = 96 + 6n

6n = 402

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Solution:

Smallest multiple of 6 greater than 100: 102.

Largest multiple of 6 less than 500: 498.

a1 = 102, d= 6, and an= 498.

Solving for n,

498 = 102 + (n−1)6 498 = 96 + 6n

6n = 402

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Solution:

Smallest multiple of 6 greater than 100: 102. Largest multiple of 6 less than 500:

498.

a1 = 102, d= 6, and an= 498.

Solving for n,

498 = 102 + (n−1)6 498 = 96 + 6n

6n = 402

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Solution:

a1 = 102, d= 6, and an= 498.

Solving for n,

498 = 102 + (n−1)6 498 = 96 + 6n

6n = 402

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Solution:

a1 = 102, d= 6, and an= 498.

Solving for n,

498 = 102 + (n−1)6 498 = 96 + 6n

6n = 402

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Solution:

a1 = 102, d= 6, and an= 498.

Solving for n,

498 = 102 + (n−1)6 498 = 96 + 6n

6n = 402

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Solution:

a1 = 102, d= 6, and an= 498.

Solving for n,

498 = 102 + (n−1)6

498 = 96 + 6n

6n = 402

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Solution:

a1 = 102, d= 6, and an= 498.

Solving for n,

498 = 102 + (n−1)6 498 = 96 + 6n

6n = 402

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Solution:

a1 = 102, d= 6, and an= 498.

Solving for n,

498 = 102 + (n−1)6 498 = 96 + 6n

6n = 402

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Solution:

a1 = 102, d= 6, and an= 498.

Solving for n,

498 = 102 + (n−1)6 498 = 96 + 6n

6n = 402

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Sum of First

n

Terms of an Arithmetic Sequence

Suppose we want to find the sum 1 + 2 + 3 +· · ·+ 1000.

A clever way to solve this:

S1000 = 1 + 2 +· · ·+ 999 + 1000

S1000 = 1000 + 999 +· · ·+ 2 + 1

Adding the two equations:

2S1000 = (1 + 1000) + (2 + 999) +...+ (999 + 2) + (1000 + 1)

2S1000 = 1001 + 1001 +. . .+ 1001 + 1001

| {z }

1000 addends

Thus,

S1000 =

(1000)(1001)

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Sum of First

n

S1000 = 1 + 2 +· · ·+ 999 + 1000

S1000 = 1000 + 999 +· · ·+ 2 + 1

2S1000 = (1 + 1000) + (2 + 999) +...+ (999 + 2) + (1000 + 1)

2S1000 = 1001 + 1001 +. . .+ 1001 + 1001

| {z }

1000 addends

Thus,

S1000 =

(1000)(1001)

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Sum of First

n

S1000 = 1 + 2 +· · ·+ 999 + 1000

S1000 = 1000 + 999 +· · ·+ 2 + 1

2S1000 = (1 + 1000) + (2 + 999) +...+ (999 + 2) + (1000 + 1)

2S1000 = 1001 + 1001 +. . .+ 1001 + 1001

| {z }

1000 addends

Thus,

S1000 =

(1000)(1001)

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Sum of First

n

S1000 = 1 + 2 +· · ·+ 999 + 1000

S1000 = 1000 + 999 +· · ·+ 2 + 1

2S1000 = (1 + 1000) + (2 + 999) +...+ (999 + 2) + (1000 + 1)

2S1000 = 1001 + 1001 +. . .+ 1001 + 1001

| {z }

1000 addends

Thus,

S1000 =

(1000)(1001)

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Sum of First

n

S1000 = 1 + 2 +· · ·+ 999 + 1000

S1000 = 1000 + 999 +· · ·+ 2 + 1

2S1000 =

(1 + 1000) + (2 + 999) +...+ (999 + 2) + (1000 + 1) 2S1000 = 1001 + 1001 +. . .+ 1001 + 1001

| {z }

1000 addends

Thus,

S1000 =

(1000)(1001)

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Sum of First

n

S1000 = 1 + 2 +· · ·+ 999 + 1000

S1000 = 1000 + 999 +· · ·+ 2 + 1

2S1000 = (1 + 1000) + (2 + 999) +...+ (999 + 2) + (1000 + 1)

2S1000 = 1001 + 1001 +. . .+ 1001 + 1001

| {z }

1000 addends

Thus,

S1000 =

(1000)(1001)

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Sum of First

n

S1000 = 1 + 2 +· · ·+ 999 + 1000

S1000 = 1000 + 999 +· · ·+ 2 + 1

2S1000 = (1 + 1000) + (2 + 999) +...+ (999 + 2) + (1000 + 1)

2S1000 = 1001 + 1001 +. . .+ 1001 + 1001

| {z }

1000 addends

Thus,

S1000 =

(1000)(1001)

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Sum of First

n

S1000 = 1 + 2 +· · ·+ 999 + 1000

S1000 = 1000 + 999 +· · ·+ 2 + 1

2S1000 = (1 + 1000) + (2 + 999) +...+ (999 + 2) + (1000 + 1)

2S1000 = 1001 + 1001 +. . .+ 1001 + 1001

| {z }

1000 addends

Thus,

S1000 =

(1000)(1001)

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Sum of First

n

S1000 = 1 + 2 +· · ·+ 999 + 1000

S1000 = 1000 + 999 +· · ·+ 2 + 1

2S1000 = (1 + 1000) + (2 + 999) +...+ (999 + 2) + (1000 + 1)

2S1000 = 1001 + 1001 +. . .+ 1001 + 1001

| {z }

1000 addends

Thus,

S1000 =

(1000)(1001)

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Sum of First

n

S1000 = 1 + 2 +· · ·+ 999 + 1000

S1000 = 1000 + 999 +· · ·+ 2 + 1

2S1000 = (1 + 1000) + (2 + 999) +...+ (999 + 2) + (1000 + 1)

2S1000 = 1001 + 1001 +. . .+ 1001 + 1001

| {z }

1000 addends

Thus,

S1000 =

(1000)(1001)

2 =

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Sum of First

n

S1000 = 1 + 2 +· · ·+ 999 + 1000

S1000 = 1000 + 999 +· · ·+ 2 + 1

2S1000 = (1 + 1000) + (2 + 999) +...+ (999 + 2) + (1000 + 1)

2S1000 = 1001 + 1001 +. . .+ 1001 + 1001

| {z }

1000 addends

Thus,

S1000 =

(1000)(1001)

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Sum of First

n

LetSn=a1+a2+· · ·+an be the sum of the first n terms of an

arithmetic sequence.

Sn= a1 + (a1+d) +· · ·+ [a1+ (n−1)d]

Sn= [a1+ (n−1)d] + [a1+ (n−2)d] +· · ·+ a1

Adding these two equations,

2Sn= [2a1+ (n−1)d] + [2a1+ (n−1)d] +· · ·+ [2a1 + (n−1)d]

| {z }

naddends

Therefore, 2Sn= n[2a1+ (n−1)d] =n[a1 +a1 + (n−1)d] =

n(a1+an)

The sum of the firstn terms of an arithmetic sequence is

Sn =

n

2[2a1+ (n−1)d] =

n

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Sum of First

n

Sn= a1 + (a1+d) +· · ·+ [a1+ (n−1)d]

Sn= [a1+ (n−1)d] + [a1+ (n−2)d] +· · ·+ a1

2Sn= [2a1+ (n−1)d] + [2a1+ (n−1)d] +· · ·+ [2a1 + (n−1)d]

| {z }

naddends

n(a1+an)

Sn =

n

2[2a1+ (n−1)d] =

n

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Sum of First

n

Sn= a1 + (a1+d) +· · ·+ [a1+ (n−1)d]

Sn= [a1+ (n−1)d] + [a1+ (n−2)d] +· · ·+ a1

2Sn= [2a1+ (n−1)d] + [2a1+ (n−1)d] +· · ·+ [2a1 + (n−1)d]

| {z }

naddends

n(a1+an)

Sn =

n

2[2a1+ (n−1)d] =

n

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Sum of First

n

Sn= a1 + (a1+d) +· · ·+ [a1+ (n−1)d]

Sn= [a1+ (n−1)d] + [a1+ (n−2)d] +· · ·+ a1

2Sn= [2a1+ (n−1)d] + [2a1+ (n−1)d] +· · ·+ [2a1 + (n−1)d]

| {z }

naddends

n(a1+an)

Sn =

n

2[2a1+ (n−1)d] =

n

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Sum of First

n

Sn= a1 + (a1+d) +· · ·+ [a1+ (n−1)d]

Sn= [a1+ (n−1)d] + [a1+ (n−2)d] +· · ·+ a1

2Sn= [2a1+ (n−1)d] + [2a1+ (n−1)d] +· · ·+ [2a1 + (n−1)d]

| {z }

naddends

n(a1+an)

Sn =

n

2[2a1+ (n−1)d] =

n

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Sum of First

n

Sn= a1 + (a1+d) +· · ·+ [a1+ (n−1)d]

Sn= [a1+ (n−1)d] + [a1+ (n−2)d] +· · ·+ a1

2Sn= [2a1+ (n−1)d] + [2a1+ (n−1)d] +· · ·+ [2a1 + (n−1)d]

| {z }

naddends

Therefore, 2Sn=

n[2a1+ (n−1)d] =n[a1 +a1 + (n−1)d] =

n(a1+an)

Sn =

n

2[2a1+ (n−1)d] =

n

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Sum of First

n

Sn= a1 + (a1+d) +· · ·+ [a1+ (n−1)d]

Sn= [a1+ (n−1)d] + [a1+ (n−2)d] +· · ·+ a1

2Sn= [2a1+ (n−1)d] + [2a1+ (n−1)d] +· · ·+ [2a1 + (n−1)d]

| {z }

naddends

Therefore, 2Sn=n[2a1+ (n−1)d] =

n[a1 +a1 + (n−1)d] =

n(a1+an)

Sn =

n

2[2a1+ (n−1)d] =

n

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Sum of First

n

Sn= a1 + (a1+d) +· · ·+ [a1+ (n−1)d]

Sn= [a1+ (n−1)d] + [a1+ (n−2)d] +· · ·+ a1

2Sn= [2a1+ (n−1)d] + [2a1+ (n−1)d] +· · ·+ [2a1 + (n−1)d]

| {z }

naddends

Therefore, 2Sn=n[2a1+ (n−1)d] =n[a1 +a1+ (n−1)d] =

n(a1+an)

Sn =

n

2[2a1+ (n−1)d] =

n

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Sum of First

n

Sn= a1 + (a1+d) +· · ·+ [a1+ (n−1)d]

Sn= [a1+ (n−1)d] + [a1+ (n−2)d] +· · ·+ a1

2Sn= [2a1+ (n−1)d] + [2a1+ (n−1)d] +· · ·+ [2a1 + (n−1)d]

| {z }

naddends

n(a1+an)

Sn =

n

2[2a1+ (n−1)d] =

n

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Sum of First

n

Sn= a1 + (a1+d) +· · ·+ [a1+ (n−1)d]

Sn= [a1+ (n−1)d] + [a1+ (n−2)d] +· · ·+ a1

2Sn= [2a1+ (n−1)d] + [2a1+ (n−1)d] +· · ·+ [2a1 + (n−1)d]

| {z }

naddends

n(a1+an)

Sn =

n

2[2a1+ (n−1)d] =

n

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Example: Find 16 + 10 + 4−2− · · · −50.

Solution:

16,10,4,−2,· · · ,−50 is an arithmetic sequence with d =−6.

First, find the number of terms.

−50 = 16 + (n−1)(−6) −66 = (n−1)(−6)

11 = n−1

n = 12

Hence, we are looking forS12.

S12=

12

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Example: Find 16 + 10 + 4−2− · · · −50.

Solution:

−50 = 16 + (n−1)(−6) −66 = (n−1)(−6)

11 = n−1

n = 12

S12=

12

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Example: Find 16 + 10 + 4−2− · · · −50.

Solution:

−50 = 16 + (n−1)(−6) −66 = (n−1)(−6)

11 = n−1

n = 12

S12=

12

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Example: Find 16 + 10 + 4−2− · · · −50.

Solution:

−50 = 16 + (n−1)(−6)

−66 = (n−1)(−6) 11 = n−1

n = 12

S12=

12

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Example: Find 16 + 10 + 4−2− · · · −50.

Solution:

−50 = 16 + (n−1)(−6) −66 = (n−1)(−6)

11 = n−1

n = 12

S12=

12

(90)

Example: Find 16 + 10 + 4−2− · · · −50.

Solution:

−50 = 16 + (n−1)(−6) −66 = (n−1)(−6)

11 = n−1

n = 12

S12=

12

(91)

Example: Find 16 + 10 + 4−2− · · · −50.

Solution:

−50 = 16 + (n−1)(−6) −66 = (n−1)(−6)

11 = n−1

n = 12

S12=

12

(92)

Example: Find 16 + 10 + 4−2− · · · −50.

Solution:

−50 = 16 + (n−1)(−6) −66 = (n−1)(−6)

11 = n−1

n = 12

S12=

12

(93)

Example: Find 16 + 10 + 4−2− · · · −50.

Solution:

−50 = 16 + (n−1)(−6) −66 = (n−1)(−6)

11 = n−1

n = 12

S12=

12

(94)

Example: Find 16 + 10 + 4−2− · · · −50.

Solution:

−50 = 16 + (n−1)(−6) −66 = (n−1)(−6)

11 = n−1

n = 12

S12=

12

2 [16 + (−50)] =

(95)

Example: Find 16 + 10 + 4−2− · · · −50.

Solution:

−50 = 16 + (n−1)(−6) −66 = (n−1)(−6)

11 = n−1

n = 12

S12=

12

(96)

Example: On January 1, James saved P10 from his allowance. He saved P12 on January 2, P14 on January 3, and so on in

arithmetic progression. How much money, in total, has he saved before the first day of February?

Solution:

Here, a1 = 10 and d = 2. FindS31.

S31=

31

2 [2(10) + (30)2] = 31

2 (80) = 1240

(97)

Solution:

Here, a1 =

10 andd = 2. FindS31.

S31=

31

2 [2(10) + (30)2] = 31

2 (80) = 1240

(98)

Solution:

Here, a1 = 10 and d=

2. Find S31.

S31=

31

2 [2(10) + (30)2] = 31

2 (80) = 1240

(99)

Solution:

Here, a1 = 10 and d= 2.

FindS31.

S31=

31

2 [2(10) + (30)2] = 31

2 (80) = 1240

(100)

Solution:

Here, a1 = 10 and d= 2. Find S31.

S31=

31

2 [2(10) + (30)2] = 31

2 (80) = 1240

(101)

Solution:

S31=

31

2 [2(10) + (30)2] =

31

2 (80) = 1240

(102)

Solution:

S31=

31

2 [2(10) + (30)2] = 31

2 (80) =

1240

(103)

Solution:

S31=

31

2 [2(10) + (30)2] = 31

2 (80) = 1240

(104)

Solution:

S31=

31

2 [2(10) + (30)2] = 31

2 (80) = 1240

(105)

An arena has 45 rows, with the number of seats of each row following an arithmetic progression. The fifth row has 97 seats while the last row has 177 seats. Find the total number of seats from the fourteenth row up to the thirty-second row.

Solution:

Letai be the number of seats in the ith row. Here, a5 = 97 and

a45= 177. We need to find

S32−S13=

32

2 (2a1+ 31d)− 13

2 (2a1+ 12d) To find a1 and d,

(

a1+ 4d= 97 (1)

a1+ 44d= 177 (2)

Subtracting (1) from (2): 40d= 80 ⇒d= 2. Then we get

a1 = 89.

(106)

Solution:

Letai be the number of seats in the ith row.

Here, a5 = 97 and

S32−S13=

32

2 (2a1+ 31d)− 13

(

a1+ 4d= 97 (1)

a1+ 44d= 177 (2)

a1 = 89.

(107)

Solution:

a45= 177.

We need to find

S32−S13=

32

2 (2a1+ 31d)− 13

(

a1+ 4d= 97 (1)

a1+ 44d= 177 (2)

a1 = 89.

(108)

An arena has 45 rows, with the number of seats of each row following an arithmetic progression. The fifth row has 97 seats while the last row has 177 seats. Find thetotal number of seats from the fourteenth row up to the thirty-second row.

Solution:

S32−S13

= 32

2 (2a1+ 31d)− 13

(

a1+ 4d= 97 (1)

a1+ 44d= 177 (2)

a1 = 89.

(109)

Solution:

S32−S13=

32

2 (2a1+ 31d)

− 13

(

a1+ 4d= 97 (1)

a1+ 44d= 177 (2)

a1 = 89.

(110)

Solution:

S32−S13=

32

2 (2a1+ 31d)− 13

2 (2a1+ 12d)

To find a1 and d, (

a1+ 4d= 97 (1)

a1+ 44d= 177 (2)

a1 = 89.

(111)

Solution:

S32−S13=

32

2 (2a1+ 31d)− 13

(

a1+ 4d= 97 (1)

a1+ 44d= 177 (2)

a1 = 89.

(112)

Solution:

S32−S13=

32

2 (2a1+ 31d)− 13

(

a1+ 4d= 97 (1)

a1+ 44d= 177 (2)

Subtracting (1) from (2): 40d= 80

⇒d= 2. Then we get

a1 = 89.

(113)

Solution:

S32−S13=

32

2 (2a1+ 31d)− 13

(

a1+ 4d= 97 (1)

a1+ 44d= 177 (2)

Subtracting (1) from (2): 40d= 80 ⇒d= 2.

Then we get

a1 = 89.

(114)

Solution:

S32−S13=

32

2 (2a1+ 31d)− 13

(

a1+ 4d= 97 (1)

a1+ 44d= 177 (2)

a1 = 89.

(115)

Solution:

S32−S13=

32

2 (2a1+ 31d)− 13

2 (2a1+ 12d)

To find a1 and d, (

a1+ 4d= 97 (1)

a1+ 44d= 177 (2)

a1 = 89.

S32−S13

(116)

Solution:

S32−S13=

32

2 (2a1+ 31d)− 13

2 (2a1+ 12d)

To find a1 and d, (

a1+ 4d= 97 (1)

a1+ 44d= 177 (2)

a1 = 89.

S32−S13= 19a1−418d=

(117)

Solution:

S32−S13=

32

2 (2a1+ 31d)− 13

(

a1+ 4d= 97 (1)

a1+ 44d= 177 (2)

a1 = 89.

S32−S13= 19a1−418d= 19(89)−418(2)

(118)

Solution:

S32−S13=

32

2 (2a1+ 31d)− 13

(

a1+ 4d= 97 (1)

a1+ 44d= 177 (2)

a1 = 89.

S32−S13= 19a1−418d= 19(89)−418(2) = 1691−836

(119)

Solution:

S32−S13=

32

2 (2a1+ 31d)− 13

(

a1+ 4d= 97 (1)

a1+ 44d= 177 (2)

a1 = 89.

(120)

Geometric Sequences

Definition

A geometric sequenceor geometric progression is a sequence such that there existsr∈_R such that an+1

an

=r for all n ≥1. The

constantr is called thecommon ratio of the sequence.

Examples: The following are geometric sequences:

1 1,3,9,27,81, where r= 3

2 1

2,− 1 3,

2 9,−

4

27, where r=− 2 3

(121)

Geometric Sequences

Definition

an

1 1,3,9,27,81,

where r= 3

2 1

2,− 1 3,

2 9,−

4

27, where r=− 2 3

(122)

Geometric Sequences

Definition

an

1 1,3,9,27,81, where r= 3

2 1

2,− 1 3,

2 9,−

4

27, where r=− 2 3

(123)

Geometric Sequences

Definition

an

1 1,3,9,27,81, where r= 3

2 1

2,− 1 3,

2 9,−

4 27,

where r=−2 3

(124)

Geometric Sequences

Definition

an

1 1,3,9,27,81, where r= 3

2 1

2,− 1 3,

2 9,−

4

27, where r=− 2 3

(125)

Geometric Sequences

Definition

an

1 1,3,9,27,81, where r= 3

2 1

2,− 1 3,

2 9,−

4

27, where r=− 2 3

3 e4,1, e−4, e−8,

(126)

Geometric Sequences

Definition

an

1 1,3,9,27,81, where r= 3

2 1

2,− 1 3,

2 9,−

4

27, where r=− 2 3

(127)

Example: Find the value of k so that k+ 1, −k, k−2 form a geometric progression.

Solution:

a2

a1

= a3

a2

−k k+ 1 =

k−2 −k

k2 = (k+ 1)(k−2)

k2 ₌ _k2₋_k₋₂

k = −2

(128)

Solution:

a2

a1

= a3

a2

−k k+ 1 =

k−2 −k

k2 = (k+ 1)(k−2)

k2 ₌ _k2₋_k₋₂

k = −2

(129)

Solution:

a2

a1

= a3

a2

−k k+ 1 =

k−2 −k

k2 = (k+ 1)(k−2)

k2 ₌ _k2₋_k₋₂

k = −2

(130)

Solution:

a2

a1

= a3

a2

−k k+ 1 =

k−2 −k

k2 = (k+ 1)(k−2)

k2 ₌ _k2₋_k₋₂

k = −2

(131)

Solution:

a2

a1

= a3

a2

−k k+ 1 =

k−2 −k

k2 = (k+ 1)(k−2)

k2 ₌ _k2₋_k₋₂

k = −2

(132)

Solution:

a2

a1

= a3

a2

−k k+ 1 =

k−2 −k

k2 = (k+ 1)(k−2)

k2 ₌ _k2₋_k₋₂

k = −2

(133)

Solution:

a2

a1

= a3

a2

−k k+ 1 =

k−2 −k

k2 = (k+ 1)(k−2)

k2 ₌ _k2₋_k₋₂

k = −2

(134)

The

n

th Term of a Geometric Sequence

Given a geometric sequence with first terma1 and common ratio

r,

a2 = a1r

a3 = a2r= (a1r)r =a1r2

a4 = a3r= (a1r2)r=a1r3

a5 = a4r= (a4r3)r=a1r4

.. .

The nth term of a geometric sequence is given by

(135)

The

n

th Term of a Geometric Sequence

r,

a2 = a1r

a3 =

a2r= (a1r)r =a1r2

a4 = a3r= (a1r2)r=a1r3

a5 = a4r= (a4r3)r=a1r4

.. .

(136)

The

n

th Term of a Geometric Sequence

r,

a2 = a1r

a3 = a2r=

(a1r)r =a1r2

a4 = a3r= (a1r2)r=a1r3

a5 = a4r= (a4r3)r=a1r4

.. .

(137)

The

n

r,

a2 = a1r

a3 = a2r= (a1r)r =

a1r2

a4 = a3r= (a1r2)r=a1r3

a5 = a4r= (a4r3)r=a1r4

.. .

(138)

The

n

th Term of a Geometric Sequence

r,

a2 = a1r

a3 = a2r= (a1r)r =a1r2

a4 = a3r= (a1r2)r=a1r3

a5 = a4r= (a4r3)r=a1r4

.. .

(139)

The

n

r,

a2 = a1r

a3 = a2r= (a1r)r =a1r2

a4 = a3r=

(a1r2)r=a1r3

a5 = a4r= (a4r3)r=a1r4

.. .

(140)

The

n

th Term of a Geometric Sequence

r,

a2 = a1r

a3 = a2r= (a1r)r =a1r2 a4 = a3r= (a1r2)r=

a1r3

a5 = a4r= (a4r3)r=a1r4

.. .

(141)

The

n

r,

a2 = a1r

a3 = a2r= (a1r)r =a1r2

a4 = a3r= (a1r2)r=a1r3

a5 = a4r= (a4r3)r=a1r4

.. .

(142)

The

n

th Term of a Geometric Sequence

r,

a2 = a1r

a3 = a2r= (a1r)r =a1r2

a4 = a3r= (a1r2)r=a1r3

a5 = a4r= (a4r3)r=a1r4

.. .

(143)

The

n

r,

a2 = a1r

a3 = a2r= (a1r)r =a1r2

a4 = a3r= (a1r2)r=a1r3

a5 = a4r= (a4r3)r=a1r4

.. .

(144)

The

n

th Term of a Geometric Sequence

r,

a2 = a1r

a3 = a2r= (a1r)r =a1r2

a4 = a3r= (a1r2)r=a1r3

a5 = a4r= (a4r3)r=a1r4

.. .

(145)

The

n

r,

a2 = a1r

a3 = a2r= (a1r)r =a1r2

a4 = a3r= (a1r2)r=a1r3

a5 = a4r= (a4r3)r=a1r4

.. .

(146)

Example: What is the 41st term in the geometric sequence

e−6, e−3,1, . . .?

Solution:

Here, a1 =e−6 and r= e3. So,

(147)

e−6, e−3,1, . . .?

Solution: Here, a1 =e−6

and r= e3_{. So,}

(148)

e−6, e−3,1, . . .?

Solution:

Here, a1 =e−6 and r=

e3_{. So,}

(149)

e−6, e−3,1, . . .?

Solution:

(150)

e−6, e−3,1, . . .?

Solution:

a41=

(151)

e−6, e−3,1, . . .?

Solution:

a41= (e−6)(e3)40 =

(152)

e−6, e−3,1, . . .?

Solution:

a41= (e−6)(e3)40 =e−6e120=

(153)

e−6, e−3,1, . . .?

Solution:

(154)

Example: The sequence 2, a, b, c, d,10 forms a geometric sequence. Findc.

Solution:

Here, a1 = 2 and a6 = 10.

a6 = a1r5

10 = 2r5

5 = r5

r =√5 5

Therefore,

c=a4 =a1r3 = 2(

5 √

(155)

Solution: Here, a1 = 2

and a6 = 10.

a6 = a1r5

10 = 2r5

5 = r5

r =√5 5

Therefore,

c=a4 =a1r3 = 2(

5 √

(156)

Solution:

Here, a1 = 2 and a6 = 10.

a6 = a1r5

10 = 2r5

5 = r5

r =√5 5

Therefore,

c=a4 =a1r3 = 2(

5 √

(157)

Solution:

Here, a1 = 2 and a6 = 10.

a6 = a1r5

10 = 2r5

5 = r5

r =√5 5

Therefore,

c=a4 =a1r3 = 2(

5 √

(158)

Solution:

Here, a1 = 2 and a6 = 10.

a6 = a1r5

10 = 2r5

5 = r5

r =√5 5

Therefore,

c=a4 =a1r3 = 2(

5 √

(159)

Solution:

Here, a1 = 2 and a6 = 10.

a6 = a1r5

10 = 2r5

5 = r5

r =√5 5

Therefore,

c=a4 =a1r3 = 2(

5 √

(160)

Solution:

Here, a1 = 2 and a6 = 10.

a6 = a1r5

10 = 2r5

5 = r5

r =√5 5

Therefore,

c=a4 =a1r3 = 2(

5 √

(161)

Solution:

Here, a1 = 2 and a6 = 10.

a6 = a1r5

10 = 2r5

5 = r5

r =√5 5

Therefore,

c=

a4 =a1r3 = 2(

5 √

(162)

Solution:

Here, a1 = 2 and a6 = 10.

a6 = a1r5

10 = 2r5

5 = r5

r =√5 5

Therefore,

c=a4 =

a1r3 = 2(

5 √

(163)

Solution:

Here, a1 = 2 and a6 = 10.

a6 = a1r5

10 = 2r5