Mathematics Chapter 3 – Sequence and Series | Nepal eNotes

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MATHEMATICS • CHAPTER 3

Chapter 3: Sequence and Series

Arithmetic Progression, Arithmetic Series, Geometric Progression and Geometric Series — reconstructed from the supplied 85-page handwritten notes.

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Source note: The uploaded filename says “Chapter 3 – Linear Programming 6.3,” but page 1 of the handwritten PDF is headed Lesson 3: Sequence and Series. The page content is therefore used as the source of truth.

Arithmetic Progression (A.P.) – Formulae

Common differenced = t₂ − t₁

n-th termtₙ = a + (n−1)d

Last terml = a + (n−1)d

Sum of n termsSₙ = n/2 [2a + (n−1)d]

Alternative sumSₙ = n/2(a+l)

PDF Pages 1–22

Exercise 3.1 – Arithmetic Progression

2, 5, 8, 11, …

a=2, d=3

t₅=14, t₆=17, t₇=20

tₙ = 3n−1

18, 15, 12, 9, …

a=18, d=−3

t₅=6, t₆=3, t₇=0

1, −2, −5, −8, …

d=−3

t₆=−14, t₇=−17

tₙ=4−3n

½, 1, 3/2, 2, …

a=1/2, d=1/2

t₅=5/2, t₆=3, t₇=7/2

tₙ=n/2

Find the 14th term of 6, 4, 2, …

a=6, d=−2

t₁₄=6+13(−2)

∴ t₁₄=−20

First term 9, common difference 4. Find the 13th term.

t₁₃=9+12×4

∴ 57

5, 11, 17, … to 25 terms

a=5, d=6

l=5+24×6

∴ 149

First term 24 and 9th term 36

36=24+8d

∴ d=3/2

21, 26, … , 136

136=21+(n−1)5

∴ n=24

7, 2, … , −43

−43=7+(n−1)(−5)

∴ n=11

3, 6, … , 96

96=3+(n−1)3

∴ n=32

5, 9, … , 77

77=5+(n−1)4

∴ n=19

If t₇=62 and t₁₉=2

a+6d=62

a+18d=2

12d=−60 ⇒ d=−5

a=92

If t₄=75 and t₁₀=117

a+3d=75

a+9d=117

6d=42 ⇒ d=7

a=54

If t₁₁=23 and t₁₅=3

a+10d=23

a+14d=3

d=−5, a=73

Compare the terms of two A.P.s

For the sequences shown on pages 18–19, the source writes their n-th terms and equates them.

One comparison gives n=11; another gives n=6.

Pages 20–22 contain additional algebraic relations between the p-th, q-th and r-th terms of an A.P.; the original symbolic derivation is retained in the embedded PDF.

PDF Pages 23–48

Exercise 3.3 – Arithmetic Series

1+3+5+… to 25 terms

S₂₅=25/2[2+24×2]

∴ 625

40+35+30+… to 20 terms

a=40, d=−5

S₂₀=20/2[80−95]

∴ −150

−5−3−1+…+27

a=−5, d=2, l=27 ⇒ n=17

S₁₇=17/2(−5+27)

∴ 187

½+3/2+5/2+… to 16 terms

S₁₆=16/2[1+15]

∴ 128

1+4+7+…+37

a=1, d=3, l=37 ⇒ n=13

S₁₃=13/2(38)

∴ 247

2−9−20−…−130

a=2, d=−11, l=−130 ⇒ n=13

S₁₃=13/2(−128)

∴ −832

Sigma notation examples

Σ(3n−2), n=4 to 7 = 58

Σ(3n−1), n=0 to 4 = 25

Σ(n²+1), n=3 to 7 = 140

Σ(4n+5), n=2 to 6 = 105

Σ(n²−3), n=2 to 5 = 42

First term 16, common difference 4. Find S₅.

S₅=5/2[32+16]

∴ 120

a=3, d=2. Find S₁₀.

S₁₀=10/2[6+18]

∴ 120

d=−3 and S₇=0

0=7/2[2a−18]

∴ a=9

d=4 and S₅=120

120=5/2[2a+16]

∴ a=16

Series 2,4,6,… has sum 420

420=n/2[4+2(n−1)]

420=n(n+1)

(n−20)(n+21)=0

∴ n=20

First term 18, second term 15, sum 45

a=18, d=−3

The resulting quadratic gives n=3 or n=10.

S₇=49 and S₁₇=289

2a+6d=14

2a+16d=34

d=2, a=1

S₁₀=100

10th term 39 and 17th term 67

a+9d=39

a+16d=67

d=4, a=3

25th term=99

S₂₆=1378

Pages 46–48 contain application problems involving regularly increasing amounts/ages/payments. The handwritten source applies the same A.P. term and sum formulae.

Geometric Progression (G.P.) – Formulae

Common ratior=t₂/t₁

n-th termtₙ=arⁿ⁻¹

Finite sumSₙ=a(rⁿ−1)/(r−1)

Alternative sumSₙ=a(1−rⁿ)/(1−r)

PDF Pages 49–66

Exercise 3.4 – Geometric Progression

1,3,9,27,…

a=1, r=3

t₅=81, t₆=243, t₇=729

64,32,16,8,…

a=64, r=1/2

t₅=4, t₆=2, t₇=1

−3/2,3,−6,12,…

a=−3/2, r=−2

t₅=−24, t₆=48, t₇=−96

243,−81,27,−9,…

a=243, r=−1/3

t₅=3, t₆=−1, t₇=1/3

First term 3 and 7th term 192

192=3r⁶

r⁶=64

∴ r=2

First term 2 and 3rd term 242

242=2r²

r²=121

∴ r=±11

Common ratio 3 and 3rd term 36. Find 5th term.

36=9a ⇒ a=4

t₅=4×81

∴ 324

First term 16 and r=1/2

Sequence: 16, 8, 4, 2, …

1,2,4,…,256

256=2ⁿ⁻¹=2⁸

∴ n=9

96,−48,…,−3/8

a=96, r=−1/2

The source compares powers and obtains n=9.

2nd term 9 and 5th term 243

ar=9, ar⁴=243

r³=27 ⇒ r=3

a=3

7th term=2187

3rd term 2/3 and 6th term 2/81

ar²=2/3, ar⁵=2/81

r³=1/27 ⇒ r=1/3

a=6

5th term 81 and 8th term 2187

ar⁴=81, ar⁷=2187

r³=27 ⇒ r=3

a=1

4th term 54 and 6th term 24

r²=24/54=4/9

r=2/3

a=729/4

x+2, 2x+4, 3x+11 are in G.P.

(2x+4)/(x+2)=(3x+11)/(2x+4)

x²−x−6=0

(x−3)(x+2)=0

x=−2 rejected

∴ x=3

PDF Pages 67–85

Exercise 3.6 – Geometric Series

5+10+20+… to 7 terms

S₇=5(2⁷−1)

∴ 635

4+2+1+… to 8 terms

a=4, r=1/2

∴ S₈=255/32

1−2+4−8+… to 8 terms

a=1, r=−2

∴ S₈=−85

81−27+9−… to 8 terms

a=81, r=−1/3

Source result: S₈=1640/27

3+6+12+24+…+384

a=3, r=2, l=384

Sₙ=(lr−a)/(r−1)

∴ 765

3−9+27−…−729

a=3, r=−3, l=−729

Sₙ=(2187−3)/(−4)

∴ −546

Sigma notation

Σ3ᵏ, k=1 to 5 = 363

Σ2·3ᵏ, k=1 to 6 = 2184

Σ3(−2)ⁿ, n=1 to 7 = −258

Common ratio 2, last term 768, sum 1533

1533=(768×2−a)/(2−1)

∴ a=3

a=6, l=384, Sₙ=762

762=(384r−6)/(r−1)

∴ r=2

a=7, l=189, r=3

Sₙ=(189×3−7)/(3−1)

∴ 280

a=3, r=2, Sₙ=3069

3069=3(2ⁿ−1)

2ⁿ=1024

∴ n=10

a=6, r=2, Sₙ=1530

1530=6(2ⁿ−1)

2ⁿ=256

∴ n=8

How many terms of 1+3+9+… make 1093?

1093=(3ⁿ−1)/2

3ⁿ=2187=3⁷

∴ n=7

First term 2, common ratio 3 and sum 728

728=2(3ⁿ−1)/(3−1)

3ⁿ=729

∴ n=6

3rd term 12 and 7th term 192. Find S₁₀.

ar²=12, ar⁶=192

r⁴=16 ⇒ r=2

a=3

∴ S₁₀=3069

3rd term 12 and 6th term 96. Find S₈.

ar²=12, ar⁵=96

r³=8 ⇒ r=2

a=3

∴ S₈=765

S₄=40 and sum of first two terms = 4

a(r+1)(r²+1)=40

a(r+1)=4

r²+1=10 ⇒ r=3

a=1

S₈=(3⁸−1)/2

∴ S₈=3280