Class 10 Mathematics Unit 6 – Sequence and Series | Nepal eNotes

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CLASS 10 MATHEMATICS • UNIT 6

Unit 6: Sequence and Series

Arithmetic Sequence & Series and Geometric Sequence & Series — WordPress-ready notes reconstructed from the supplied 47-page handwritten PDF.

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Source note: Although the uploaded file name says “Cylinder, Sphere, Hemisphere and Cone,” the actual 47 scanned pages are handwritten notes for Lesson 6: Sequence and Series. The page content has therefore been used as the source of truth.

Arithmetic Progression (A.P.) – Important Formulae

Arithmetic Mean

A.M. = (a + b) / 2

n-th Term

t_n = a + (n − 1)d

Last Term

l = a + (n − 1)d

Sum of n Terms

S_n = n/2 [2a + (n − 1)d]

Alternative Sum Formula

S_n = n/2 (a + l)

PDF Pages 1–28

Exercise 6.1 – Arithmetic Sequence and Series

PDF Pages 1–2

1. Arithmetic Mean

(a) Between 6 and 10

A.M. = (6 + 10)/2

∴ A.M. = 8

(b) Between −2 and 2

A.M. = (−2 + 2)/2

∴ A.M. = 0

(c) Between −9 and 8

A.M. = (−9 + 8)/2

∴ A.M. = −1/2

(d) Between (a+b) and (a−b)

A.M. = [(a+b) + (a−b)]/2

= 2a/2

∴ A.M. = a

PDF Pages 2–4

2. Insert Arithmetic Means

(a) Four arithmetic means between 5 and 20

Total terms = 6

20 = 5 + (6−1)d

15 = 5d

d = 3

Required means: 8, 11, 14, 17

(b) Six arithmetic means between 70 and 14

Total terms = 8

14 = 70 + 7d

d = −8

Required means: 62, 54, 46, 38, 30, 22

(c) Six arithmetic means between 5 and −9

Total terms = 8

−9 = 5 + 7d

d = −2

Required means: 3, 1, −1, −3, −5, −7

PDF Pages 5–6

3. Find the Unknown Value in an A.P.

(a) 5, n, 9 are in A.P.

Arithmetic mean of first and last terms:

n = (5 + 9)/2

∴ n = 7

(b) n+1, n+5, 3n+1 are in A.P.

n + 5 = [(n+1) + (3n+1)] / 2

2n + 10 = 4n + 2

8 = 2n

∴ n = 4

(c) n+2, 3n, 9n+1 are in A.P.

3n = [(n+2) + (9n+1)] / 2

6n = 10n + 3

Source working concludes n = 3.

The handwritten algebra in part (c) is internally inconsistent, so the source result is preserved as written rather than silently replacing it.

PDF Pages 6–10

4. Sum of Arithmetic Series

(a) 7 + 11 + 15 + 19 + … to 20 terms

a = 7, d = 4, n = 20

S₂₀ = 20/2 [2×7 + 19×4]

= 10(14 + 76)

∴ S₂₀ = 900

(b) 9 + 4 − 1 − 6 − … to 20 terms

a = 9, d = −5, n = 20

S₂₀ = 20/2 [18 + 19(−5)]

= 10(18 − 95)

∴ S₂₀ = −770

(c) 1/2 + 3/2 + 5/2 + … to 16 terms

a = 1/2, d = 1, n = 16

S₁₆ = 16/2 [1 + 15]

= 8 × 16

∴ S₁₆ = 128

(d) 5 + 10 + 15 + … + 65

a = 5, d = 5, l = 65

65 = 5 + (n−1)5

n = 13

S₁₃ = 13/2(5 + 65)

∴ S = 455

(e) −64, −48, −32, … , 32

a = −64, d = 16, l = 32

32 = −64 + (n−1)16

n = 7

S₇ = 7/2(−64 + 32)

∴ S = −112

PDF Pages 10–13

5. Sums of Natural Numbers

(a) Sum of first 10 odd numbers

Series: 1 + 3 + 5 + …

a = 1, d = 2, n = 10

S₁₀ = 10/2 [2 + 18]

∴ S = 100

(b) Sum of first 100 natural numbers

1 + 2 + 3 + … + 100

S₁₀₀ = 100/2 [2 + 99]

∴ S = 5,050

(c) Sum of natural numbers from 50 to 100

a = 50, l = 100, n = 51

S = 51/2(50 + 100)

∴ S = 3,825

PDF Pages 13–15

6. Given First Term, Last Term and Sum

(a) a = 1, l = 50, S_n = 204

204 = n/2(1 + 50)

408 = 51n

n = 8

50 = 1 + 7d

d = 7

(b) a = 17, l = −99/8, S_n = 907/16

The source applies S_n = n/2(a+l) to find n, then uses l = a+(n−1)d.

Source result: d = −99/8 (as written).

The fractions on pages 14–15 are difficult to read consistently; the method is clear, but some final handwritten fractions are retained as source-work rather than independently reconstructed.

PDF Pages 15–17

7. Find the First Term

(a) d = −3, n = 10, S₁₀ = 325

325 = 10/2 [2a + 9(−3)]

65 = 2a − 27

2a = 92

∴ a = 46

(b) d = 9, n = 9, S₉ = 108

108 = 9/2 [2a + 8×9]

24 = 2a + 72

2a = −48

∴ a = −24

(c) d = 3, n = 10, S₁₀ = 155

155 = 10/2 [2a + 9×3]

31 = 2a + 27

2a = 4

∴ a = 2

PDF Pages 17–20

8. Find the Number of Terms

(a) 4 + 10 + 16 + 22 + … has sum 379

a = 4, d = 6

379 = n/2 [8 + 6(n−1)]

379 = n(3n + 1)

3n² + n − 379 = 0

Source factorization gives n = 11.

(b) First term = 36, d = 9, sum = 540

540 = n/2 [72 + 9(n−1)]

1,080 = n(63 + 9n)

n² + 7n − 120 = 0

(n−8)(n+15)=0

∴ n = 8

PDF Pages 26–28

Application Problems of A.P.

Factory production

A worker produced 1,000 caps in the first year and increased production by 100 caps each year for 10 years.

a = 1000, d = 100, n = 10

S₁₀ = 10/2 [2×1000 + 9×100]

= 5(2,900)

∴ Total production = 14,500 caps

Therefore, the worker cannot produce 15,000 caps in 10 years.

Salary-increment problem

Pages 27–28 contain a salary problem where the monthly salary increases by a fixed amount every year and the total earnings are given.

The handwritten total and some zeroes are faint/overwritten. The source uses the A.P. sum formula and solves a quadratic equation to obtain approximately 12 years.

PDF Pages 20–26

A.P. from Given Terms or Partial Sums

9(a). 3rd term = −15 and 8th term = 10. Find S₁₆.

a + 2d = −15 … (i)

a + 7d = 10 … (ii)

Subtracting: 5d = 25 ⇒ d = 5

a + 10 = −15 ⇒ a = −25

S₁₆ = 16/2 [2(−25) + 15×5]

= 8(−50 + 75)

∴ S₁₆ = 200

9(b). 5th term = 10 and 11th term = 22. Find S₂₀.

a + 4d = 10

a + 10d = 22

6d = 12 ⇒ d = 2

a = 10 − 8 = 2

S₂₀ = 20/2 [4 + 19×2]

= 10(42)

∴ S₂₀ = 420

10(a). S₆ = 75 and S₁₂ = 390. Find S₂₀.

From S₆: 2a + 5d = 25 … (i)

From S₁₂: 2a + 11d = 65 … (ii)

Subtracting gives 6d = 40

Source obtains d = 20/3 and then finds a from (i).

The handwritten final result for S₂₀ is 3,550/3.

10(b). Sum of first 7 terms = 21 and sum of first 12 terms = 126

From S₇: 2a + 6d = 6 … (i)

From S₁₂: 2a + 11d = 21 … (ii)

5d = 15 ⇒ d = 3

2a + 18 = 6 ⇒ a = −6

12th term = −6 + 11×3 = 27

Geometric Progression (G.P.) – Important Formulae

Geometric Mean

G.M. = √(ab)

n-th Term

t_n = arⁿ⁻¹

Common Ratio

r = t_n+1 / t_n

Sum of n Terms

S_n = a(rⁿ − 1)/(r − 1), r ≠ 1

Alternative Form

S_n = a(1 − rⁿ)/(1 − r)

PDF Pages 29–47

Exercise 6.2 – Geometric Sequence and Series

PDF Pages 29–30

1. Geometric Mean

(a) Between −4 and −64

G.M. = √[(−4)(−64)]

= √256

∴ G.M. = 16

(b) Between 1/5 and 125

G.M. = √[(1/5)×125]

= √25

∴ G.M. = 5

(c) Between 7 and 343

G.M. = √(7×343)

= √2401

∴ G.M. = 49

PDF Pages 30–32

2. Insert Geometric Means

(a) Four geometric means between 6 and 192

Total intervals = 5

r = (192/6)^1/5 = 32^1/5 = 2

Required means: 12, 24, 48, 96

(b) Three geometric means between 5 and 405

r = (405/5)^1/4 = 81^1/4 = 3

Required means: 15, 45, 135

(c) Three geometric means between 9/4 and 4/9

r = [(4/9)/(9/4)]^1/4

= (16/81)^1/4

= 2/3

Required means: 3/2, 1, 2/3

PDF Pages 32–33

3. Find the Unknown Value in a G.P.

(a) 9, n, 36 are in G.P.

n² = 9×36

n² = 324

∴ n = 18 (positive geometric mean)

(b) Another three-term G.P. problem

The middle page contains a faint/ambiguous fraction sequence. The handwritten setup uses equality of successive ratios, but the first term is not legible enough to state confidently.

(c) 5, 25, n+1 are in G.P.

25/5 = (n+1)/25

5 = (n+1)/25

n+1 = 125

∴ n = 124

PDF Pages 34–38

4. Sum of Geometric Series

(a) 2 + 4 + 8 + 16 + … to 6 terms

a = 2, r = 2, n = 6

S₆ = 2(2⁶−1)/(2−1)

= 2(64−1)

∴ S₆ = 126

(b) 1/9 + 1/3 + 1 + … to 5 terms

a = 1/9, r = 3, n = 5

S₅ = (1/9)(3⁵−1)/(3−1)

= (1/9)(242/2)

∴ S₅ = 121/9

(c) −1/4 + 1/2 − 1 + … to 6 terms

a = −1/4, r = −2, n = 6

S₆ = a(1−r⁶)/(1−r)

= (−1/4)(1−64)/3

∴ S₆ = 21/4

(d) 16 + 8 + 4 + … + 1/16

a = 16, r = 1/2

1/16 = 16(1/2)ⁿ⁻¹

Source determines n and evaluates the finite sum.

Handwritten result: 511/16

(e) 1 + 1/3 + 1/9 + … + 1/729

a = 1, r = 1/3

1/729 = (1/3)⁶, so n = 7

S₇ = [1 − (1/3)⁷] / [1 − 1/3]

∴ S₇ = 1093/729

PDF Pages 38–40

5. Find the Common Ratio from First Term, Last Term and Sum

(a) First term = 2, last term = 486, sum = 728

Using S_n = (lr − a)/(r−1):

728 = (486r − 2)/(r − 1)

728r − 728 = 486r − 2

242r = 726

∴ r = 3

(b) First term = 5, last term = 1215, sum = 1820

1820 = (1215r − 5)/(r−1)

Source solution gives:

∴ r = 3

(c) First term = 3, last term = 768, sum = 1533

1533 = (768r − 3)/(r−1)

1533r − 1533 = 768r − 3

765r = 1530

∴ r = 2

PDF Pages 43–45

6. Find the Number of Terms in a G.P.

(a) 32 + 48 + 72 + … has sum 665

a = 32, r = 3/2

665 = 32[(3/2)ⁿ − 1] / (1/2)

665 = 64[(3/2)ⁿ − 1]

(3/2)ⁿ = 729/64 = (3/2)⁶

∴ n = 6

(b) 6 − 12 + 24 − 48 + … has sum −2046

a = 6, r = −2

−2046 = 6[1−(−2)ⁿ] / 3

−1023 = 1 − (−2)ⁿ

(−2)ⁿ = 1024

∴ n = 10

PDF Pages 46–47

7. Installment Problem

Sophia borrowed Rs 43,680 and repaid it in installments. The first installment was Rs 120 and each installment was three times the previous installment.

Total sum = S₆ = 43,680

a = 120, r = 3

6th installment:

t₆ = 120 × 3⁵

= 120 × 243

= Rs 29,160

Difference between first and last installments:

29,160 − 120 = Rs 29,040