Chapter-5 (Sequence and Series) Exercise Solutions

Sequence and Series: Exercise Solutions

Mathematics • Grade 10 • Exercises 6.1 and 6.2

Notes: Misprints in the book or answer key are pointed out in small print, and the solution shown is the one that is mathematically consistent with the key.

Exercise 6.1: Arithmetic Means and Sum of Arithmetic Series

1. (a) What are arithmetic means? (b) If a, m, b are in arithmetic sequence, express m in terms of a and b. (c) Find the mid value of 12 and 18.

(a) The terms lying between the first and last terms of an arithmetic sequence are arithmetic means.
(b) m − a = b − m ⇒ m = (a + b)/2
(c) (12 + 18)/2 = 15

2. Find the arithmetic mean between (a) 6 and 10 (b) −2 and 2 (c) −4 and 8 (d) (a + b) and (a − b).

(a) (6 + 10)/2 = 8
(b) (−2 + 2)/2 = 0
(c) (−4 + 8)/2 = 2 (the key prints 0)
(d) [(a + b) + (a − b)]/2 = 2a/2 = a

3. Find the arithmetic means: (a) 4 means between 5 and 20 (b) 6 means between 70 and 14 (c) 6 means between 5 and −9.

(a) n = 6: 20 = 5 + 5d ⇒ d = 3. Means: 8, 11, 14, 17
(b) n = 8: 14 = 70 + 7d ⇒ d = −8. Means: 62, 54, 46, 38, 30, 22
(c) n = 8: −9 = 5 + 7d ⇒ d = −2. Means: 3, 1, −1, −3, −5, −7

4. The sequences are arithmetic. Find x: (a) 5, x, 9, … (b) x + 1, x + 5, 3x + 1, … (c) x + 2, 3x, 4x + 1, …

(a) x − 5 = 9 − x ⇒ x = 7
(b) (x + 5) − (x + 1) = (3x + 1) − (x + 5) ⇒ 4 = 2x − 4 ⇒ x = 4
(c) 3x − (x + 2) = (4x + 1) − 3x ⇒ 2x − 2 = x + 1 ⇒ x = 3

5. Find the sum: (a) 7 + 11 + 15 + 19 + … to 20 terms (b) 4 − 1 − 6 − 11 − 16 − … to 7 terms (c) ½ + 3/2 + 5/2 + … to 16 terms (d) 5 + 10 + 15 + … + 65 (e) −64 − 48 − 32 − … + 32 (f) sum of first 10 odd numbers (g) sum of first 100 natural numbers (h) sum of natural numbers from 50 to 100.

(a) a = 7, d = 4: S₂₀ = 10 [14 + 76] = 900
(b) a = 4, d = −5: S₇ = 7/2 [8 − 30] = −77
(c) a = ½, d = 1: S₁₆ = 8 [1 + 15] = 128
(d) d = 5, 65 = 5 + (n − 1)5 ⇒ n = 13. S = 13/2 (5 + 65) = 455
(e) d = 16, 32 = −64 + (n − 1)16 ⇒ n = 7. S = 7/2 (−64 + 32) = −112
(f) 1 + 3 + 5 + …: a = 1, d = 2, n = 10: S = 5 [2 + 18] = 100
(g) S = 100/2 (1 + 100) = 5050
(h) n = 51: S = 51/2 (50 + 100) = 3825

6. Find the common difference. (a) first term 1, last term 50, sum 204 (b) first term 160, last term 20, sum 440 (c) first term 17, last term −99/8, sum 407/16.

(a) 204 = n/2 (51) ⇒ n = 8. 50 = 1 + 7d ⇒ d = 7
(b) As printed (a = 160), n = 880/180 is not a whole number. The key (d = −4) fits a first term of 60: 440 = n/2 (80) ⇒ n = 11; 20 = 60 + 10d ⇒ d = −4
(c) a + l = 17 − 99/8 = 37/8. n = (2 × 407/16) ÷ (37/8) = 11. −99/8 = 17 + 10d ⇒ d = (−235/8)/10 = −47/16

7. Find the first term. (a) d = −3, n = 10, sum 325 (b) d = 9, n = 9, sum 108 (c) d = 3, n = 10, sum 155.

(a) 325 = 5 [2a − 27] ⇒ 2a = 92 ⇒ a = 46
(b) 108 = 9/2 [2a + 72] ⇒ 24 = 2a + 72 ⇒ a = −24
(c) 155 = 5 [2a + 27] ⇒ 31 = 2a + 27 ⇒ a = 2

8. (a) How many terms of 4 + 10 + 16 + 22 + … have sum 374? (b) The first term is 36 and d = 9; how many terms have sum 540?

(a) a = 4, d = 6: 374 = n/2 [8 + 6(n − 1)] ⇒ 3n² + n − 374 = 0 ⇒ (3n + 34)(n − 11) = 0 ⇒ n = 11
(b) 540 = n/2 [72 + 9(n − 1)] ⇒ 9n² + 63n − 1080 = 0 ⇒ n² + 7n − 120 = 0 ⇒ (n + 15)(n − 8) = 0 ⇒ n = 8

9. (a) Sum of first 16 terms if t₃ = −15 and t₈ = 10. (b) Sum of first 20 terms if t₅ = 10 and t₁₁ = 22.

(a) t₈ − t₃ = 5d = 25 ⇒ d = 5; a = −15 − 10 = −25. S₁₆ = 8 [−50 + 75] = 200
(b) 6d = 12 ⇒ d = 2; a = 10 − 8 = 2. S₂₀ = 10 [4 + 38] = 420

10. (a) The sum of first 6 terms is 75 and of first 12 terms is 390; find the sum of first 20 terms. (b) The sum of first 7 terms is 21 and of first 12 terms is 126; find the series.

(a) Corrected question: the sum of the first 6 terms is 105 (the key's 1050 requires this). 3(2a + 5d) = 105 ⇒ 2a + 5d = 35; 6(2a + 11d) = 390 ⇒ 2a + 11d = 65. Subtracting: 6d = 30 ⇒ d = 5, a = 5. S₂₀ = 10 [10 + 95] = 1050
(b) 7/2 (2a + 6d) = 21 ⇒ a + 3d = 3; 6(2a + 11d) = 126 ⇒ 2a + 11d = 21. 2(3 − 3d) + 11d = 21 ⇒ 5d = 15 ⇒ d = 3, a = −6. Series: −6 − 3 + 0 + 3 + …

11. (a) A worker produced 1000 Nepali caps in the first year and increased production by 100 every year. Can he produce 15,000 caps in 10 years? (b) A new employee earns Rs. 60,000 initially and the pay rises by Rs. 24,000 every year. In how many years will he earn Rs. 1 crore in total?

(a) a = 1000, d = 100, n = 10: S₁₀ = 5 [2000 + 900] = 14,500. No, he produces 500 caps less.
(b) Yearly income a = 60,000 × 12 = 7,20,000 and, as in the key, it rises by Rs. 24,000 each year: n/2 [14,40,000 + 24,000(n − 1)] = 1,00,00,000 ⇒ 3n² + 177n − 2500 = 0 ⇒ n = 11.8 ⇒ nearly 12 years. (If the monthly pay itself rose by Rs. 24,000 each year, it would take about 6.6 years.)

Exercise 6.2: Geometric Means and Sum of Geometric Series

1. (a) What is meant by geometric mean? (b) If positive numbers a, m, b are in geometric sequence, express m in terms of a and b. (c) What is the geometric mean between 3 and 27?

(a) The terms lying between the first and last terms of a geometric sequence are geometric means.
(b) m/a = b/m ⇒ m² = ab ⇒ m = √(ab)
(c) √(3 × 27) = √81 = 9

2. Find the geometric mean between (a) −4 and −64 (b) 1/5 and 125 (c) 7 and 343.

(a) m² = 256 ⇒ m = ±16. With a positive common ratio the sequence −4, −16, −64 has r = 4, so m = −16.
(b) √(1/5 × 125) = √25 = 5
(c) √(7 × 343) = √2401 = 49

3. Find the geometric means: (a) 4 means between 6 and 192 (b) 3 means between 5 and 405 (c) 3 means between 9/4 and 4/9.

(a) n = 6: 192 = 6r⁵ ⇒ r⁵ = 32 ⇒ r = 2. Means: 12, 24, 48, 96
(b) n = 5: 405 = 5r⁴ ⇒ r = 3. Means: 15, 45, 135
(c) n = 5: 4/9 = (9/4)r⁴ ⇒ r⁴ = 16/81 ⇒ r = 2/3. Means: 3/2, 1, 2/3

4. Find x in the geometric sequences: (a) 4, x, 9 (b) x, 4, 8 (c) 5, 25, x + 1.

(a) x² = 36 ⇒ x = 6
(b) 4² = 8x ⇒ x = 2
(c) 25² = 5(x + 1) ⇒ x + 1 = 125 ⇒ x = 124

5. Find the sum: (a) 2 + 4 + 8 + 16 + … to 6 terms (b) 1/9 + 1/3 + 1 + … to 5 terms (c) −1/4 + 1/2 − 1 + … to 6 terms (d) 16 + 8 + 4 + … + 1/16 (e) 1 + 1/3 + 1/9 + … + 1/729.

(a) r = 2: S₆ = 2(2⁶ − 1)/(2 − 1) = 126
(b) a = 1/9, r = 3: S₅ = (1/9)(243 − 1)/2 = 121/9
(c) a = −1/4, r = −2: S₆ = (−1/4)(64 − 1)/(−3) = 21/4
(d) r = ½, l = 1/16: S = (lr − a)/(r − 1) = (1/32 − 16)/(−1/2) = 511/16 = 31 15/16
(e) r = 1/3, l = 1/729 = (1/3)⁶ (7 terms): S = (lr − a)/(r − 1) = (1/2187 − 1)/(−2/3) = 1093/729 (the key prints 1364/729, which is too large: the sum can never exceed 3/2)

6. Find the common ratio: (a) first term 2, last term 486, sum 728 (b) first term 5, last term 1215, sum 1820 (c) first term 3, last term 768, sum 1533.

Use S = (lr − a)/(r − 1).
(a) 728(r − 1) = 486r − 2 ⇒ 242r = 726 ⇒ r = 3
(b) 1820(r − 1) = 1215r − 5 ⇒ 605r = 1815 ⇒ r = 3
(c) 1533(r − 1) = 768r − 3 ⇒ 765r = 1530 ⇒ r = 2

7. (a) Sum of first 10 terms if t₂ = 4 and t₇ = 128. (b) Sum of first 7 terms if t₃ = 3 and t₅ = 81.

(a) t₇ ÷ t₂ = r⁵ = 32 ⇒ r = 2; a = 4 ÷ 2 = 2. S₁₀ = 2(2¹⁰ − 1)/(2 − 1) = 2046
(b) The key (1093) fits t₃ = 9: r² = 81 ÷ 9 = 9 ⇒ r = 3; a = 9 ÷ 9 = 1. S₇ = (3⁷ − 1)/2 = 1093. (With t₃ = 3 as printed, r² = 27 and r is irrational.)

8. (a) How many terms of 32 + 48 + 72 + … have sum 665? (b) How many terms of 6 − 12 + 24 − 48 + … have sum −2046?

(a) r = 3/2: 665 = 32((3/2)ⁿ − 1)/(1/2) ⇒ (3/2)ⁿ = 665/64 + 1 = 729/64 = (3/2)⁶ ⇒ n = 6
(b) r = −2: −2046 = 6((−2)ⁿ − 1)/(−3) ⇒ (−2)ⁿ − 1 = 1023 ⇒ (−2)ⁿ = 1024 ⇒ n = 10

9. Sarita borrowed Rs. 43,680 from Garima and pays it back in 6 instalments, each three times the previous one. Find the difference of the first and last instalments.

r = 3, n = 6: 43,680 = a(3⁶ − 1)/2 = 364a ⇒ a = 120. Last = 120 × 3⁵ = 29,160. Difference = 29,160 − 120 = Rs. 29,040

Corrections to the Textbook

Question	Book says	Correct
6.1 Q2(c)	0	2
6.1 Q6(b)	first term 160	first term 60 (d = −4)
6.1 Q10(a)	S₆ = 75	S₆ = 105 (S₂₀ = 1050)
6.1 Q11(b)	pay rises Rs. 24,000 a year	yearly income rises by Rs. 24,000 (about 12 years)
6.2 Q5(e)	1364/729	1093/729
6.2 Q7(b)	t₃ = 3	t₃ = 9 (S₇ = 1093)
Notes, AM Example 3	t₉ = 23	t₉ = 43

Sequence and Series • Grade 10 Mathematics