Exercise 5.2 answers: Arithmetic Progressions

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(i)

1. The row gives a = 7, d = 3, n = 8.
2. Use aₙ = a + (n−1)d.
3. a₈ = 7 + (8−1)×3 = 7 + 21 = 28.

Answeraₙ = 28

(ii)

1. The row gives a = −18, n = 10, aₙ = 0.
2. Use aₙ = a + (n−1)d with aₙ = 0.
3. 0 = −18 + (10−1)d, so 9d = 18, d = 2.

Answerd = 2

(iii)

1. The row gives d = −3, n = 18, aₙ = −5.
2. Use aₙ = a + (n−1)d with aₙ = −5.
3. −5 = a + (18−1)(−3) = a − 51, so a = 46.

Answera = 46

(iv)

1. The row gives a = −18.9, d = 2.5, aₙ = 3.6.
2. Use aₙ = a + (n−1)d with aₙ = 3.6.
3. 3.6 = −18.9 + (n−1)×2.5, so (n−1)×2.5 = 22.5, n−1 = 9, n = 10.

Answern = 10

(v)

1. The row gives a = 3.5, d = 0, n = 105.
2. Use aₙ = a + (n−1)d with d = 0.
3. aₙ = 3.5 + (105−1)×0 = 3.5 (the list never moves).

Answeraₙ = 3.5

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(i) 30th term of the AP: 10, 7, 4

1. a = 10, d = 7 − 10 = −3.
2. a₃₀ = a + 29d = 10 + 29×(−3) = 10 − 87 = −77.
3. This matches option (C).

Answer−77 (option C)

(ii) 11th term of the AP: −3, −1/2, 2

1. a = −3, d = −1/2 − (−3) = 5/2.
2. a₁₁ = a + 10d = −3 + 10×(5/2) = −3 + 25 = 22.
3. This matches option (B).

Answer22 (option B)

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(i) 2, ▢, 26

1. This is an AP of 3 terms: a₁ = 2, a₃ = 26.
2. d = (26 − 2)/2 = 12.
3. Missing term a₂ = 2 + 12 = 14.

Answer14

(ii) ▢, 13, ▢, 3

1. This is an AP of 4 terms: a₂ = 13, a₄ = 3.
2. 2d = 3 − 13 = −10, so d = −5.
3. a₁ = 13 − d = 18. a₃ = 13 + d = 8.

Answera₁ = 18, a₃ = 8

(iii) 5, ▢, ▢, 9 1/2

1. This is an AP of 4 terms: a₁ = 5, a₄ = 9 1/2.
2. 3d = 9.5 − 5 = 4.5, so d = 1.5.
3. a₂ = 5 + 1.5 = 6.5, a₃ = 6.5 + 1.5 = 8.

Answera₂ = 13/2 = 6 1/2, a₃ = 8

(iv) −4, ▢, ▢, ▢, ▢, 6

1. This is an AP of 6 terms: a₁ = −4, a₆ = 6.
2. 5d = 6 − (−4) = 10, so d = 2.
3. a₂ = −2, a₃ = 0, a₄ = 2, a₅ = 4.

Answera₂ = −2, a₃ = 0, a₄ = 2, a₅ = 4

(v) ▢, 38, ▢, ▢, ▢, −22

1. This is an AP of 6 terms: a₂ = 38, a₆ = −22.
2. 4d = −22 − 38 = −60, so d = −15.
3. a₁ = 38 − d = 53. a₃ = 38 + d = 23. a₄ = 8. a₅ = −7.

Answera₁ = 53, a₃ = 23, a₄ = 8, a₅ = −7

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1. a = 3, d = 8 − 3 = 5.
2. Set a + (n−1)d = 78: 3 + (n−1)×5 = 78.
3. (n−1)×5 = 75, so n − 1 = 15, n = 16.

AnswerThe 16th term is 78.

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(i) 7, 13, 19, …, 205

1. a = 7, d = 13 − 7 = 6, last term = 205.
2. 205 = 7 + (n−1)×6, so (n−1)×6 = 198, n−1 = 33.
3. n = 34.

Answer34 terms

(ii) 18, 15 1/2, 13, …, −47

1. a = 18, d = 15 1/2 − 18 = −2.5, last term = −47.
2. −47 = 18 + (n−1)×(−2.5), so (n−1)×(−2.5) = −65, n−1 = 26.
3. n = 27.

Answer27 terms

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1. a = 11, d = 8 − 11 = −3.
2. Set a + (n−1)d = −150: 11 + (n−1)×(−3) = −150.
3. (n−1)×(−3) = −161, so n − 1 = 161/3, which is not a whole number.

AnswerNo, −150 is not a term of this AP.

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1. a + 10d = 38 and a + 15d = 73.
2. Subtracting: 5d = 35, so d = 7. Then a = 38 − 70 = −32.
3. a₃₁ = a + 30d = −32 + 210 = 178.

Answer178

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1. a + 2d = 12 and a + 49d = 106 (the 50th term is the last term).
2. Subtracting: 47d = 94, so d = 2. Then a = 12 − 4 = 8.
3. a₂₉ = a + 28d = 8 + 56 = 64.

Answer64

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1. a + 2d = 4 and a + 8d = −8.
2. Subtracting: 6d = −12, so d = −2. Then a = 4 − 2×(−2) = 8.
3. Set a + (n−1)d = 0: 8 + (n−1)×(−2) = 0, so n − 1 = 4, n = 5.

AnswerThe 5th term is zero.

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1. a₁₇ − a₁₀ = (a + 16d) − (a + 9d) = 7d.
2. This gap is given as 7, so 7d = 7.
3. d = 1.

Answerd = 1

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1. a = 3, d = 15 − 3 = 12.
2. aₙ − a₅₄ = (n − 54)×d, and this must equal 132.
3. (n − 54)×12 = 132, so n − 54 = 11, n = 65.

AnswerThe 65th term.

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1. Let the two APs have first terms a₁, a₂ and the same common difference d.
2. 100th terms: (a₁+99d) − (a₂+99d) = a₁ − a₂ = 100. The d's cancel.
3. 1000th terms: (a₁+999d) − (a₂+999d) = a₁ − a₂ as well — the same 100, since d cancels again.

Answer100 — the difference stays the same at every position, because the equal common differences always cancel.

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1. The smallest 3-digit multiple of 7 is 105; the largest is 994.
2. These form an AP: a = 105, d = 7, last term = 994.
3. 994 = 105 + (n−1)×7, so n − 1 = 127, n = 128.

Answer128

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1. The first multiple of 4 after 10 is 12; the last one before 250 is 248.
2. These form an AP: a = 12, d = 4, last term = 248.
3. 248 = 12 + (n−1)×4, so n − 1 = 59, n = 60.

Answer60

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1. First AP: a = 63, d = 2, so its nth term is 63 + (n−1)×2.
2. Second AP: a = 3, d = 7, so its nth term is 3 + (n−1)×7.
3. Setting them equal: 63 + 2(n−1) = 3 + 7(n−1), so 5(n−1) = 60, n − 1 = 12, n = 13.

Answern = 13

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1. a₇ − a₅ = 2d, and this equals 12, so d = 6.
2. a₃ = a + 2d = 16, so a = 16 − 12 = 4.
3. The AP is 4, 10, 16, 22, …

Answera = 4, d = 6

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1. Read the AP backwards: it starts at 253 and steps by −5 each time (the reverse of +5).
2. The 20th term from the last is the 20th term of this reversed AP.
3. 253 + (20−1)×(−5) = 253 − 95 = 158.

Answer158

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1. a₄+a₈ = (a+3d)+(a+7d) = 2a+10d = 24, so a+5d = 12.
2. a₆+a₁₀ = (a+5d)+(a+9d) = 2a+14d = 44, so a+7d = 22.
3. Subtracting: 2d = 10, d = 5. Then a = 12 − 25 = −13.
4. First three terms: −13, −8, −3.

Answer−13, −8, −3

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1. His salary forms an AP: a = 5000, d = 200.
2. Set a + (n−1)d = 7000: 5000 + (n−1)×200 = 7000, so n − 1 = 10, n = 11.
3. The 11th year of work, starting from 1995, is 1995 + 10 = 2005.

Answer2005, his 11th year counting 1995 as the first (NCERT's answer says "11th year").

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1. Her savings form an AP: a = 5, d = 1.75.
2. Set a + (n−1)d = 20.75: 5 + (n−1)×1.75 = 20.75, so (n−1)×1.75 = 15.75.
3. n − 1 = 9, so n = 10.

Answern = 10

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