Exercise 5.3 answers: Arithmetic Progressions

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(i) 2, 7, 12, …, to 10 terms

1. a = 2, d = 5, n = 10.
2. Sₙ = n/2 × [2a + (n−1)d] = 10/2 × [4 + 9×5] = 5 × 49 = 245.

Answer245

(ii) −37, −33, −29, …, to 12 terms

1. a = −37, d = 4, n = 12.
2. S₁₂ = 12/2 × [2×(−37) + 11×4] = 6 × [−74 + 44] = 6 × (−30) = −180.

Answer−180

(iii) 0.6, 1.7, 2.8, …, to 100 terms

1. a = 0.6, d = 1.1, n = 100.
2. S₁₀₀ = 100/2 × [2×0.6 + 99×1.1] = 50 × [1.2 + 108.9] = 50 × 110.1 = 5505.

Answer5505

(iv) 1/15, 1/12, 1/10, …, to 11 terms

1. a = 1/15, d = 1/12 − 1/15 = 1/60, n = 11.
2. S₁₁ = 11/2 × [2×(1/15) + 10×(1/60)] = 11/2 × [2/15 + 1/6] = 11/2 × 3/10 = 33/20.

Answer33/20

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(i) 7 + 10½ + 14 + . . . + 84

1. This is an AP: first term a = 7, common difference d = 10½ − 7 = 7/2, last term l = 84.
2. Use l = a + (n − 1)d to find how many terms there are: 84 = 7 + (n − 1)×(7/2), which gives n = 23.
3. Sum = n/2 × (a + l) = 23/2 × (7 + 84) = 23/2 × 91 = 2093/2.

Answer2093/2 (that is, 1046½)

(ii) 34 + 32 + 30 + . . . + 10

1. First term a = 34, common difference d = 32 − 34 = −2, last term l = 10.
2. Find n: 10 = 34 + (n − 1)×(−2), which gives n = 13.
3. Sum = n/2 × (a + l) = 13/2 × (34 + 10) = 13/2 × 44 = 286.

Answer286

(iii) –5 + (–8) + (–11) + . . . + (–230)

1. First term a = −5, common difference d = −8 − (−5) = −3, last term l = −230.
2. Find n: −230 = −5 + (n − 1)×(−3), which gives n = 76.
3. Sum = n/2 × (a + l) = 76/2 × (−5 + (−230)) = 38 × (−235) = −8930.

Answer−8930

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(i) given a = 5, d = 3, aₙ = 50, find n and Sₙ.

1. aₙ = a + (n − 1)d gives 50 = 5 + (n − 1) × 3, so (n − 1) × 3 = 45, n − 1 = 15 and n = 16.
2. Sₙ = n/2 × (a + aₙ) = 16/2 × (5 + 50) = 8 × 55 = 440.

Answern = 16, S₁₆ = 440

(ii) given a = 7, a₁₃ = 35, find d and S₁₃.

1. a₁₃ = a + 12d gives 35 = 7 + 12d, so 12d = 28 and d = 28/12 = 7/3.
2. S₁₃ = 13/2 × (a + a₁₃) = 13/2 × (7 + 35) = 13/2 × 42 = 273.

Answerd = 7/3, S₁₃ = 273

(iii) given a₁₂ = 37, d = 3, find a and S₁₂.

1. a₁₂ = a + 11d gives 37 = a + 11 × 3 = a + 33, so a = 4.
2. S₁₂ = 12/2 × (a + a₁₂) = 6 × (4 + 37) = 6 × 41 = 246.

Answera = 4, S₁₂ = 246

(iv) given a₃ = 15, S₁₀ = 125, find d and a₁₀.

1. a₃ = a + 2d = 15 … (1)
2. S₁₀ = 10/2 × (2a + 9d) = 125, so 2a + 9d = 25 … (2)
3. From (1), 2a = 30 − 4d. Put this into (2): 30 − 4d + 9d = 25, so 5d = −5 and d = −1.
4. Then a = 15 − 2 × (−1) = 17, and a₁₀ = a + 9d = 17 − 9 = 8.

Answerd = −1, a₁₀ = 8

(v) given d = 5, S₉ = 75, find a and a₉.

1. S₉ = 9/2 × (2a + 8 × 5) = 75, so 2a + 40 = 75 × 2/9 = 50/3.
2. 2a = 50/3 − 40 = 50/3 − 120/3 = −70/3, so a = −35/3.
3. a₉ = a + 8d = −35/3 + 40 = −35/3 + 120/3 = 85/3.

Answera = −35/3, a₉ = 85/3

(vi) given a = 2, d = 8, Sₙ = 90, find n and aₙ.

1. Sₙ = n/2 × (2a + (n − 1)d) = n/2 × (4 + 8(n − 1)) = n/2 × (8n − 4) = n(4n − 2).
2. So n(4n − 2) = 90, which gives 4n² − 2n − 90 = 0, or 2n² − n − 45 = 0.
3. This factorises as (n − 5)(2n + 9) = 0, so n = 5 or n = −9/2. A number of terms must be a positive whole number, so n = 5.
4. aₙ = a₅ = a + 4d = 2 + 4 × 8 = 34.

Answern = 5, a₅ = 34

(vii) given a = 8, aₙ = 62, Sₙ = 210, find n and d.

1. Sₙ = n/2 × (a + aₙ) gives 210 = n/2 × (8 + 62) = 35n, so n = 6.
2. aₙ = a + (n − 1)d gives 62 = 8 + 5d, so 5d = 54 and d = 54/5.

Answern = 6, d = 54/5

(viii) given aₙ = 4, d = 2, Sₙ = –14, find n and a.

1. aₙ = a + (n − 1)d gives 4 = a + 2(n − 1), so a = 6 − 2n.
2. Sₙ = n/2 × (a + aₙ) = n/2 × (6 − 2n + 4) = n/2 × (10 − 2n) = n(5 − n).
3. So n(5 − n) = −14, which gives n² − 5n − 14 = 0, that is (n − 7)(n + 2) = 0.
4. n = −2 is not possible for a number of terms, so n = 7, and a = 6 − 2 × 7 = −8.

Answern = 7, a = −8

(ix) given a = 3, n = 8, S = 192, find d.

1. S = n/2 × (2a + (n − 1)d) gives 192 = 8/2 × (6 + 7d) = 4 × (6 + 7d).
2. So 6 + 7d = 48, 7d = 42 and d = 6.

Answerd = 6

(x) given l = 28, S = 144, and there are total 9 terms

1. S = n/2 × (a + l) gives 144 = 9/2 × (a + 28).
2. So a + 28 = 144 × 2/9 = 32, and a = 4.

Answera = 4

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1. AP: a = 9, d = 17 − 9 = 8.
2. Sₙ = n/2×(2×9+(n−1)×8) = 636 gives 4n²+5n−636=0.
3. Solving the quadratic gives n = 12 (the other root is negative, so it is rejected).

Answer12 terms

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1. Sₙ = n/2×(a+l) gives 400 = n/2×(5+45) = 25n, so n = 16.
2. l = a+(n−1)d gives 45 = 5+15d, so d = 8/3.

Answer16 terms, common difference 8/3

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1. l = a+(n−1)d gives 350 = 17+(n−1)×9, so n = 38.
2. Sₙ = n/2×(a+l) = 38/2×(17+350) = 19×367 = 6973.

Answer38 terms, sum 6973

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1. a₂₂ = a+21d gives 149 = a+147, so a = 2.
2. S₂₂ = 22/2×(2×2+21×7) = 11×151 = 1661.

Answer1661

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1. d = a₃ − a₂ = 18−14 = 4, and a = a₂−d = 14−4 = 10.
2. S₅₁ = 51/2×(2×10+50×4) = 51/2×220 = 5610.

Answer5610

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1. Given: S₇ = 49, S₁₇ = 289
2. Use Sₙ = n/2 × (2a + (n − 1)d)
3. For n = 7: 49 = 7/2 × (2a + 6d)
4. 14 = 2a + 6d
5. a + 3d = 7 ... (1)
6. For n = 17: 289 = 17/2 × (2a + 16d)
7. 34 = 2a + 16d
8. a + 8d = 17 ... (2)
9. Subtracting (1) from (2): 5d = 10, d = 2
10. From (1): a + 6 = 7, a = 1
11. Find Sₙ = n/2 × (2 × 1 + (n − 1) × 2)
12. Sₙ = n/2 × (2 + 2n − 2) = n/2 × 2n = n²

AnswerSₙ = n²

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(i) aₙ = 3 + 4n

1. aₙ₊₁ − aₙ = (3+4(n+1)) − (3+4n) = 4, a constant, so it is an AP with d = 4.
2. First term a₁ = 3+4 = 7.
3. S₁₅ = 15/2×(2×7+14×4) = 15/2×70 = 525.

Answerd = 4, S₁₅ = 525

(ii) aₙ = 9 – 5n

1. aₙ₊₁ − aₙ = (9−5(n+1)) − (9−5n) = −5, a constant, so it is an AP with d = −5.
2. First term a₁ = 9−5 = 4.
3. S₁₅ = 15/2×(2×4+14×(−5)) = 15/2×(−62) = −465.

Answerd = −5, S₁₅ = −465

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1. S₁ = 4 × 1 − 1² = 3. The sum of the first one term is just the first term, so the first term is 3.
2. S₂ = 4 × 2 − 2² = 8 − 4 = 4, so the sum of the first two terms is 4.
3. Second term = S₂ − S₁ = 4 − 3 = 1.
4. S₃ = 4 × 3 − 3² = 12 − 9 = 3, so the 3rd term = S₃ − S₂ = 3 − 4 = −1.
5. S₁₀ = 4 × 10 − 10² = −60 and S₉ = 4 × 9 − 9² = −45, so the 10th term = S₁₀ − S₉ = −60 − (−45) = −15.
6. In general, aₙ = Sₙ − Sₙ₋₁ = (4n − n²) − (4(n − 1) − (n − 1)²) = 4n − n² − (4n − 4 − n² + 2n − 1) = 4n − n² − 4n + 4 + n² − 2n + 1 = 5 − 2n.

Answerfirst term 3, sum of first two terms 4, second term 1, 3rd term −1, 10th term −15, aₙ = 5 − 2n

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1. The positive integers divisible by 6 are 6, 12, 18, . . ., an AP with a = 6, d = 6.
2. S₄₀ = 40/2×(2×6+39×6) = 20×246 = 4920.

Answer4920

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1. The multiples of 8 are 8, 16, 24, . . ., an AP with a = 8, d = 8.
2. S₁₅ = 15/2×(2×8+14×8) = 15/2×128 = 960.

Answer960

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1. The odd numbers are 1, 3, 5, . . ., 49: a = 1, d = 2, l = 49.
2. l = a+(n−1)d gives 49 = 1+(n−1)×2, so n = 25.
3. S₂₅ = 25/2×(1+49) = 25×25 = 625.

Answer625

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1. Penalties: ₹200, ₹250, ₹300, . . ., an AP with a = 200, d = 50.
2. S₃₀ = 30/2×(2×200+29×50) = 15×1850 = 27750.

Answer₹27750

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1. Let the first (largest) prize be ₹a. Each prize is ₹20 less than the one before, so the prizes form an AP with d = −20 and n = 7.
2. S₇ = 7/2 × (2a + 6 × (−20)) = 700, so 2a − 120 = 700 × 2/7 = 200.
3. 2a = 320, so a = 160.
4. The prizes are ₹160, ₹140, ₹120, ₹100, ₹80, ₹60 and ₹40 (check: they add up to ₹700).

Answer₹160, ₹140, ₹120, ₹100, ₹80, ₹60, ₹40

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1. One section of Class I to Class XII plants 1, 2, 3, . . ., 12 trees, an AP with a = 1, d = 1, n = 12.
2. Sum for one section = 12/2×(1+12) = 78.
3. There are 3 sections of each class, so total trees = 3×78 = 234.

Answer234 trees

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1. Length of a semicircle of radius r is πr, so total length = π × (sum of the 13 radii).
2. Radii are 0.5, 1.0, 1.5, . . ., an AP with a = 0.5, d = 0.5, n = 13.
3. Sum of radii = 13/2×(2×0.5+12×0.5) = 13/2×7 = 45.5 cm.
4. Total length = 22/7×45.5 = 143 cm.

Answer143 cm

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1. Logs per row form an AP: a = 20, d = −1.
2. Sₙ = n/2×(40−(n−1)) = 200 gives n²−41n+400 = 0.
3. This gives n = 16 or n = 25; n = 25 would need the top row to have 20−24 = −4 logs, which is impossible, so n = 16.
4. Top row = a+(16−1)(−1) = 20−15 = 5 logs.

Answer16 rows, 5 logs in the top row

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1. Distance for potato 1 (there and back) = 2×5 = 10 m.
2. Distance for potato 2 = 2×(5+3) = 16 m, and so on: an AP with a = 10, d = 6, n = 10.
3. Total = 10/2×(2×10+9×6) = 5×74 = 370 m.

Answer370 m

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