Exercise 5.4 (Optional) answers: Arithmetic Progressions

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1. a = 121, d = 117−121 = −4.
2. For aₙ to be negative: 121+(n−1)×(−4) must be less than 0, which works out to n greater than 31.25.
3. Since n must be a whole number, try n = 32: a₃₂ = 121−31×4 = −3, which is negative.
4. Check n = 31: a₃₁ = 121−30×4 = 1, still positive. So the first negative term is the 32nd.

Answerthe 32nd term

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1. Let the first term be a and the common difference d. Then a₃ = a + 2d and a₇ = a + 6d.
2. Sum: (a + 2d) + (a + 6d) = 6, so 2a + 8d = 6, that is a + 4d = 3 … (1)
3. Product: (a + 2d)(a + 6d) = 8 … (2)
4. From (1), a = 3 − 4d. Then a + 2d = 3 − 2d and a + 6d = 3 + 2d, so (2) becomes (3 − 2d)(3 + 2d) = 8, that is 9 − 4d² = 8.
5. So 4d² = 1, d² = 1/4, and d = 1/2 or d = −1/2.
6. If d = 1/2: a = 3 − 4 × 1/2 = 1 (so a₃ = 2, a₇ = 4). S₁₆ = 16/2 × (2 × 1 + 15 × 1/2) = 8 × 19/2 = 76.
7. If d = −1/2: a = 3 + 4 × 1/2 = 5 (so a₃ = 4, a₇ = 2). S₁₆ = 16/2 × (2 × 5 + 15 × (−1/2)) = 8 × 5/2 = 20.
8. Both APs satisfy both conditions, so the question has two answers.

AnswerS₁₆ = 76 or S₁₆ = 20 (there are two possible APs)

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1. Top and bottom rungs are 2½ m = 250 cm apart, and rungs are 25 cm apart, so number of rungs = 250/25+1 = 11.
2. Rung lengths form an AP from a = 45 cm down to l = 25 cm, with 11 terms.
3. Total length = 11/2×(45+25) = 11/2×70 = 385 cm.

Answer385 cm

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1. Sum of house numbers 1 to (x−1) = (x−1)x/2.
2. Sum of house numbers (x+1) to 49 = sum of 1 to 49 minus sum of 1 to x = 1225 − x(x+1)/2.
3. Setting the two equal, (x−1)x/2 = 1225 − x(x+1)/2, gives x² = 1225, so x = 35 (taking the positive root, since a house number cannot be negative).

Answerx = 35

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1. Volume of the first (bottom) step = rise × tread × length = ¼×½×50 = 25/4 m³.
2. Each step above adds one more ¼ m of height over the same tread and length, so the step volumes are an AP: 25/4, 2×25/4, 3×25/4, . . ., an AP with a = 25/4, d = 25/4, n = 15.
3. Total volume = 15/2×(2×25/4+14×25/4) = 15/2×100 = 750 m³.

Answer750 m³

Watch this explained “The jar, answered”, 10:32 into Gauss's pairing trick, and why it generalises