Class 10 Maths Chapter 5: Arithmetic Progressions worksheet with answer key
A 20-mark worksheet on Arithmetic Progressions for about 40 minutes of class or homework: 5 multiple-choice, 3 very short answer and 3 short answer questions, written from the NCERT chapter and checked by a second reviewer. Print the PDF for the class; the answer key carries the marking scheme. Teachers may photocopy it freely.
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Answer: (c) −6
d = a₂ − a₁ = 11 − 17 = −6. (6 subtracts the wrong way round.)
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Answer: (c) −35
d = 5 − 9 = −4, so a₁₂ = 9 + 11 × (−4) = −35. (−39 is a + 12d; 53 drops the sign of d.)
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Answer: (b) 20th term
The kth term carries (k − 1) lots of d, so k − 1 = 19 and k = 20.
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Answer: (a) its first term and its common difference
The terms are a, a + d, a + 2d, . . ., so both a and d are needed; either alone fixes nothing else.
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Answer: (b) −100
Here a = −4 and d = 0, so S₂₅ = 25/2 [2(−4) + 24 × 0] = −100.
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The middle term is the mean of the other two: 2(2k − 1) = (k + 9) + (2k + 7), so 4k − 2 = 3k + 16 and k = 18. Check: 27, 35, 43 have d = 8.
- sets up 2(2k − 1) = (k + 9) + (2k + 7) (1)
- k = 18 (1)
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a + 3d = 11 and a + 8d = 31. Subtracting, 5d = 20, so d = 4 and a = 11 − 12 = −1. The AP is −1, 3, 7, 11, . . .
- writes both equations and gets d = 4 (1)
- a = −1 and the AP −1, 3, 7, 11, . . . (1)
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√20 = 2√5, √45 = 3√5, √80 = 4√5, √125 = 5√5. Differences: √5, √5, √5, all equal, so the list is an AP with d = √5. Next term = 6√5 = √180.
- reduces each surd to a multiple of √5 and shows three equal differences √5 (1)
- next term 6√5 (= √180) (1)
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The daily outputs form an AP with d = 300, n = 15, S₁₅ = 42000. 15/2 [2a + 14 × 300] = 42000 gives 15(a + 2100) = 42000, a + 2100 = 2800, a = 700. Fifteenth day: 700 + 14 × 300 = 4900 metres.
- sets up 15/2 [2a + 14 × 300] = 42000 (1)
- first day 700 m (1)
- fifteenth day 4900 m (1)
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The multiples run 110, 121, . . ., 396 (the first after 100, the last before 400). 396 = 110 + (n − 1) × 11 gives n − 1 = 26, n = 27. S = 27/2 (110 + 396) = 27 × 253 = 6831.
- first term 110 and last term 396 (1)
- n = 27 (1)
- sum = 6831 (1)
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aₙ₊₁ − aₙ = [11 − 4(n + 1)] − (11 − 4n) = −4, fixed, so it is an AP with a = 7, d = −4. 11 − 4n = −89 gives n = 25. S₂₅ = 25/2 (7 + (−89)) = 25/2 × (−82) = −1025.
- shows aₙ₊₁ − aₙ = −4, so an AP (1)
- −89 is the 25th term (1)
- S₂₅ = −1025 (1)