Class 10 Maths Chapter 5 Arithmetic Progressions Practice Paper 1

Class: X Subject: Mathematics Session: 2024-25 Chapter: 05 - Arithmetic Progression Time: 1½ Hours Max. Marks: 40

General Instructions:

1. All questions are compulsory.

2. This question paper contains 20 questions divided into five sections A, B, C, D and E.

3. Section A contains 10 MCQs of 1 mark each.

4. Section B contains 4 questions of 2 marks each.

5. Section C contains 3 questions of 3 marks each.

6. Section D contains 1 question of 5 marks.

7. Section E contains 2 Case Study Based questions of 4 marks each.

8. Use of calculators is not permitted.

SECTION A - Multiple Choice Questions (1 mark each)

1. If the first term of an AP is 5 and the common difference is –2, then the 10th term is:
(a) –13
(b) –15
(c) –17
(d) –23

2. The 11th term from the end of the A.P. 10, 7, 4, ..., –62 is:
(a) –20
(b) –32
(c) –35
(d) –38

3. If the sum of first 7 terms of an A.P. is 49 and that of 17 terms is 289, then the sum of first n terms is:
(a) n²
(b) 2n²
(c) n² – 1
(d) n² + 1

4. The 10th term from the end of the A.P. 4, 9, 14, ..., 254 is:
(a) 204
(b) 209
(c) 214
(d) 219

5. If m times the mth term of an A.P. is equal to n times its nth term, then the (m + n)th term is:
(a) 0
(b) 1
(c) m + n
(d) mn

6. The sum of all two-digit odd numbers is:
(a) 2475
(b) 2530
(c) 4905
(d) 5049

7. How many three-digit numbers are divisible by 7?
(a) 127
(b) 128
(c) 129
(d) 130

8. If the first term of an A.P. is –5 and the common difference is 3, then the sum of first 10 terms is:
(a) 85
(b) 95
(c) 100
(d) 135

9. Assertion
(a) : The 10th term of the A.P. 5, 8, 11, 14, ... is 32. Reason (R): The nth term of an A.P. is given by a = a + (n – 1)d. n
(a) Both assertion
(a) and reason (R) are true and reason (R) is the correct explanation of assertion
(a) .
(b) Both assertion
(a) and reason (R) are true but reason (R) is not the correct explanation of assertion
(a) .
(c) Assertion
(a) is true but reason (R) is false.
(d) Assertion
(a) is false but reason (R) is true.

10. Assertion
(a) : The sum of first n natural numbers is n(n + 1)/2. Reason (R): The sum of first n terms of an A.P. is S n = n/2[2a + (n – 1)d].
(a) Both assertion
(a) and reason (R) are true and reason (R) is the correct explanation of assertion
(a) .
(b) Both assertion
(a) and reason (R) are true but reason (R) is not the correct explanation of assertion
(a) .
(c) Assertion
(a) is true but reason (R) is false.
(d) Assertion
(a) is false but reason (R) is true.

SECTION B - Short Answer Questions (2 marks each)

11. Find the sum of first 30 terms of an A.P. whose second term is 2 and seventh term is 22.

12. The sum of three numbers in A.P. is 27 and their product is 504. Find the numbers.

13. If the 9th term of an A.P. is zero, prove that its 29th term is twice its 19th term.

14. Find the sum of all natural numbers between 100 and 500 which are divisible by 7.

SECTION C - Short Answer Questions (3 marks each)

15. The sum of the 4th and 8th terms of an A.P. is 24 and the sum of 6th and 10th terms is 44. Find the first three terms of the A.P.

16. In an A.P., the sum of first ten terms is –150 and the sum of its next ten terms is –550. Find the A.P.

17. A sum of ₹700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is ₹20 less than its preceding prize, find the value of each of the prizes.

SECTION D - Long Answer Question (5 marks)

18. A manufacturer of TV sets produced 600 sets in the third year and 700 sets in the seventh year. Assuming that the production increases uniformly by a fixed number every year, find: (i) The production in the 1st year (ii) The production in the 10th year (iii) The total production in first 7 years

SECTION E - Case Study Based Questions (4 marks each)

19. A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows: ₹200 for the first day, ₹250 for the second day, ₹300 for the third day, etc., the penalty for each succeeding day being ₹50 more than for the preceding day.
(a) What is the penalty for 10th day? (1 mark)
(b) What is the penalty for delay of 15 days? (1 mark)
(c) How many days delay would mean a penalty of ₹4950? (2 marks) OR
(c) If the contractor paid a total penalty of ₹7200, for how many days was the work delayed? (2 marks)

20. In a flower bed, there are 43 rose plants in the first row, 41 in the second, 39 in the third, and so on. There are 11 rose plants in the last row.
(a) How many rows are there in the flower bed? (1 mark)
(b) What is the total number of rose plants in the flower bed? (1 mark)
(c) If the first row had 50 plants and the pattern continued, how many rows would there be if the last row has 10 plants? (2 marks) OR
(c) If each alternate row starting from the first has white roses and others have red roses, how many white rose plants are there in total? (2 marks) DETAILED ANSWER KEY - PAPER 01

SECTION A - Answers to MCQs

1.
(a) –13

Solution:

Given: a = 5, d = –2 a 10 = a + 9d = 5 + 9(–2) = 5 – 18 = –13

2.
(b) –32

Solution:

Given A.P.: 10, 7, 4, ..., –62 a = 10, d = –3, last term = –62 11th term from end = Last term – 10d = –62 – 10(–3) = –62 + 30 = –32

3.
(a) n²

Solution:

Given: S = 49 and S = 289 7 17 S = 7/2[2a + 6d] = 49 7 2a + 6d = 14 ... (i) S = 17/2[2a + 16d] = 289 17 2a + 16d = 34 ... (ii) Subtracting (i) from (ii): 10d = 20, d = 2 From (i): 2a + 12 = 14, a = 1 S n = n/2[2(1) + (n – 1)2] S = n/2[2 + 2n – 2] n S = n/2(2n) = n² n

4.
(b) 209

Solution:

Given A.P.: 4, 9, 14, ..., 254 a = 4, d = 5, last term = 254 10th term from end = Last term – 9d = 254 – 9(5) = 254 – 45 = 209

5.
(a) 0

Solution:

Given: m × a = n × a m n m[a + (m – 1)d] = n[a + (n – 1)d] ma + m(m – 1)d = na + n(n – 1)d (m – n)a + [m(m – 1) – n(n – 1)]d = 0 (m – n)a + (m² – m – n² + n)d = 0 (m – n)a + (m – n)(m + n – 1)d = 0 (m – n)[a + (m + n – 1)d] = 0 Since m ≠ n, we have: a + (m + n – 1)d = 0 This is the (m + n)th term = 0

6.
(a) 2475

Solution:

Two-digit odd numbers: 11, 13, 15, ..., 99 a = 11, d = 2, last term = 99 Finding n: 99 = 11 + (n – 1)2 88 = 2(n – 1), n = 45 S = 45/2(11 + 99) = 45/2 × 110 = 2475 45

7.
(c) 129

Solution:

Three-digit numbers divisible by 7: 105, 112, 119, ..., 994 a = 105, d = 7, last term = 994 Finding n: 994 = 105 + (n – 1)7 889 = 7(n – 1) n – 1 = 127 n = 128... Wait, let me recalculate: 994 = 105 + 7(n – 1) 889 = 7(n – 1) 127 = n – 1 n = 128 Actually, checking: 7 × 142 = 994 ✓ 7 × 15 = 105 ✓ Number of terms = 142 – 15 + 1 = 128 Note: Correct answer should be
(b) 128

8.
(a) 85

Solution:

Given: a = –5, d = 3, n = 10 S = n/2[2a + (n – 1)d] 10 S 10 = 10/2[2(–5) + 9(3)] S = 5[–10 + 27] 10 S = 5 × 17 = 85 10

9.
(a)

Solution:

Given A.P.: 5, 8, 11, 14, ... a = 5, d = 3 a = a + 9d = 5 + 9(3) = 5 + 27 = 32 ✓ 10 Assertion is true and uses the formula stated in Reason. Both are true and R correctly explains A.

10.
(a)

Solution:

Natural numbers 1, 2, 3, ..., n form an A.P. with a = 1, d = 1 S n = n/2[2(1) + (n – 1)(1)] S = n/2[2 + n – 1] n S = n/2(n + 1) = n(n + 1)/2 ✓ n Assertion is true and directly follows from Reason's formula. Both are true and R correctly explains A.

SECTION B - Answers to Short Answer Questions

11.

Solution:

Given: a = 2, a = 22 2 7 a + d = 2 ... (i) a + 6d = 22 ... (ii) Subtracting (i) from (ii): 5d = 20, d = 4 From (i): a = 2 – 4 = –2 S = 30/2[2(–2) + 29(4)] 30 S = 15[–4 + 116] 30 S = 15 × 112 = 1680 30 Answer: 1680 12.

Solution:

Let the three numbers be (a – d), a, (a + d) Sum = (a – d) + a + (a + d) = 3a = 27 a = 9 Product = (a – d) × a × (a + d) = a(a² – d²) = 504 9(81 – d²) = 504 81 – d² = 56 d² = 25 d = ±5 When d = 5: Numbers are 4, 9, 14 When d = –5: Numbers are 14, 9, 4 Answer: The numbers are 4, 9, 14 13.

Solution:

Given: a = 0 9 a + 8d = 0 a = –8d a = a + 28d = –8d + 28d = 20d 29 a 19 = a + 18d = –8d + 18d = 10d a 29 = 20d = 2(10d) = 2 × a 19 Hence proved: a = 2 × a 29 19 14.

Solution:

Natural numbers between 100 and 500 divisible by 7: First term: 105, Last term: 497 A.P.: 105, 112, 119, ..., 497 a = 105, d = 7, last term = 497 Finding n: 497 = 105 + (n – 1)7 392 = 7(n – 1) n – 1 = 56 n = 57 S = 57/2(105 + 497) 57 S = 57/2 × 602 57 S 57 = 57 × 301 = 17,157 Answer: 17,157

SECTION C - Answers to Short Answer Questions

15.

Solution:

Given: a 4 + a 8 = 24 and a 6 + a 10 = 44 (a + 3d) + (a + 7d) = 24 2a + 10d = 24 a + 5d = 12 ... (i) (a + 5d) + (a + 9d) = 44 2a + 14d = 44 a + 7d = 22 ... (ii) Subtracting (i) from (ii): 2d = 10, d = 5 From (i): a + 25 = 12, a = –13 First three terms: a = –13 1 a = –13 + 5 = –8 2 a = –8 + 5 = –3 3 Answer: –13, –8, –3 16.

Solution:

Given: S = –150 10 Sum of next 10 terms = S – S = –550 20 10 S 20 = –150 – 550 = –700 S 10 = 10/2[2a + 9d] = –150 2a + 9d = –30 ... (i) S = 20/2[2a + 19d] = –700 20 2a + 19d = –70 ... (ii) Subtracting (i) from (ii): 10d = –40, d = –4 From (i): 2a – 36 = –30, a = 3 Answer: The A.P. is 3, –1, –5, –9, ... 17.

Solution:

Let the seven prizes be: a, (a – 20), (a – 40), ..., (a – 120) This forms an A.P. with first term = a, d = –20, n = 7 Sum = 700 S 7 = 7/2[2a + 6(–20)] = 700 7/2[2a – 120] = 700 2a – 120 = 200 2a = 320 a = 160 Seven prizes: ₹160, ₹140, ₹120, ₹100, ₹80, ₹60, ₹40 Answer: The prizes are ₹160, ₹140, ₹120, ₹100, ₹80, ₹60, and ₹40

SECTION D - Answer to Long Answer Question

18.

Solution:

Let production in 1st year = a Annual increase = d Production in 3rd year = a + 2d = 600 ... (i) Production in 7th year = a + 6d = 700 ... (ii) Subtracting (i) from (ii): 4d = 100 d = 25 From (i): a + 50 = 600 a = 550 (i) Production in 1st year = 550 sets (ii) Production in 10th year: a = a + 9d = 550 + 9(25) = 550 + 225 = 775 sets 10 (iii) Total production in first 7 years: S = 7/2[2(550) + 6(25)] 7 S 7 = 7/2[1100 + 150] S = 7/2 × 1250 7 S = 4375 sets 7

SECTION E - Answers to Case Study Based Questions

19.
(a) Penalty for 10th day: Penalty forms A.P.: 200, 250, 300, ... a = 200, d = 50 a = 200 + 9(50) = 200 + 450 = 650 10 Answer: ₹650
(b) Total penalty for 15 days delay: S = 15/2[2(200) + 14(50)] 15 S = 15/2[400 + 700] 15 S 15 = 15/2 × 1100 S = 8250 15 Answer: ₹8,250
(c) Days for penalty of ₹4950: This represents the nth day's penalty. a = 4950 n 200 + (n – 1)50 = 4950 (n – 1)50 = 4750 n – 1 = 95 n = 96 Answer: 96 days
(c) OR - Days for total penalty ₹7200: S = 7200 n n/2[2(200) + (n – 1)50] = 7200 n/2[400 + 50n – 50] = 7200 n(350 + 50n) = 14400 50n² + 350n – 14400 = 0 n² + 7n – 288 = 0 Using quadratic formula:

n = [–7 ± √(49 + 1152)]/2 = [–7 ± √1201]/2 n = [–7 ± 34.66]/2 ≈ 13.83 Since we need integer days, checking n = 14: S = 14/2[400 + 13(50)] = 7[400 + 650] = 7350 (too high) 14 Checking n = 13: S = 13/2[400 + 12(50)] = 13/2 × 1000 = 6500 (too low) 13 Actually, let me solve properly: n² + 7n – 288 = 0 (n + 23)(n – 16) = ... Let me use the formula: n = [–7 + √(49 + 1152)]/2 = [–7 + √1201]/2 Hmm, √1201 ≈ 34.66, so n ≈ 13.83 But we need exact. Let me check if 288 factors nicely: n² + 7n – 288 = 0 Trying n = 16: 256 + 112 – 288 = 80 (not zero) Trying n = 12: 144 + 84 – 288 = –60 Let me recalculate from S = 7200:

n n/2[400 + 50n – 50] = 7200 n[350 + 50n] = 14400 50n² + 350n – 14400 = 0 n² + 7n – 288 = 0 Actually, (n – 16)(n + 18) = n² + 18n – 16n – 288 = n² + 2n – 288 ≠ Let's try (n – 16)(n + 23) = n² + 23n – 16n – 368 ≠ I'll use formula: n = [–7 + √(49 + 1152)]/2 = [–7 + √1201]/2 Hmm √1201 is not a perfect square. Wait, let me check my algebra. From: n[400 + 50(n – 1)] = 14400 n[400 + 50n – 50] = 14400 n[350 + 50n] = 14400 350n + 50n² = 14400 50n² + 350n – 14400 = 0 n² + 7n – 288 = 0 Using formula: n = (–7 ± √(49 + 1152))/2 = (–7 ± √1201)/2 Since 34² = 1156 and 35² = 1225, √1201 is between 34 and 35 More precisely, √1201 ≈ 34.66 n = (–7 + 34.66)/2 ≈ 13.83 This doesn't give a whole number. Let me verify the problem setup...

Actually, for case study problems, sometimes the number works out. Let me try n = 12: S = 12/2[400 + 11(50)] = 6[400 + 550] = 6 × 950 = 5700 12 Try n = 16: S = 16/2[400 + 15(50)] = 8[400 + 750] = 8 × 1150 = 9200 16 Hmm, 7200 is between these. The problem might have a typo, or I should leave it with the quadratic solution. Answer: Using n² + 7n – 288 = 0, solving gives n ≈ 13.8 days. Since payment is for complete days, approximately 14 days. 20.
(a) Number of rows: A.P.: 43, 41, 39, ..., 11 a = 43, d = –2, last term = 11 11 = 43 + (n – 1)(–2) 11 – 43 = –2(n – 1) –32 = –2(n – 1) n – 1 = 16 n = 17 Answer: 17 rows
(b) Total rose plants:

S = 17/2(43 + 11) 17 S = 17/2 × 54 17 S 17 = 17 × 27 = 459 Answer: 459 rose plants
(c) Rows if first row has 50 plants, last has 10: a = 50, last term = 10, d = –2 10 = 50 + (n – 1)(–2) –40 = –2(n – 1) n – 1 = 20 n = 21 Answer: 21 rows
(c) OR - White rose plants (alternate rows starting from first): White roses are in rows: 1, 3, 5, 7, 9, 11, 13, 15, 17 These are 9 rows with plants: 43, 39, 35, 31, 27, 23, 19, 15, 11 This forms A.P. with a = 43, d = –4, n = 9 S = 9/2(43 + 11) 9 S 9 = 9/2 × 54 S = 9 × 27 = 243 9 Answer: 243 white rose plants