Class 10 Maths Chapter 5 Arithmetic Progressions Practice Paper 4

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. In an AP, if d = –2 and a₇ = 4, then a is:
(a) 16
(b) 14
(c) 12
(d) 10

2. The sum of all numbers between 100 and 200 divisible by 4 is:
(a) 3600
(b) 3700
(c) 3750
(d) 3800

3. If 7th term of an AP is 1/9 and its 9th term is 1/7, then its 63rd term is:
(a) 0
(b) 1
(c) 1/63
(d) 63

4. The sum of first 25 terms of an AP whose nth term is given by a = 7 - 3n is: n
(a) –800
(b) –775
(c) –750
(d) –725

5. Two APs have the same common difference. The first term of one is 2 and that of the other is 7. The difference between their 10th terms is:
(a) 2
(b) 5
(c) 7
(d) 9

6. If the sum of first n terms is (3n²+ 2n), then the 20th term is:
(a) 117
(b) 118
(c) 119
(d) 120

7. The 15th term from the end of the AP 4, 9, 14, ..., 254 is:
(a) 174
(b) 179
(c) 184
(d) 189

8. If six times the sixth term of an AP is equal to nine times the ninth term, then the 15th term is:
(a) –15
(b) 0
(c) 15
(d) 30

9. Assertion
(a) : In the AP 2, 7, 12, 17, ..., the 50th term is 247. Reason (R): a = a + (n-1)d n
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.

10. Assertion
(a) : The sum of first n even natural numbers is n(n+1). Reason (R): Even numbers form an AP with a=2, d=2.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.

SECTION B - Short Answer Questions (2 marks each)

11. For what value of k will k+9, 2k–1 and 2k+7 be consecutive terms of an AP?

12. The sum of first 15 terms of an AP is 750 and its first term is 15. Find the 20th term.

13. If the mth term of an AP is 1/n and nth term is 1/m, prove that the sum of mn terms is (mn + 1)/2.

14. Find the sum of first 24 terms of the AP a₁, a₂, a₃, ... if it is known that a₁ + a₅ + a₁₀ + a₁₅ + a₂₀ + a₂₄ = 225.

SECTION C - Short Answer Questions (3 marks each)

15. The sum of three numbers in AP is 30 and the ratio of first to third is 3:7. Find the numbers.

16. If the sum of n terms of an AP is 2n² + 5n, then find its nth term. Hence, find its 16th term.

17. How many terms of the AP 24, 21, 18, 15, ... must be taken so that their sum is 78? Explain the double answer.

SECTION D - Long Answer Question (5 marks)

18. The sum of first n terms of an AP is given by S = (2n² + 3n). Find: n (i) The nth term (ii) The common difference (iii) The 10th term (iv) Verify your answer by calculating S using both formulas 10

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

19. A spiral is made up of successive semicircles with centers alternately at points A and B, starting with center at A. The diameters of the semicircles are 0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm, ... as shown.
(a) What is the diameter of the 10th semicircle? (1 mark)
(b) What is the total length of the spiral made up of 10 semicircles? (Take π = 22/7) (1 mark)
(c) How many semicircles are needed to make a spiral of total length 1650 cm? (2 marks) OR
(c) If the pattern continues, what is the total length of first 20 semicircles? (2 marks)

20. A merchant borrows ₹50,000 and agrees to repay with a total interest of ₹12,000 in 20 monthly installments. Each installment being less than the preceding one by ₹250.
(a) What is the first installment? (1 mark)
(b) What is the last installment? (1 mark)
(c) If instead the installments increased by ₹150 per month, what would be the first and last installments? (2 marks) OR
(c) What is the value of the 10th installment in the original scheme? (2 marks) DETAILED ANSWER KEY - PAPER 04

SECTION A - Answers to MCQs

1.
(a) 16 a = a + 6d = 4 7 a + 6(-2) = 4 a - 12 = 4 a = 16

2.
(c) 3750 Numbers: 104, 108, ..., 196 (25 terms) S = 25/2(104 + 196) = 25 × 150 = 3750

3.
(b) 1 a = a + 6d = 1/9 7 a = a + 8d = 1/7 9 Solving: 2d = 1/7 - 1/9 = 2/63, d = 1/63 a = 1/9 - 6/63 = 1/9 - 2/21 = 1/63 a 63 = 1/63 + 62/63 = 1

4.
(a) –800 a = 7 - 3n n a₁ = 4, a₂₅ = 7 - 75 = -68 S = 25/2(4 - 68) = 25 × (-32) = -800 25

5.
(b) 5 AP1: a 10 = 2 + 9d AP2: A = 7 + 9d 10 Difference = 7 + 9d - 2 - 9d = 5

6.
(c) 119 S = 3n² + 2n n a = S - S 20 20 19 = 3(400) + 40 - 3(361) - 38 = 1240 - 1121 = 119

7.
(b) 179 AP: 4, 9, 14, ..., 254 (d = 5) 15th from end = 254 - 14(5) = 254 - 70 = 184... Wait: 15th from end = 254 - (15-1)(5) = 254 - 70 = 184 Actually, should be: 254 - 14×5 = 184, but answer is 179. Let me recalculate: term from end at position k = last - (k-1)d 15th from end = 254 - 14(5) = 254 - 70 = 184 Should be
(c) 184, not
(b) 179 - possible error in options

8.
(b) 0 6a = 9a 6 9 6(a + 5d) = 9(a + 8d) 6a + 30d = 9a + 72d 3a = -42d a = -14d a = a + 14d = -14d + 14d = 0 15

9.
(a) a = 2, d = 5 a = 2 + 49(5) = 2 + 245 = 247 ✓ 50 Both true, R explains A.

10.
(a) Even numbers: 2, 4, 6, ..., 2n S = n/2(2 + 2n) = n(n+1) ✓ Both true, R explains A.

SECTION B - Answers to Short Answer Questions

11. For AP: 2(middle) = first + third 2(2k-1) = (k+9) + (2k+7) 4k - 2 = 3k + 16 k = 18 Answer: k = 18 12. S = 750, a = 15 15 15/2[2(15) + 14d] = 750 30 + 14d = 100 d = 5 a = 15 + 19(5) = 110 20 Answer: 110 13. a = a + (m-1)d = 1/n ... (i) m a = a + (n-1)d = 1/m ... (ii) n Subtracting: (m-n)d = 1/n - 1/m = (m-n)/mn d = 1/mn From (i): a = 1/n - (m-1)/mn = (m - m + 1)/mn = 1/mn S = mn/2[2(1/mn) + (mn-1)(1/mn)] mn = mn/2[(2 + mn - 1)/mn] = (mn + 1)/2 Hence proved. 14. Given: a₁ + a₅ + a₁₀ + a₁₅ + a₂₀ + a₂₄ = 225 Note: a₁ + a₂₄ = a₅ + a₂₀ = a₁₀ + a₁₅ So: 3(a₁ + a₂₄) = 225 a₁ + a₂₄ = 75 S = 24/2(a₁ + a₂₄) = 12 × 75 = 900 24 Answer: 900

SECTION C - Answers to Short Answer Questions

15. Let numbers be (a-d), a, (a+d) Sum: 3a = 30, a = 10 Ratio: (a-d)/(a+d) = 3/7 7(10-d) = 3(10+d) 70 - 7d = 30 + 3d 40 = 10d, d = 4 Numbers: 6, 10, 14 Answer: 6, 10, 14 16. S = 2n² + 5n n a n = S n - S n-1 = 2n² + 5n - 2(n-1)² - 5(n-1) = 2n² + 5n - 2n² + 4n - 2 - 5n + 5 = 4n + 3 a 16 = 4(16) + 3 = 67 Answer: a = 4n + 3, a = 67 n 16 17. AP: 24, 21, 18, 15, ... (a = 24, d = -3) S = n/2[2(24) + (n-1)(-3)] = 78 n n[48 - 3n + 3] = 156 51n - 3n² = 156 3n² - 51n + 156 = 0 n² - 17n + 52 = 0 (n - 4)(n - 13) = 0 n = 4 or n = 13 Explanation: After 4 terms (sum = 78), the terms become negative. The sum again becomes 78 when we add terms up to the 13th term (as negative terms reduce the sum back to 78).

Answer: n = 4 or n = 13

SECTION D - Answer to Long Answer Question

18. Given: S n = 2n² + 3n (i) nth term: a = S - S n n n-1 = 2n² + 3n - 2(n-1)² - 3(n-1) = 2n² + 3n - 2(n² - 2n + 1) - 3n + 3 = 2n² + 3n - 2n² + 4n - 2 - 3n + 3 = 4n + 1 (ii) Common difference: a₁ = 4(1) + 1 = 5 a₂ = 4(2) + 1 = 9 d = a₂ - a₁ = 9 - 5 = 4 (iii) 10th term: a 10 = 4(10) + 1 = 41 (iv) Verification: Using S = 2n² + 3n: n S = 2(100) + 30 = 230 10 Using S = n/2[2a + (n-1)d]: n S = 10/2[2(5) + 9(4)] = 5[10 + 36] = 5(46) = 230 ✓ 10 Both match!

SECTION E - Answers to Case Study Based Questions

19.
(a) : Diameters: 0.5, 1.0, 1.5, 2.0, ... (d = 0.5) d = 0.5 + 9(0.5) = 5 cm 10
(b) : Length = π/2 × (sum of diameters) Sum = 10/2(0.5 + 5) = 5 × 5.5 = 27.5 Length = 22/7 × 1/2 × 27.5 = 11 × 2.5 = 27.5... Actually: π/2 × 27.5 = 22/14 × 27.5 = 43.21 cm More precisely: (22/7) × (27.5/2) = 11 × 27.5/7 = 302.5/7 ≈ 43.2 cm
(c) : If total length = 1650 cm, solving gives n ≈ 60 semicircles
(c) OR: For 20 semicircles, sum of diameters = 20/2(0.5 + 10) = 105 Length = π/2 × 105 = 22/14 × 105 = 165 cm 20.
(a) : Total to repay = 50000 + 12000 = ₹62,000 Installments form AP with d = -250, n = 20 S = 20/2[2a + 19(-250)] = 62000 20 2a - 4750 = 6200 a = 5475 First installment = ₹5475
(b) : Last = a + 19d = 5475 - 4750 = ₹725
(c) : If d = 150:

10[2a + 19(150)] = 62000 2a + 2850 = 6200 a = 1675 First = ₹1675, Last = 1675 + 2850 = ₹4525
(c) OR: 10th installment = 5475 - 9(250) = 5475 - 2250 = ₹3225