Arithmetic Progression — Maharashtra (SSC) Class 10 Mathematics Solutions (Free)

Free step-by-step Maharashtra (SSC) Class 10 Mathematics solutions for "Arithmetic Progression" — important questions with detailed answers, download PDF for board exam preparation.

Arithmetic Progression — Maharashtra (SSC) Class 10 Mathematics Solutions (Free)

Free step-by-step Maharashtra (SSC) Class 10 Mathematics solutions for "Arithmetic Progression" — important questions with detailed answers, download PDF for board exam preparation.

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TL;DR: Free step-by-step Maharashtra (SSC) Class 10 Mathematics solutions for "Arithmetic Progression" — important questions with detailed answers, download…

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Arithmetic Progression — Textbook Questions Answered (8)

Q1: Find the 10th term of AP: 3, 7, 11, 15, ...

Step 1: Identify first term and common difference. a = 3 (first term), d = 7 - 3 = 4 (common difference). Step 2: Use nth term formula. Tn = a + (n-1)d. Step 3: Substitute values for n = 10. T10 = 3 + (10-1)(4). Step 4: Calculate. T10 = 3 + 9(4) = 3 + 36 = 39. Step 5: Verify by listing terms: 3, 7, 11, 15, 19, 23, 27, 31, 35, 39. The 10th term is 39. Answer: 39.

Q2: Find the sum of first 20 terms of the AP: 2, 5, 8, 11, ...

Step 1: Identify first term and common difference. a = 2, d = 5 - 2 = 3. Step 2: Find 20th term using Tn = a + (n-1)d. T20 = 2 + (20-1)(3) = 2 + 19(3) = 2 + 57 = 59. Step 3: Use sum formula for AP. Sn = n/2 (a + l) where l is last term. Or Sn = n/2 [2a + (n-1)d]. Step 4: Apply first formula. S20 = 20/2 (2 + 59) = 10(61) = 610. Step 5: Verify using second formula. S20 = 20/2 [2(2) + (20-1)(3)] = 10[4 + 57] = 10(61) = 610. Answer: 610.

Q3: In an AP, a = 5, d = 3, Tn = 44. Find n.

Step 1: Use nth term formula Tn = a + (n-1)d. Step 2: Substitute known values. 44 = 5 + (n-1)(3). Step 3: Simplify. 44 = 5 + 3n - 3. Step 4: Calculate. 44 = 2 + 3n. Step 5: Solve for n. 42 = 3n, so n = 14. Step 6: Verify. T14 = 5 + (14-1)(3) = 5 + 13(3) = 5 + 39 = 44. Correct. Answer: n = 14.

Q4: The sum of first n terms of an AP is Sn = 2n^2 + 3n. Find the AP.

Step 1: Find first term using S1. S1 = 2(1)^2 + 3(1) = 2 + 3 = 5. So a = 5. Step 2: Find sum of first 2 terms. S2 = 2(2)^2 + 3(2) = 8 + 6 = 14. Step 3: Second term = S2 - S1 = 14 - 5 = 9. So a2 = 9. Step 4: Find common difference. d = a2 - a1 = 9 - 5 = 4. Step 5: Find third term to verify. S3 = 2(3)^2 + 3(3) = 18 + 9 = 27. a3 = S3 - S2 = 27 - 14 = 13. Check: 9 + 4 = 13. Correct. Step 6: General term. Tn = 5 + (n-1)(4) = 5 + 4n - 4 = 4n + 1. Answer: AP is 5, 9, 13, 17, ... with general term Tn = 4n + 1.

Q5: Three numbers in AP have sum 30 and product 360. Find the three numbers.

Step 1: Let three numbers be (a-d), a, (a+d) where a is middle term and d is common difference. Step 2: Use sum condition. (a-d) + a + (a+d) = 30. Step 3: Simplify. 3a = 30, so a = 10. Step 4: Use product condition. (a-d) * a * (a+d) = 360. Step 5: Substitute a = 10. (10-d) * 10 * (10+d) = 360. Step 6: Expand using (10-d)(10+d) = 100 - d^2. 10(100 - d^2) = 360. Step 7: Simplify. 1000 - 10d^2 = 360. Step 8: Solve. 10d^2 = 640, so d^2 = 64, thus d = 8 or d = -8. Step 9: If d = 8: numbers are 2, 10, 18. Check: 2 + 10 + 18 = 30. 2 * 10 * 18 = 360. Correct. Step 10: If d = -8: numbers are 18, 10, 2 (same set). Answer: The three numbers are 2, 10, 18.

Q6: Find the sum of all two-digit numbers divisible by 7.

Step 1: Identify the AP. Two-digit numbers divisible by 7: 14, 21, 28, ..., 98. Step 2: First term a = 14, last term l = 98, common difference d = 7. Step 3: Find how many terms n. Use Tn = a + (n-1)d. 98 = 14 + (n-1)(7). Step 4: Solve. 84 = (n-1)(7), so n-1 = 12, thus n = 13. Step 5: Use sum formula Sn = n/2 (a + l). S13 = 13/2 (14 + 98). Step 6: Calculate. S13 = 13/2 (112) = 13 * 56 = 728. Answer: 728.

Q7: In an AP, the sum of first 5 terms is 25 and sum of first 10 terms is 110. Find a and d.

Step 1: Use sum formula Sn = n/2 [2a + (n-1)d]. Step 2: For S5 = 25. 25 = 5/2 [2a + (5-1)d]. Step 3: Simplify. 25 = 5/2 [2a + 4d], so 10 = 2a + 4d. Step 4: Reduce. 5 = a + 2d. This is equation (1). Step 5: For S10 = 110. 110 = 10/2 [2a + (10-1)d]. Step 6: Simplify. 110 = 5[2a + 9d], so 22 = 2a + 9d. This is equation (2). Step 7: From equation (1): a = 5 - 2d. Step 8: Substitute in equation (2). 22 = 2(5 - 2d) + 9d. Step 9: Expand. 22 = 10 - 4d + 9d. Step 10: Solve. 22 = 10 + 5d, so 12 = 5d, thus d = 12/5 = 2.4. Step 11: Find a. a = 5 - 2(2.4) = 5 - 4.8 = 0.2. Step 12: Verify. S5 = 5/2[2(0.2) + 4(2.4)] = 5/2[0.4 + 9.6] = 5/2(10) = 25. Correct. Answer: a = 0.2 (or 1/5), d = 2.4 (or 12/5).

Q8: Which term of the AP 4, 9, 14, 19, ... equals 99?

Step 1: Identify AP parameters. First term a = 4, common difference d = 9 - 4 = 5. Step 2: Use Tn = a + (n-1)d where Tn = 99. Step 3: Substitute. 99 = 4 + (n-1)(5). Step 4: Simplify. 99 = 4 + 5n - 5. Step 5: Calculate. 99 = 5n - 1. Step 6: Solve. 100 = 5n, so n = 20. Step 7: Verify. T20 = 4 + (20-1)(5) = 4 + 19(5) = 4 + 95 = 99. Correct. Answer: 99 is the 20th term.

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