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.
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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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 = …
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…
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