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πŸ“˜SCERT 10th Maths Chapter 1 - Arithmetic Sequence πŸ”’- Revision - Q&A | GrowWisePath πŸ’‘

πŸ“˜ SCERT 10th Maths Chapter 1 – Arithmetic Sequence πŸ”’ | Revision – Q&A | GrowWisePath πŸ’‘

πŸ“˜ SCERT 10th Maths Chapter 1 – Arithmetic Sequence | Revision with Q&A | GrowWisePath πŸ’‘

Master the chapter with key concepts and exam-focused questions. Tap the buttons below to toggle sections.

πŸ’‘ Important Points for Revision

  • Definition: An Arithmetic Sequence is a sequence of numbers in which the difference between any two consecutive terms is constant.
  • Common Difference (d): The fixed number added to each term to get the next term.
    Formula: d = Tn - Tn-1
  • General Term Formula: Tn = a + (n - 1) × d
    Where:
    • Tn = nth term
    • a = first term
    • n = term position
    • d = common difference
  • Change in Term vs. Position: Ξ”T = Ξ”n × d
  • Types:
    • Increasing sequence (d > 0)
    • Decreasing sequence (d < 0)
    • Constant sequence (d = 0)
  • Examples:
    • Even numbers: 2, 4, 6, ... (d = 2)
    • Odd numbers: 1, 3, 5, ... (d = 2)
    • Multiples of 5: 5, 10, 15, ... (d = 5)
    • Decreasing: 100, 95, 90, ... (d = -5)
  • Not Arithmetic: Squares (1, 4, 9...), Powers (2, 4, 8...), Reciprocals (1, 1/2, 1/3...)

❓ Important Questions and Answers

  1. Q1. Find the 25th term of the sequence: 1, 11, 21, ...

    a = 1, d = 10
    T25 = a + (25 - 1) × d = 1 + 24 × 10 = 241

  2. Q2. The 3rd term is 37 and the 7th term is 73. Find the first five terms.

    d = (73 - 37) ÷ (7 - 3) = 9
    a = 37 - 2 × 9 = 19
    Sequence: 19, 28, 37, 46, 55

  3. Q3. The 4th term is 45 and the 5th term is 56. Find the first term.

    d = 11
    a = 45 - 3 × 11 = 12
    Sequence: 12, 23, 34, 45, 56

  4. Q4. What is the 21st term of the sequence: 100, 95, 90, ...?

    a = 100, d = -5
    T21 = 100 + 20 × (-5) = 0

  5. Q5. The 3rd term is 12 and the 7th term is 32. Find the 15th term.

    d = (32 - 12) ÷ 4 = 5
    T15 = 32 + 8 × 5 = 72

  6. Q6. Write an arithmetic sequence with 1 and 11 as the first and third terms.

    T3 = a + 2d → 11 = 1 + 2d → d = 5
    Sequence: 1, 6, 11, 16, 21

  7. Q7. The 4th term is 65 and the 5th term is 54. Find the first term.

    d = -11
    a = 65 + 3 × 11 = 98
    Sequence: 98, 87, 76, 65, 54

❓CourseBook Solved Questions on Arithmetic Sequences

  1. Q1. Write four arithmetic sequences with sum of the first four terms equal to 100.

    Let first term be a and common difference be d.
    Sum of first 4 terms = a + (a + d) + (a + 2d) + (a + 3d) = 4a + 6d
    Set: 4a + 6d = 100
    Choose different values of a and solve for d:

    • If a = 5 → 4×5 + 6d = 100 → 20 + 6d = 100 → d = 80 ÷ 6 ≈ 13.33 → Sequence: 5, 18.33, 31.66, 44.99
    • If a = 10 → 4×10 + 6d = 100 → 40 + 6d = 100 → d = 60 ÷ 6 = 10 → Sequence: 10, 20, 30, 40
    • If a = 20 → 4×20 + 6d = 100 → 80 + 6d = 100 → d = 20 ÷ 6 ≈ 3.33 → Sequence: 20, 23.33, 26.66, 29.99
    • If a = 25 → 4×25 + 6d = 100 → 100 + 6d = 100 → d = 0 → Sequence: 25, 25, 25, 25

  2. Q2. The 1st term of an arithmetic sequence is 5 and the sum of the first 6 terms is 105. Calculate the first six terms.

    Given: a = 5, S6 = 105
    Formula: Sn = n/2 × [2a + (n - 1)d]
    S6 = 6/2 × [2×5 + 5d] = 3 × (10 + 5d) = 105
    → 10 + 5d = 35 → d = 5
    Sequence: 5, 10, 15, 20, 25, 30

  3. Q3. The sum of the 7th and 8th terms of an arithmetic sequence is 50. Calculate the sum of the first 14 terms.

    Let a = first term, d = common difference
    T7 = a + 6d, T8 = a + 7d
    T7 + T8 = 2a + 13d = 50
    So, 2a + 13d = 50 → Equation (1)
    Sum of first 14 terms: S14 = 14/2 × [2a + 13d] = 7 × (2a + 13d)
    From (1), 2a + 13d = 50 → S14 = 7 × 50 = 350

  4. Q4. Write the first three terms of each arithmetic sequence:

    • (i) 1st term = 30, Sum of first 3 terms = 300
      Let terms be: 30, 30 + d, 30 + 2d
      Sum = 30 + (30 + d) + (30 + 2d) = 90 + 3d = 300 → d = 70
      Sequence: 30, 100, 170
    • (ii) 1st term = 30, Sum of first 4 terms = 300
      Terms: 30, 30 + d, 30 + 2d, 30 + 3d → Sum = 120 + 6d = 300 → d = 30
      Sequence: 30, 60, 90, 120
    • (iii) 1st term = 30, Sum of first 5 terms = 300
      Sum = 5/2 × [2×30 + 4d] = 2.5 × (60 + 4d) = 300 → 60 + 4d = 120 → d = 15
      Sequence: 30, 45, 60, 75, 90
    • (iv) 1st term = 30, Sum of first 6 terms = 300
      Sum = 6/2 × [2×30 + 5d] = 3 × (60 + 5d) = 300 → 60 + 5d = 100 → d = 8
      Sequence: 30, 38, 46, 54, 62, 70

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