π SCERT 10th Maths Chapter 1 – Arithmetic Sequence | Revision with Q&A | GrowWisePath π‘
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π‘ 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: ΞT = Ξn × 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
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Q1. Find the 25th term of the sequence: 1, 11, 21, ...
a = 1, d = 10
T25 = a + (25 - 1) × d = 1 + 24 × 10 = 241 -
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 -
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 -
Q4. What is the 21st term of the sequence: 100, 95, 90, ...?
a = 100, d = -5
T21 = 100 + 20 × (-5) = 0 -
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 -
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 -
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
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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
-
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 -
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 -
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
- (i) 1st term = 30, Sum of first 3 terms = 300
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