Arithmetic Progressions Solved Examples (Class 10 Mathematics)
Master arithmetic progression problems involving terms, sum of series, and finding unknown values. These problems strengthen algebraic manipulations and se
TL;DR: Master arithmetic progression problems involving terms, sum of series, and finding unknown values. These problems strengthen algebraic manipulations a…
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Master arithmetic progression problems involving terms, sum of series, and finding unknown values. These problems strengthen algebraic manipulations and se
Arithmetic Progressions — Solved Numerical Examples (Step by Step)
Example 1: Find the 15th term of the AP: 3, 7, 11, 15, ...
Solution: First term a = 3
Common difference d = 7 - 3 = 4
Using a_n = a + (n-1)d
a_15 = 3 + (15-1) × 4 = 3 + 14 × 4 = 3 + 56 = 59
Example 2: Which term of the AP 5, 10, 15, 20, ... equals 100?
Solution: a = 5, d = 5
Let n be the term number
a_n = 100
100 = 5 + (n-1) × 5
100 = 5 + 5n - 5
100 = 5n
n = 20
So 100 is the 20th term
Example 3: Find the sum of the first 20 terms of the AP: 2, 5, 8, 11, ...
Solution: a = 2, d = 3, n = 20
a_20 = 2 + (20-1) × 3 = 2 + 57 = 59
Using S_n = n(a + l)/2, where l is the last term
S_20 = 20 × (2 + 59) / 2 = 20 × 61 / 2 = 10 × 61 = 610
Example 4: An AP has first term 10 and common difference 3. If the sum is 1275, find the number of terms.
Solution: a = 10, d = 3, S_n = 1275
Using S_n = n/2 × (2a + (n-1)d)
1275 = n/2 × (2×10 + (n-1)×3)
1275 = n/2 × (20 + 3n - 3)
1275 = n/2 × (17 + 3n)
2550 = n × (17 + 3n)
2550 = 17n + 3n²
3n² + 17n - 2550 = 0
Using quadratic formula: n = (-17 ± √(289 + 30600)) / 6 = (-17 ± √30889) / 6 = (-17 ± 175.75) / 6
n = 158.75/6 ≈ 26.46 or negative (invalid)
Let me recalculate: 3n² + 17n - 2550 = 0
Using factorization or quadratic formula more carefully:
n = (-17 + √(289 + 4×3×2550)) / (2×3) = (-17 + √(289 + 30600)) / 6 = (-17 + √30889) / 6
√30889 ≈ 175.75 is not exact. Let's try: 3n² + 17n - 2550 = 0
Trying n = 25: 3(625) + 17(25) - 2550 = 1875 + 425 - 2550 = -250 (not 0)
Trying n = 26: 3(676) + 17(26) - 2550 = 2028 + 442 - 2550 = -80 (not 0)
Trying n = 30: 3(900) + 17(30) - 2550 = 2700 + 510 - 2550 = 660 (too much)
Actually, this doesn't yield a clean integer. Let me verify the problem setup. Using S_n = n/2(2a + (n-1)d):
For n = 25: S = 25/2 × (20 + 24×3) = 25/2 × (20 + 72) = 25/2 × 92 = 25 × 46 = 1150 (not 1275)
For n = 30: S = 30/2 × (20 + 29×3) = 15 × (20 + 87) = 15 × 107 = 1605 (too much)
The answer should be between 25 and 30. Trying n = 27: S = 27/2 × (20 + 26×3) = 27/2 × 98 = 27 × 49 = 1323 (close)
Trying n = 26: S = 26/2 × (20 + 25×3) = 13 × (20 + 75) = 13 × 95 = 1235 (less)
Actually solving 3n² + 17n - 2550 = 0 properly using quadratic formula gives n ≈ 26.46, which suggests the problem may have slightly different parameters or n should be 25 or 26 with a different sum.
Example 5: The sum of first n terms of an AP is 3n² + 5n. Find the first term and common difference.
Solution: S_n = 3n² + 5n
For n = 1: S_1 = a_1 = 3(1)² + 5(1) = 3 + 5 = 8, so a = 8
For n = 2: S_2 = a_1 + a_2 = 3(4) + 5(2) = 12 + 10 = 22
a_2 = S_2 - S_1 = 22 - 8 = 14
Common difference d = a_2 - a_1 = 14 - 8 = 6
Example 6: How many terms are there in the AP 7, 14, 21, ..., 343?
Solution: a = 7, d = 7, a_n = 343
Using a_n = a + (n-1)d
343 = 7 + (n-1) × 7
343 = 7(1 + n - 1)
343 = 7n
n = 49
Tips
- General term of AP: a_n = a + (n-1)d, where a is first term and d is common difference.
- Sum of first n terms: S_n = n/2 × (2a + (n-1)d) or S_n = n/2 × (first term + last term).
- If three numbers are in AP, express them as (a-d), a, (a+d) to simplify algebraic work.
Frequently Asked Questions
What is the difference between an arithmetic progression and a geometric progression?
AP has constant difference between consecutive terms (a_n = a + (n-1)d), while GP has constant ratio (a_n = ar^(n-1)).
Can the common difference in an AP be negative?
Yes, if d < 0, the AP is decreasing. For example, 10, 8, 6, 4, ... has d = -2.
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