Arithmetic and Geometric Progression
Arithmetic Progression (AP)
Definition
A sequence in which the difference between consecutive terms is constant is called an Arithmetic Progression. The constant difference is called the common difference (d).
a, a + d, a + 2d, a + 3d, ...
nth Term of an AP
Tₙ = a + (n − 1)d
Where:
- a = first term
- d = common difference
- n = term number
Sum of n Terms of an AP
Sₙ = n/2 [2a + (n − 1)d]
or
Sₙ = n/2 (a + l)
Where l = last term = a + (n − 1)d.
Arithmetic Mean (AM)
If three numbers are in AP, the middle term is the arithmetic mean.
If a, b, c are in AP, then b = (a + c) / 2
Geometric Progression (GP)
Definition
A sequence in which the ratio between consecutive terms is constant is called a Geometric Progression. The constant ratio is called the common ratio (r).
a, ar, ar², ar³, ...
nth Term of a GP
Tₙ = arⁿ⁻¹
Sum of n Terms of a GP
Sₙ = a(1 − rⁿ) / (1 − r), when r ≠ 1
Sₙ = na, when r = 1
Geometric Mean (GM)
If three numbers are in GP, the middle term is the geometric mean.
If a, b, c are in GP, then b² = ac, i.e., b = ±√(ac)
Worked Examples — AP
Example 1: Finding nth Term
Find the 15th term of the AP: 3, 7, 11, 15, ...
Solution: a = 3, d = 7 − 3 = 4 T₁₅ = 3 + (15 − 1) × 4 = 3 + 56 = 59
Example 2: Finding Sum of n Terms
Find the sum of first 20 terms of the AP: 2, 5, 8, 11, ...
Solution: a = 2, d = 3, n = 20 S₂₀ = 20/2 [2(2) + (20 − 1)3] = 10[4 + 57] = 10 × 61 = 610
Example 3: Finding Number of Terms
How many terms of the AP: 4, 8, 12, ... sum to 400?
Solution: a = 4, d = 4, Sₙ = 400 n/2 [2(4) + (n − 1)4] = 400 n/2 [8 + 4n − 4] = 400 n/2 [4n + 4] = 400 n(2n + 2) = 400 2n² + 2n − 400 = 0 n² + n − 200 = 0 (n + ?)(n − ?) — using formula: n = [−1 ± √(1 + 800)] / 2 = [−1 ± √801] / 2
Since √801 ≈ 28.3, n = (−1 + 28.3) / 2 ≈ 13.65 So n = 13 terms (sum is close to 400; let us verify: S₁₃ = 13/2[8 + 48] = 13/2 × 56 = 364, S₁₄ = 14/2[8 + 52] = 7 × 60 = 420)
Wait, let me redo this more carefully.
Sₙ = n/2 [2(4) + (n − 1)4] = n/2 [8 + 4n − 4] = n/2 [4n + 4] = 2n² + 2n
Setting equal to 400: 2n² + 2n = 400 2n² + 2n − 400 = 0 n² + n − 200 = 0
Using quadratic formula: n = [−1 ± √(1 + 800)] / 2 = [−1 ± √801] / 2
n = (−1 + 28.32) / 2 ≈ 13.66
Since n must be an integer, the sum of 13 terms is: S₁₃ = 13/2 [2(4) + 12(4)] = 13/2 [8 + 48] = 13/2 × 56 = 13 × 28 = 364
So 13 terms sum to 364, and 14 terms would sum to 420.
Example 4: Three Numbers in AP
Three numbers in AP have sum 15 and product 80. Find the numbers.
Solution: Let the numbers be a − d, a, a + d. Sum = (a − d) + a + (a + d) = 3a = 15 → a = 5 Product = (5 − d)(5)(5 + d) = 80 25 − d² = 16 d² = 9 → d = ±3
Numbers: 2, 5, 8 (if d = 3) or 8, 5, 2 (if d = −3)
Worked Examples — GP
Example 1: Finding Terms of GP
Find the 5th term of the GP: 2, 6, 18, ...
Solution: a = 2, r = 6/2 = 3 T₅ = 2 × 3⁴ = 2 × 81 = 162
Example 2: Sum of GP Terms
Find the sum of the first 6 terms of the GP: 3, 6, 12, 24, ...
Solution: a = 3, r = 2, n = 6 S₆ = 3(1 − 2⁶) / (1 − 2) = 3(1 − 64) / (−1) = 3(−63) / (−1) = 189
Example 3: Three Numbers in GP
Find three numbers in GP whose sum is 14 and product is 64.
Solution: Let numbers be a/r, a, ar. Product = a/r × a × ar = a³ = 64 → a = 4 Sum = 4/r + 4 + 4r = 14 4/r + 4r = 10 4 + 4r² = 10r 4r² − 10r + 4 = 0 2r² − 5r + 2 = 0 (2r − 1)(r − 2) = 0 r = 1/2 or r = 2
Numbers: 8, 4, 2 or 2, 4, 8
Comparison: AP vs GP
| Feature | Arithmetic Progression | Geometric Progression |
|---|---|---|
| Definition | Constant difference (d) | Constant ratio (r) |
| nth term | a + (n − 1)d | arⁿ⁻¹ |
| Sum of n terms | n/2 [2a + (n − 1)d] | a(1 − rⁿ)/(1 − r) |
| Middle term (mean) | (a + c)/2 | ±√(ac) |
| Growth | Linear | Exponential |
Common Mistakes and Fixes
| Mistake | Fix |
|---|---|
| Confusing d and r | d is added, r is multiplied |
| Using wrong sum formula in GP | If r < 1 use (1 − rⁿ), if r > 1 use (rⁿ − 1) |
| Forgetting that n is the term number | In Tₙ = a + (n − 1)d, n starts at 1 |
| Incorrect sign in AP sum formula | Write Sₙ = n/2 [2a + (n − 1)d] |
| Using GP sum when r = 1 | Use Sₙ = na when r = 1 (not the formula with denominator) |
ICSE Exam Focus
AP and GP carry 10–14 marks combined in ICSE exams. Questions include:
- Finding nth term and sum of n terms.
- Word problems involving AP (salary increments, savings).
- Three numbers in AP or GP.
- Finding a, d, r, n given conditions.
Marks Blueprint:
| Topic | Marks |
|---|---|
| AP: nth term and sum | 4 |
| AP: word problems | 3–4 |
| GP: nth term and sum | 3 |
| GP: three numbers | 2 |
| Mixed AP-GP concepts | 2 |
Self-Test Questions
-
Find the 20th term of the AP: 5, 9, 13, 17, ...
-
The sum of three numbers in AP is 24 and their product is 440. Find the numbers.
-
Find the sum of the first 8 terms of the GP: 1, 3, 9, 27, ...
-
The 4th term of a GP is 8 and the 7th term is 64. Find the common ratio and the first term.
-
A man saves ₹500 in the first month, ₹550 in the second, ₹600 in the third, and so on. How much does he save in 2 years?
-
Distinguish between an arithmetic progression and a geometric progression, giving examples.
In ICSE exams, AP word problems often appear in the context of savings, salaries, and instalments. GP problems are less frequent but test conceptual clarity.
