An arithmetic sequence is a sequence where each term is obtained by adding a fixed number (common difference d) to the previous term.
a = first term, d = common difference, n = term number
Find the 15th term of 3, 7, 11, 15, ...
a = 3, d = 4, n = 15
T₁₅ = 3 + (15−1)×4 = 3 + 56 = 59
Which term of 5, 8, 11, ... is 50?
50 = 5 + (n−1)×3 → 45 = 3(n−1) → n−1 = 15 → n = 16
where l = last term = a + (n−1)d
Find the sum of the first 20 terms of 4, 7, 10, 13, ...
a = 4, d = 3, n = 20
S₂₀ = 20/2 [2(4) + (20−1)(3)] = 10[8 + 57] = 10 × 65 = 650
A worker earns Ksh 8,000 in January and gets a Ksh 500 raise each month. Total earnings in 12 months?
a = 8000, d = 500, n = 12
S₁₂ = 12/2[2(8000) + 11(500)] = 6[16000 + 5500] = 6 × 21500 = Ksh 129,000
A geometric sequence has a constant ratio r between consecutive terms.
a = first term, r = common ratio, n = term number
Find the 6th term of 3, 6, 12, 24, ...
a = 3, r = 2, n = 6
T₆ = 3 × 2⁵ = 3 × 32 = 96
In a GP, T₂ = 6 and T₅ = 48. Find a and r.
ar = 6 ... (i) ar⁴ = 48 ... (ii)
(ii)÷(i): r³ = 8 → r = 2 → a = 3
Find the sum of first 8 terms of 2, 6, 18, ...
a = 2, r = 3, n = 8
S₈ = 2(3⁸ − 1)/(3−1) = 2(6561−1)/2 = 6560
A town has 5,000 people. Population grows by 10% per year. Total population over 5 years?
a = 5000, r = 1.1, n = 5
S₅ = 5000(1.1⁵−1)/(1.1−1) = 5000(0.6105)/0.1 = 30,526
When |r| < 1, a geometric series converges (approaches a finite sum as n → ∞).
Find S∞ of 16, 8, 4, 2, ...
a = 16, r = 0.5
S∞ = 16/(1−0.5) = 16/0.5 = 32
Express 0.333... as a fraction using GP.
0.333... = 0.3 + 0.03 + 0.003 + ... → a = 0.3, r = 0.1
S∞ = 0.3/(1−0.1) = 0.3/0.9 = 1/3
| Feature | Arithmetic (AP) | Geometric (GP) |
|---|---|---|
| Pattern | Add common difference d | Multiply by common ratio r |
| nth term | a + (n−1)d | arⁿ⁻¹ |
| Sum Sₙ | n/2[2a+(n−1)d] | a(rⁿ−1)/(r−1) |
| Sum to ∞ | Does not exist | a/(1−r) if |r|<1 |
| Test | T₂−T₁ = T₃−T₂ | T₂/T₁ = T₃/T₂ |
Differences: 5, 10, 20 — NOT constant → not AP
Ratios: 10/5=2, 20/10=2, 40/20=2 — constant → GP, r=2
Ksh 10,000 invested at 12% compound interest per year. Amount after 5 years?
a = 10,000, r = 1.12, n = 5
T₅ = 10,000 × 1.12⁴ = 10,000 × 1.5735 = Ksh 15,735
Starting salary Ksh 30,000, annual increment Ksh 2,000. Salary in 10th year?
T₁₀ = 30,000 + (10−1)×2,000 = 30,000 + 18,000 = Ksh 48,000
A ball dropped from 10 m bounces to 60% of previous height. Total distance travelled?
First drop: 10 m. Each bounce up then down: 10×0.6 + 10×0.6 = 2×10×0.6
Total = 10 + 2×[10×0.6/(1−0.6)] = 10 + 2×15 = 40 m
The 5th term of an AP is 21 and the 10th term is 41. Find a, d, and S₁₅.
T₅ = a + 4d = 21 ... (i) T₁₀ = a + 9d = 41 ... (ii)
(ii)−(i): 5d = 20 → d = 4, a = 5
S₁₅ = 15/2[2(5)+14(4)] = 15/2[10+56] = 15/2×66 = 495
A GP has first term 3 and sum to infinity 12. Find r and the 4th term.
S∞ = a/(1−r) → 12 = 3/(1−r) → 1−r = 0.25 → r = 0.75
T₄ = 3 × (0.75)³ = 3 × 0.4219 = 1.266
1. The nth term of an AP is Tₙ = .
2. For a GP, the common ratio r = .
3. The sum to infinity of a GP exists only when |r| .
4. The 8th term of 2, 5, 8, 11, ... is .
5. S∞ = a/(1−r) applies to a series.
6. (4 marks) The 3rd term of a GP is 12 and the 6th term is 96. Find a, r, and S₆.
7. (3 marks) An AP has a=7, d=3. Find n if Sₙ = 255.
8. (3 marks) Express 0.272727... as a fraction using GP sum to infinity.