Sequences and series show up everywhere in aptitude — salary increments, population growth, bouncing balls, installment payments. The good news is there are really only two types we need to master: Arithmetic Progression (AP) and Geometric Progression (GP).
In simple language, an AP is a sequence where we add the same number each time (like 2, 5, 8, 11…), and a GP is where we multiply by the same number each time (like 3, 6, 12, 24…).
Arithmetic Progression (AP)
An AP has a constant difference between consecutive terms. That difference is called d (common difference).
The alternate sum formula Sₙ = (n/2)(first + last) is often faster because we just need the first term, last term, and count. Think of it like this — the average of an AP is always the average of the first and last terms, and total = average × count.
Example 1: Find the nth term
Find the 20th term of the AP: 3, 7, 11, 15…
a = 3, d = 7 - 3 = 4.
a₂₀ = 3 + (20 - 1)(4) = 3 + 76 = 79
Example 2: Sum of terms
Find the sum of the first 15 terms of the AP: 5, 9, 13…
a = 5, d = 4, n = 15.
S₁₅ = (15/2)[2(5) + (14)(4)] = (15/2)[10 + 56] = (15/2)(66) = 495
The Middle Term Property
This is a neat shortcut. In any three consecutive AP terms, the middle one is the average of the other two. So if we have terms a, b, c in AP, then b = (a + c)/2.
This extends: in an AP with an odd number of terms, the sum = middle term × number of terms.
Example 3: Using middle term
The sum of three numbers in AP is 27, and their product is 648. Find them.
Let the three numbers be (a - d), a, (a + d). This is the standard trick for three AP terms — it automatically centers on a.
Sum: (a - d) + a + (a + d) = 3a = 27 → a = 9
Product: (9 - d)(9)(9 + d) = 648
- 9(81 - d²) = 648
- 81 - d² = 72
- d² = 9 → d = 3
Numbers: 6, 9, 12
Geometric Progression (GP)
A GP has a constant ratio between consecutive terms. That ratio is called r (common ratio).
Example 4: Sum to infinity
A ball is dropped from 80 meters. Each time it bounces, it rises to 3/4 of the height it fell from. Find the total distance traveled.
Going down: 80, 60, 45, … (GP with a = 80, r = 3/4) Going up: 60, 45, … (GP with a = 60, r = 3/4)
Total down = 80 / (1 - 3/4) = 80 / (1/4) = 320 Total up = 60 / (1 - 3/4) = 60 / (1/4) = 240
Total distance = 320 + 240 = 560 meters
Alternative shortcut: Total distance = first drop × (1 + r)/(1 - r) = 80 × (1 + 3/4)/(1 - 3/4) = 80 × (7/4)/(1/4) = 80 × 7 = 560.
Example 5: Finding terms
The 3rd term of a GP is 12 and the 6th term is 96. Find the GP.
a₃ = ar² = 12 a₆ = ar⁵ = 96
Divide: ar⁵/ar² = r³ = 96/12 = 8 → r = 2
From ar² = 12: a(4) = 12 → a = 3
GP: 3, 6, 12, 24, 48, 96…
Harmonic Progression (HP)
An HP is a sequence whose reciprocals form an AP. We rarely solve HP directly — instead, we convert to AP, solve, and convert back.
If a, b, c are in HP, then 1/a, 1/b, 1/c are in AP, which means 1/b = (1/a + 1/c)/2.
The harmonic mean of a and b = 2ab/(a + b).
The relationship between means: AM ≥ GM ≥ HM (with equality when all terms are equal). This inequality is super useful for optimization problems.
Word Problem Patterns
Salary increment (AP)
“Starting salary is 30,000. Annual increment of 2,000. What’s the salary in the 10th year?”
a = 30000, d = 2000, n = 10. a₁₀ = 30000 + 9(2000) = 48,000
Population growth (GP)
“Population of a city is 50,000 and grows at 10% per year. What will it be after 3 years?”
a = 50000, r = 1.10, n = 3. Population = 50000 × (1.10)³ = 50000 × 1.331 = 66,550
Installments
“Total amount is 1000. Paid in 5 installments increasing by 50 each time. Find the first installment.”
AP with n = 5, d = 50, S₅ = 1000. (5/2)[2a + 4(50)] = 1000 2a + 200 = 400 → a = 100
Useful Shortcuts
- Sum of first n natural numbers: n(n+1)/2
- Sum of squares of first n natural numbers: n(n+1)(2n+1)/6
- Sum of cubes of first n natural numbers: [n(n+1)/2]² (yes, it’s the square of the sum of naturals!)
- Sum of first n even numbers: n(n+1)
- Sum of first n odd numbers: n²
These come up so often that memorizing them saves a huge amount of time.
Common Exam Variations
- Find the nth term or sum of an AP/GP.
- “How many terms are needed for the sum to exceed X?”
- Three numbers in AP/GP with given sum and product.
- Insert n arithmetic/geometric means between two numbers.
- Mixed problems: “Sum of an AP equals the sum of a GP…”
- Real-world: salary, depreciation, population, bouncing ball.
Practice Problems
Problem 1: The sum of the first 10 terms of an AP is 155, and the first term is 2. Find the common difference.
Problem 2: Find the sum to infinity of the GP: 27, 9, 3, 1, …
Problem 3: The 4th and 7th terms of an AP are 11 and 23 respectively. Find the sum of the first 15 terms.
Answers
Problem 1: S₁₀ = (10/2)[2(2) + 9d] = 155 → 5[4 + 9d] = 155 → 4 + 9d = 31 → 9d = 27 → d = 3.
Problem 2: a = 27, r = 9/27 = 1/3. Since |r| < 1: S∞ = 27/(1 - 1/3) = 27/(2/3) = 40.5 (or 81/2).
Problem 3: a₄ = a + 3d = 11 and a₇ = a + 6d = 23. Subtracting: 3d = 12 → d = 4. So a = 11 - 12 = -1. S₁₅ = (15/2)[2(-1) + 14(4)] = (15/2)[−2 + 56] = (15/2)(54) = 405.