Sequence and Series: AP, GP Formulas and Tricks for Competitive Exams

Sequence and series is one of the most formula-consistent topics in competitive mathematics. Unlike topics where problem presentation varies widely, sequence and series questions always follow the same structure — identify the type, apply the formula, calculate. The entire topic reduces to two main sequences: Arithmetic Progression (AP) and Geometric Progression (GP), plus a small set of special series.

The challenge most candidates face is not memorizing the formulas — it is recognizing quickly whether a sequence is AP, GP, or something else, and then choosing the right formula among the several available for each type. This guide solves that recognition problem explicitly, alongside every formula and worked example you need for exam day.

In SSC CGL Tier 1, sequence and series contributes 2–3 questions. In Tier 2, this increases to 3–4 questions. In CAT, number series and progression questions appear in 2–3 questions under modern math. In IBPS PO, number series is a dedicated section of 5 questions in Prelims.

Part 1: Identifying the Sequence Type

Before applying any formula, identify what type of sequence you have. This takes 10 seconds and determines everything that follows.

Step 1 — Check for AP (Arithmetic Progression)

Subtract consecutive terms. If the difference is constant — it is an AP.

Example: 3, 7, 11, 15, 19 ...
Differences: 7−3=4, 11−7=4, 15−11=4 → constant difference = AP

Step 2 — Check for GP (Geometric Progression)

Divide consecutive terms. If the ratio is constant — it is a GP.

Example: 2, 6, 18, 54, 162 ...
Ratios: 6/2=3, 18/6=3, 54/18=3 → constant ratio = GP

Step 3 — Check for Special Series

If neither difference nor ratio is constant — check these patterns:

Part 2: Arithmetic Progression (AP)

Definition

An AP is a sequence where each term differs from the previous by a fixed value called the common difference (d).

General form: a, a+d, a+2d, a+3d ...

Where a = first term, d = common difference

Core AP Formulas

nth term (general term):
aₙ = a + (n−1)d

Sum of first n terms:
Sₙ = n/2 × (2a + (n−1)d)

Alternate sum formula (when first and last term are known):
Sₙ = n/2 × (a + l)

Where l = last term = aₙ

Common difference:
d = (aₙ − a) / (n−1)

Worked Example 1 — Finding nth Term

Find the 15th term of AP: 3, 7, 11, 15 ...

a = 3, d = 4, n = 15

a₁₅ = 3 + (15−1) × 4 = 3 + 56 = 59

Worked Example 2 — Sum of AP

Find the sum of first 20 terms of AP: 5, 8, 11, 14 ...

a = 5, d = 3, n = 20

S₂₀ = 20/2 × (2×5 + (20−1)×3)
= 10 × (10 + 57)
= 10 × 67 = 670

Worked Example 3 — Using Alternate Sum Formula

Sum of AP: 3, 7, 11 ... 99?

a = 3, l = 99, d = 4

First find n: 99 = 3 + (n−1)×4 → 96 = (n−1)×4 → n = 25

S₂₅ = 25/2 × (3 + 99) = 25/2 × 102 = 25 × 51 = 1275

AP Properties — Exam Shortcuts

Property 1 — Middle term is the average:
In an AP with odd number of terms, the middle term = sum / number of terms = arithmetic mean of all terms.

Worked Example 4:
Three terms of AP have sum 21 and product 280. Find the terms.

Let terms = a−d, a, a+d
Sum: (a−d) + a + (a+d) = 3a = 21 → a = 7
Product: (7−d) × 7 × (7+d) = 280
7(49 − d²) = 280
49 − d² = 40 → d² = 9 → d = 3

Terms = 4, 7, 10 ✓

Property 2 — Assume symmetric terms:

• 3 terms in AP: a−d, a, a+d
• 4 terms in AP: a−3d, a−d, a+d, a+3d
• 5 terms in AP: a−2d, a−d, a, a+d, a+2d

Using symmetric form eliminates d in sum calculations — a major time saver.

Property 3 — Sum of terms equidistant from ends:
In any AP, the sum of terms equidistant from both ends is always equal to (first term + last term).

a₁ + aₙ = a₂ + aₙ₋₁ = a₃ + aₙ₋₂ = ... = a + l

Worked Example 5:
In AP: 4, 9, 14 ... 64. Find sum of 3rd and 8th terms from the start.

a + l = 4 + 64 = 68 — same for every equidistant pair.

Finding Number of Terms in an AP

n = (l − a)/d + 1

Worked Example 6:
How many terms in AP: 7, 13, 19 ... 205?

n = (205 − 7)/6 + 1 = 198/6 + 1 = 33 + 1 = 34 terms

Part 3: Geometric Progression (GP)

Definition

A GP is a sequence where each term is obtained by multiplying the previous term by a fixed value called the common ratio (r).

General form: a, ar, ar², ar³ ...

Where a = first term, r = common ratio

Core GP Formulas

nth term:
aₙ = a × r⁽ⁿ⁻¹⁾

Sum of first n terms (r ≠ 1):

When r > 1: Sₙ = a × (rⁿ − 1) / (r − 1)

When r < 1: Sₙ = a × (1 − rⁿ) / (1 − r)

Sum of infinite GP (|r| < 1):
S∞ = a / (1 − r)

Common ratio:
r = aₙ / aₙ₋₁ (ratio of any term to its preceding term)

Worked Example 7 — Finding nth Term

Find the 8th term of GP: 3, 6, 12, 24 ...

a = 3, r = 2, n = 8

a₈ = 3 × 2⁽⁸⁻¹⁾ = 3 × 2⁷ = 3 × 128 = 384

Worked Example 8 — Sum of GP

Find sum of first 6 terms of GP: 2, 6, 18, 54 ...

a = 2, r = 3, n = 6

S₆ = 2 × (3⁶ − 1) / (3 − 1)
= 2 × (729 − 1) / 2
= 728 = 728

Worked Example 9 — Infinite GP

Sum of infinite GP: 1/2 + 1/4 + 1/8 + ...

a = 1/2, r = 1/2 (|r| < 1 ✓)

S∞ = (1/2) / (1 − 1/2) = (1/2) / (1/2) = 1

GP Properties — Exam Shortcuts

Property 1 — Symmetric GP terms:

• 3 terms in GP: a/r, a, ar
• 5 terms in GP: a/r², a/r, a, ar, ar²

Product of symmetric terms: (a/r) × a × (ar) = a³
Middle term of 3-term GP = ∛(product of all three)

Worked Example 10:
Three terms of GP have product 512 and sum 28. Find the terms.

Let terms = a/r, a, ar
Product: a³ = 512 → a = 8
Sum: 8/r + 8 + 8r = 28 → 8/r + 8r = 20 → 8r² − 20r + 8 = 0
4r² − 10r + 4 = 0 → (4r − 2)(r − 2) = 0 → r = 1/2 or r = 2

Terms (r=2): 4, 8, 16 ✓

Property 2 — GP mean:
If a, b, c are in GP → b² = ac (b is the geometric mean of a and c)

Worked Example 11:
If 4, x, 36 are in GP, find x.

x² = 4 × 36 = 144 → x = 12

Part 4: Arithmetic Mean and Geometric Mean

Arithmetic Mean (AM)

AM of two numbers a and b = (a + b) / 2

If A is the AM between a and b: A = (a + b) / 2, which means a, A, b form an AP.

Geometric Mean (GM)

GM of two numbers a and b = √(a × b)

If G is the GM between a and b: G = √(ab), which means a, G, b form a GP.

AM-GM Inequality

For any two positive numbers: AM ≥ GM

(a + b)/2 ≥ √(ab)

Equality holds when a = b.

Exam application: When a question asks for the minimum value of (x + 1/x) for x > 0 — use AM-GM:
(x + 1/x)/2 ≥ √(x × 1/x) = 1
x + 1/x ≥ 2 → minimum value = 2 (when x = 1)

Part 5: Special Series — Sum Formulas

These four sum formulas appear directly in SSC CGL and CAT. Memorize them exactly.

Sum of First n Natural Numbers

1 + 2 + 3 + ... + n = n(n+1) / 2

Worked Example 12:
Sum of first 50 natural numbers?

= 50 × 51 / 2 = 1275

Sum of Squares of First n Natural Numbers

1² + 2² + 3² + ... + n² = n(n+1)(2n+1) / 6

Worked Example 13:
1² + 2² + 3² + ... + 10²?

= 10 × 11 × 21 / 6 = 2310 / 6 = 385

Sum of Cubes of First n Natural Numbers

1³ + 2³ + 3³ + ... + n³ = [n(n+1) / 2]²

Note: Sum of cubes = (Sum of natural numbers)²

Worked Example 14:
1³ + 2³ + 3³ + ... + 8³?

= [8 × 9 / 2]² = ² = 1296

Quick check: 1+2+...+8 = 36, so 36² = 1296 ✓

Sum of First n Odd Numbers

1 + 3 + 5 + ... + (2n−1) = n²

Worked Example 15:
Sum of first 12 odd numbers?

= 12² = 144

Sum of First n Even Numbers

2 + 4 + 6 + ... + 2n = n(n+1)

Worked Example 16:
Sum of first 15 even numbers?

= 15 × 16 = 240

Part 6: Number Series for IBPS PO Prelims

IBPS PO Prelims has a dedicated number series section of 5 questions. Each question gives a series with one missing term. The approach is always the same: identify the pattern, apply it to find the missing term.

The 8 Most Common IBPS Number Series Patterns

Pattern 1 — Constant Difference (AP):
5, 11, 17, 23, _ → d = 6 → 29

Pattern 2 — Constant Ratio (GP):
3, 9, 27, 81, _ → r = 3 → 243

Pattern 3 — Differences Form an AP:
2, 3, 5, 8, 12, 17, _ → differences: 1,2,3,4,5,6 → 23

Pattern 4 — Squares:
1, 4, 9, 16, 25, _ → n² → 36

Pattern 5 — Cubes:
1, 8, 27, 64, _ → n³ → 125

Pattern 6 — Squares ± constant:
2, 5, 10, 17, 26, _ → n²+1 → 37

Pattern 7 — Two Alternating Series:
2, 3, 4, 6, 6, 12, 8, _ → Series 1: 2,4,6,8 (d=2) | Series 2: 3,6,12,_ (r=2) → 24

Pattern 8 — Multiplication + Addition:
1, 3, 7, 15, 31, _ → each term × 2 + 1 → 31×2+1 = 63

Speed Strategy for IBPS Number Series

Step 1 (5 sec): Find differences between consecutive terms.
Step 2 (5 sec): If differences are constant → AP done.
Step 3 (5 sec): If differences increase — find second-order differences.
Step 4 (5 sec): If differences not helpful — find ratios.
Step 5 (5 sec): If ratios not constant — check for squares/cubes pattern.

Target: Identify pattern within 20 seconds, answer within 45 seconds total.

Part 7: HP (Harmonic Progression) — Basic Coverage

Harmonic Progression is less frequently tested but appears occasionally in SSC CGL Tier 2 and CAT.

Definition

A sequence is in HP if the reciprocals of its terms form an AP.

1/a, 1/(a+d), 1/(a+2d) ... → corresponding AP: a, a+d, a+2d ...

Harmonic Mean

HM of two numbers a and b = 2ab / (a+b)

AM-GM-HM Relationship

For any two positive numbers: AM ≥ GM ≥ HM

AM × HM = GM²

Worked Example 17:
If AM of two numbers is 10 and GM is 8, find HM.

HM = GM² / AM = 64 / 10 = 6.4

Part 8: Common Exam Traps

Trap 1 — Confusing nth Term with Sum

aₙ = a + (n−1)d gives the nth term — a single value.
Sₙ = n/2 × (2a + (n−1)d) gives the sum of first n terms — a total.

Many candidates substitute into the wrong formula. Read the question carefully: "find the 10th term" vs "find the sum of 10 terms."

Trap 2 — GP with Negative Ratio

When r is negative, GP terms alternate between positive and negative. The sum formula still applies — just track signs carefully.

Example: 1, −2, 4, −8, 16 ... (r = −2)
a₅ = 1 × (−2)⁴ = 16 ✓

Trap 3 — Infinite GP Condition

S∞ = a/(1−r) applies ONLY when |r| < 1. If |r| ≥ 1, the infinite series diverges — sum does not exist.

Always check |r| < 1 before applying the infinite GP formula.

Trap 4 — Sum of Series Starting from a Value Other Than 1

If series starts from k instead of 1:
Sum from k to n = Sₙ − S₍ₖ₋₁₎

Worked Example 18:
Sum of 5² + 6² + 7² + ... + 12²?

= S₁₂ − S₄
= [12×13×25/6] − [4×5×9/6]
= 650 − 30 = 620

Quick Reference Formula Sheet

Exam-Wise Strategy

3-Week Practice Plan