Important Number Sequences and Formulas
The difference between a student who scores 45/50 in Quantitative Aptitude and one who scores 35/50 is often not intelligence — it is whether they have these seven series formulas instant-recalled versus slowly derived.
Definition: A number sequence (or series) is an ordered list of numbers following a specific rule. A series is the sum of the terms of a sequence.
Why Series Formulas Appear in Every Exam
SSC CGL, SSC CHSL, RRB NTPC, and banking exams test series in two ways:
- Direct computation — "Find the sum of the first 50 odd numbers."
- Missing term — "Find the missing term in: 2, 6, 12, 20, ?, 42."
Formula knowledge collapses a 2-minute calculation to a 10-second substitution.
Core Summation Formulas
Formula 1: Sum of First n Natural Numbers
S = n(n + 1) / 2
Natural numbers are 1, 2, 3, 4, … This formula is attributed to Gauss, who as a child reportedly summed 1 to 100 by pairing: (1+100), (2+99), ... = 50 pairs of 101 → 5050.
Verification: n = 4 → 1+2+3+4 = 10; formula: 4×5/2 = 10. ✓
Formula 2: Sum of Squares of First n Natural Numbers
S = n(n + 1)(2n + 1) / 6
This counts the total number of unit squares in layered arrangements — relevant in geometry and combinatorics, but tested purely computationally in exams.
Verification: n = 3 → 1+4+9 = 14; formula: 3×4×7/6 = 14. ✓
Formula 3: Sum of Cubes of First n Natural Numbers
S = [n(n + 1) / 2]²
Notice this equals the square of Formula 1. This elegant identity means: (sum of first n natural numbers)² = sum of their cubes.
Verification: n = 3 → 1+8+27 = 36; formula: [3×4/2]² = 6² = 36. ✓
Formula 4: Sum of First n Even Numbers
S = n(n + 1)
The first n even numbers are 2, 4, 6, ..., 2n.
Verification: n = 4 → 2+4+6+8 = 20; formula: 4×5 = 20. ✓
Formula 5: Sum of First n Odd Numbers
S = n²
The first n odd numbers are 1, 3, 5, ..., (2n−1).
Verification: n = 4 → 1+3+5+7 = 16; formula: 4² = 16. ✓
Real-world Indian example: The number of bricks in a square staircase arrangement — 1 on the top step, 3 on the next, 5 on the next — equals n² for n steps. This is directly the sum of first n odd numbers.
Formula 6: Sum of an Arithmetic Series (General)
S = (n/2) × (first term + last term)
or equivalently S = (n/2) × [2a + (n−1)d], where a = first term, d = common difference.
Definition: An arithmetic progression (AP) is a sequence where consecutive terms differ by a constant value called the common difference (d).
Worked Example:
Question: Find the sum of the AP: 3, 7, 11, ..., 99.
Solution:
Step 1: Identify a = 3, d = 4.
Step 2: Find n. Last term = a + (n−1)d → 99 = 3 + (n−1)×4 → n−1 = 24 → n = 25.
Step 3: S = (25/2) × (3 + 99) = (25/2) × 102 = 25 × 51 = 1275.
Conclusion: Sum = 1275.
Identifying Missing Terms in Number Series
When asked to find a missing term, you must first decode the pattern. Common pattern types:
| Pattern Type | Example | Rule |
|---|---|---|
| Constant difference (AP) | 3, 7, 11, ?, 19 | +4 each time |
| Constant ratio (GP) | 2, 6, 18, ?, 162 | ×3 each time |
| Increasing difference | 1, 2, 4, 7, 11, ? | Differences: +1, +2, +3, +4... |
| Perfect squares | 1, 4, 9, 16, ?, 36 | n² sequence |
| Perfect cubes | 1, 8, 27, ?, 125 | n³ sequence |
| Prime numbers | 2, 3, 5, 7, ?, 13 | Consecutive primes |
| Fibonacci-type | 1, 1, 2, 3, 5, ? | Each term = sum of two before |
Worked Example (Missing Term):
Question: Find the missing term: 2, 6, 12, 20, ?, 42.
Solution:
Step 1: Write differences: 6−2=4, 12−6=6, 20−12=8, ?−20=?, 42−?=?
Step 2: Differences are 4, 6, 8 — increasing by 2 each time. Next difference = 10.
Step 3: Missing term = 20 + 10 = 30. Check: 42 − 30 = 12 ✓ (next difference in sequence).
Conclusion: Missing term = 30.
Alternative recognition: The series is n(n+1): 1×2=2, 2×3=6, 3×4=12, 4×5=20, 5×6=30, 6×7=42. ✓
Applying the "Sum of Odd Numbers" Formula
Question: Find the sum 1 + 3 + 5 + 7 + ... + 99.
Solution:
Step 1: Count terms. The kth odd number = 2k−1. Set 2k−1=99 → k=50. So n=50.
Step 2: Sum = n² = 50² = 2500.
Conclusion: Sum = 2500.
| Series | Formula | Example (n=5) |
|---|---|---|
| 1+2+3+…+n | n(n+1)/2 | 15 |
| 1²+2²+…+n² | n(n+1)(2n+1)/6 | 55 |
| 1³+2³+…+n³ | [n(n+1)/2]² | 225 |
| 2+4+6+…+2n | n(n+1) | 30 |
| 1+3+5+…+(2n−1) | n² | 25 |
- ✓- Sum of first n natural numbers: n(n+1)/2 — the Gauss formula.
- ✓- Sum of first n squares: n(n+1)(2n+1)/6 — includes 2n+1 factor.
- ✓- Sum of first n cubes = square of sum of first n naturals: [n(n+1)/2]².
- ✓- Sum of first n even numbers: n(n+1) — exactly double the natural numbers formula.
- ✓- Sum of first n odd numbers: n² — clean, elegant, fast.
- ✓- For any AP: S = (n/2)(first + last) — only requires knowing endpoints and count.
- ✓- To find n for a series ending at term L: n = (L − a)/d + 1.
- ✓- Pattern identification in missing-term problems: check differences, ratios, and recognise square/cube/prime sequences.
"Never Square Cubes Evenly Or Awkwardly" → Natural, Squares, Cubes, Even, Odd, AP — the six formula types. For values: think n/2 × (first+last) works for ALL arithmetic sums — the special formulas are just fast versions of this.
- ✓- Five key formulas cover natural numbers, squares, cubes, even numbers, and odd numbers — each worth knowing as a reflex.
- ✓- The sum of first n odd numbers equals n² — elegant and instantly testable.
- ✓- Sum of first n cubes equals the square of the sum of first n naturals — a powerful identity.
- ✓- For general APs, use S = (n/2)(first + last) or S = (n/2)[2a + (n−1)d].
- ✓- Missing-term problems require identifying the underlying rule: AP (constant difference), GP (constant ratio), or a recognisable sequence like primes, squares, or Fibonacci.
- ✓- Always verify your pattern by checking whether it correctly predicts a term you already know.
Unit Digit Patterns (Cyclicity)
You will never actually compute 7⁹⁵ in an exam hall — but you can find its unit digit in under 10 seconds once you understand cyclicity, and that skill alone is worth 2–3 marks per paper.
Definition: The unit digit of a number is its ones-place digit (the digit in the rightmost position). For any power aⁿ, the unit digit depends only on the unit digit of a and the value of n — never on the full value of a.
Definition: Cyclicity refers to the fact that as n increases, the unit digit of aⁿ repeats in a fixed cycle. The length of this cycle is called the cycle length (or period).
Why Unit Digit Patterns Work
When you compute successive powers of any number, the unit digit of the result depends only on the unit digit of the base and the unit digit of the previous result. For example:
- 7¹ = 7 → unit digit 7
- 7² = 49 → unit digit 9
- 7³ = 343 → unit digit 3
- 7⁴ = 2401 → unit digit 1
- 7⁵ = 16807 → unit digit 7 ← cycle restarts!
The pattern (7, 9, 3, 1) repeats with period 4. You never need to compute the actual value.
The Five Cyclicity Cases
Case 1: Digits 0, 1, 5, 6 — Cycle Length 1
These digits are fixed point digits — raising them to any power never changes the unit digit.
| Digit | Any power | Unit digit |
|---|---|---|
| 0 | 0ⁿ | 0 |
| 1 | 1ⁿ | 1 |
| 5 | 5ⁿ | 5 |
| 6 | 6ⁿ | 6 |
Real-world example: 256¹⁰⁰ ends in 6. 5⁵⁰⁰ ends in 5. No calculation required.
Why: 5² = 25, 5³ = 125, 5⁴ = 625 — unit digit is always 5. This is because 5 × 5 = 25, always ending in 5 again. Similarly, 6 × 6 = 36, always ending in 6.
Case 2: Digit 4 — Cycle Length 2
4¹ = 4, 4² = 16, 4³ = 64, 4⁴ = 256 → cycle: (4, 6, 4, 6, ...)
Rule: odd power → unit digit 4; even power → unit digit 6.
| Power (n) | n is odd | n is even |
|---|---|---|
| Unit digit of 4ⁿ | 4 | 6 |
Case 3: Digit 9 — Cycle Length 2
9¹ = 9, 9² = 81, 9³ = 729, 9⁴ = 6561 → cycle: (9, 1, 9, 1, ...)
Rule: odd power → unit digit 9; even power → unit digit 1.
Case 4: Digits 2, 3, 7, 8 — Cycle Length 4
These are the most tested digits in exams:
| Digit | Position 1 | Position 2 | Position 3 | Position 4 |
|---|---|---|---|---|
| 2ⁿ | 2 | 4 | 8 | 6 |
| 3ⁿ | 3 | 9 | 7 | 1 |
| 7ⁿ | 7 | 9 | 3 | 1 |
| 8ⁿ | 8 | 4 | 2 | 6 |
Memory pattern for 2: 2 → 4 → 8 → 6 (even powers show 4, 6 alternating; odd powers show 2, 8 alternating).
Memory pattern for 7: Note that 7 and 3 have the same cycle elements (7,9,3,1 and 3,9,7,1) — just reversed starting points.
The Universal Method: Using Mod 4
For digits with cycle length 4 (2, 3, 7, 8), the position in the cycle is determined by n mod 4:
| n mod 4 | Position in 4-cycle |
|---|---|
| 1 | 1st element of cycle |
| 2 | 2nd element of cycle |
| 3 | 3rd element of cycle |
| 0 (i.e., divisible by 4) | 4th element of cycle |
Key rule: When n mod 4 = 0, use the 4th position (not the 0th — cycles are 1-indexed).
Step-by-Step Procedure
- Identify the unit digit of the base (ignore all other digits).
- Determine the cycle for that digit.
- If cycle length is 1: done.
- If cycle length is 2: check if exponent is odd or even.
- If cycle length is 4: compute n mod 4; if result is 0, use position 4.
- Read off the unit digit from that position.
Worked Example 1
Question: Find the unit digit of 2⁸³.
Solution:
Step 1: Unit digit of base = 2; cycle for 2 = (2, 4, 8, 6), length 4.
Step 2: 83 mod 4 = 3 (since 83 = 4×20 + 3).
Step 3: Position 3 in cycle (2, 4, 8, 6) → unit digit = 8.
Conclusion: Unit digit of 2⁸³ is 8.
Worked Example 2
Question: Find the unit digit of 3¹⁰⁰.
Solution:
Step 1: Cycle for 3 = (3, 9, 7, 1), length 4.
Step 2: 100 mod 4 = 0.
Step 3: n mod 4 = 0 → use position 4 → unit digit = 1.
Conclusion: Unit digit of 3¹⁰⁰ is 1.
Worked Example 3
Question: Find the unit digit of 9⁴⁷.
Solution:
Step 1: Cycle for 9 = (9, 1), length 2.
Step 2: 47 is odd → unit digit = 9.
Conclusion: Unit digit of 9⁴⁷ is 9.
Multi-Term Unit Digit Problems
Exams also ask for the unit digit of expressions like 2⁴³ × 3³² + 7¹¹.
Approach: Find the unit digit of each part separately, then apply the operation to just those unit digits.
Question: Unit digit of 2⁴³ × 3³²?
- 2⁴³: 43 mod 4 = 3 → position 3 in (2,4,8,6) → 8.
- 3³²: 32 mod 4 = 0 → position 4 in (3,9,7,1) → 1.
- Unit digit of product = unit digit of 8 × 1 = 8.
| Digit | Cycle | Length | Method |
|---|---|---|---|
| 0 | 0 | 1 | Always 0 |
| 1 | 1 | 1 | Always 1 |
| 5 | 5 | 1 | Always 5 |
| 6 | 6 | 1 | Always 6 |
| 4 | 4, 6 | 2 | Odd→4, Even→6 |
| 9 | 9, 1 | 2 | Odd→9, Even→1 |
| 2 | 2, 4, 8, 6 | 4 | n mod 4 |
| 3 | 3, 9, 7, 1 | 4 | n mod 4 |
| 7 | 7, 9, 3, 1 | 4 | n mod 4 |
| 8 | 8, 4, 2, 6 | 4 | n mod 4 |
- ✓- Unit digit depends only on the unit digit of the base — ignore all other digits.
- ✓- Digits 0, 1, 5, 6 always produce the same unit digit regardless of power.
- ✓- Digits 4 and 9 have cycles of length 2: parity of exponent determines the unit digit.
- ✓- Digits 2, 3, 7, 8 have cycles of length 4: use n mod 4, mapping 0 → position 4.
- ✓- For multi-term expressions, find each term's unit digit, then combine using the operation.
- ✓- The cycles for 2 and 8 are reverses of each other: (2,4,8,6) vs (8,4,2,6).
- ✓- The cycles for 3 and 7 are reverses of each other: (3,9,7,1) vs (7,9,3,1).
- ✓- Memorise cycles as 4-digit numbers: 2486, 3971, 7931, 8426.
"Zero Is Sure, Five Stays Six" → 0,1,5,6 never change. For the cycle-4 group: "Two Four Eight Six" for digit 2 (then read 3,7,8 as rotations). Remember: 49 is 4×4's friend (4's cycle is 4,6) and 9's cycle is 9 then 1.
- ✓- Cyclicity is the pattern of repeating unit digits in successive powers of a base digit.
- ✓- Digits 0,1,5,6 have cycle length 1; 4 and 9 have cycle length 2; 2,3,7,8 have cycle length 4.
- ✓- For cycle-4 digits, compute n mod 4 to find the cycle position (0 maps to position 4).
- ✓- For cycle-2 digits (4 and 9), check only whether the exponent is odd or even.
- ✓- To find the unit digit of a product or sum of powers, find each unit digit separately first.
- ✓- Memorise the four 4-digit cycles: 2→2486, 3→3971, 7→7931, 8→8426.
Worked Example: Unit Digit and Series
The best way to lock in a concept is to watch it solve a real problem — these worked examples stitch together unit digit cyclicity and odd-number series into the exact question formats SSC and RRB examiners favour.
Problem Set A: Unit Digit of a Large Power
Worked Example 1
Question: Find the unit digit of 7⁹⁵.
Solution:
Step 1: Identify the base's unit digit. The base is 7 — unit digit is 7.
Step 2: Recall the cycle for 7. Powers of 7 produce unit digits: 7, 9, 3, 1, 7, 9, 3, 1, … The cycle length is 4.
Step 3: Find the position in the cycle. Compute 95 mod 4.
95 = 4 × 23 + 3, so 95 mod 4 = 3.
Step 4: Read position 3 from the cycle (7, 9, 3, 1). Unit digit = 3.
Conclusion: The unit digit of 7⁹⁵ is 3.
Why this works: The unit digit of 7ⁿ is entirely determined by where n falls in the 4-step repeating cycle. After every 4 steps the pattern resets. 95 = 92 (divisible by 4) + 3, so 7⁹⁵ has the same unit digit as 7³ = 343, which ends in 3.
Worked Example 2
Question: Find the unit digit of 8¹⁰².
Solution:
Step 1: Cycle for 8 is (8, 4, 2, 6), length 4.
Step 2: 102 mod 4 = 2 (since 102 = 4×25 + 2).
Step 3: Position 2 in cycle (8, 4, 2, 6) → unit digit = 4.
Conclusion: The unit digit of 8¹⁰² is 4.
Worked Example 3 (Combined Expression)
Question: Find the unit digit of 3⁸⁴ + 6⁷⁵.
Solution:
Step 1 (3⁸⁴): Cycle for 3 is (3, 9, 7, 1), length 4. 84 mod 4 = 0 → position 4 → unit digit of 3⁸⁴ = 1.
Step 2 (6⁷⁵): Digit 6 always has unit digit 6 regardless of power.
Step 3: Unit digit of sum = unit digit of (1 + 6) = 7.
Conclusion: The unit digit of 3⁸⁴ + 6⁷⁵ is 7.
Common misconception corrected: When adding two large powers, you cannot add the full numbers — but you CAN add their unit digits to get the unit digit of the sum (as long as you look only at the ones place of the result). This is valid because (…a) + (…b) always ends in the unit digit of a+b.
Problem Set B: Sum of an Odd Number Series
Worked Example 4
Question: Find the sum 1 + 3 + 5 + 7 + … + 99.
Solution:
Step 1: Confirm this is a sum of consecutive odd numbers starting at 1.
Step 2: Count the number of terms. The kth odd number is 2k − 1. Set 2k − 1 = 99:
2k = 100 → k = 50.
So there are 50 terms.
Step 3: Apply the formula: Sum of first n odd numbers = n².
Sum = 50² = 2500.
Conclusion: 1 + 3 + 5 + 7 + … + 99 = 2500.
Why n² works: Pair the terms from both ends: (1 + 99) = 100, (3 + 97) = 100, … There are 25 such pairs, total = 2500. This confirms n² = 50² = 2500.
Real-world Indian analogy: Imagine laying tiles in a square courtyard. Row 1 has 1 tile, row 2 has 3 tiles (the L-shaped border), row 3 adds 5 more, and so on. After n rows the total is always n² tiles — the sum of first n odd numbers literally builds a square.
Worked Example 5 (Odd numbers not starting from 1)
Question: Find the sum of all odd numbers from 21 to 51.
Solution:
Step 1: The series is 21, 23, 25, ..., 51. This is an AP with a = 21, d = 2.
Step 2: Number of terms: n = (51 − 21)/2 + 1 = 30/2 + 1 = 16.
Step 3: Sum = (n/2) × (first + last) = (16/2) × (21 + 51) = 8 × 72 = 576.
Conclusion: Sum = 576.
Alternative method: Sum of odd numbers up to 51 = 26² = 676 (there are 26 odd numbers from 1 to 51). Sum of odd numbers up to 19 = 10² = 100 (there are 10 odd numbers from 1 to 19). Difference = 676 − 100 = 576. ✓
Problem Set C: Combined Quick-Fire Questions
Worked Example 6
Question: What is the unit digit of (3 × 7)⁴⁸?
Solution:
Step 1: 3 × 7 = 21, unit digit = 1.
Step 2: Digit 1 raised to any power always gives unit digit 1.
Conclusion: Unit digit = 1.
Shortcut: You don't even need to compute 3 × 7 fully; just note that the unit digit of 21 is 1, and 1ⁿ = 1 always.
Worked Example 7
Question: The sum of the first 20 odd numbers is how much more than the sum of the first 15 odd numbers?
Solution:
Step 1: Sum of first 20 odd numbers = 20² = 400.
Step 2: Sum of first 15 odd numbers = 15² = 225.
Step 3: Difference = 400 − 225 = 175.
Conclusion: The sum of the first 20 odd numbers exceeds the sum of the first 15 odd numbers by 175.
| Task | Direct computation | Unit digit method |
|---|---|---|
| 7⁹⁵ | Compute 7⁹⁵ — impossibly large | 95 mod 4 = 3 → cycle pos 3 → 3 |
| 3⁸⁴ + 6⁷⁵ | Impossible without a calculator | 1 + 6 = 7 (unit digit) |
| Sum 1+3+…+99 | Add 50 terms manually | n=50, 50²=2500 |
- ✓- The unit digit of 7ⁿ is found by n mod 4; the cycle is (7, 9, 3, 1).
- ✓- When n mod 4 = 0, always use position 4, not position 0.
- ✓- For combined expressions (sum/product), find each term's unit digit, then combine.
- ✓- Sum of the first n odd numbers = n²; to identify n, solve 2n−1 = last term.
- ✓- For odd-number series not starting at 1, use the AP formula S = (n/2)(first + last).
- ✓- Digit 6 always produces unit digit 6; digit 1 always produces unit digit 1.
- ✓- The "square pairing" technique (matching first with last) independently confirms n².
- ✓- Each formula is fast — the key is instant recall of the cycle or the formula.
"Seven Nines Three One" — the cycle for 7 as a spoken chant. For odd-number sums: "Count, Square, Done" — count how many odd numbers, square that count.
- ✓- Unit digit of 7⁹⁵: cycle (7,9,3,1), 95 mod 4 = 3, answer = 3.
- ✓- For digits 2, 3, 7, 8: compute n mod 4; map 0 to position 4 in the 4-step cycle.
- ✓- Sum 1+3+5+…+(2n−1) = n²; find n by solving 2n−1 = last odd term.
- ✓- For odd sums not starting at 1: use the AP sum formula (n/2)(first+last).
- ✓- Combined unit digit problems: evaluate each component separately, then combine.
- ✓- Digit 6 (and 1, 5, 0) never change their unit digit — instant answer for bases ending in these.
⚡ Speed Tricks & Shortcuts
- Sum 1..n = n(n+1)/2; sum of first n odd = n²; first n even = n(n+1).
- Sum of squares = n(n+1)(2n+1)/6; sum of cubes = [n(n+1)/2]².
- AP: nth term = a+(n−1)d; sum = (n/2)(first+last).
- GP: nth term = a·r^(n−1); spot a constant ratio to identify a GP fast.
Off-by-one in the term count: "how many terms from 7 to 47 step 5" = (47−7)/5 + 1 = 9, not 8.
Number Patterns and Series — Revision Notes
Quick-revision notes for Number Patterns and Series — the must-know points for SSC CGL Tier-I/II.
- Number series follow a rule based on differences, ratios, or algebraic patterns.
- Arithmetic progression (AP): constant difference; nth term = a + (n−1)d.
- Geometric progression (GP): constant ratio; nth term = a·r^(n−1).
- Sum of first n natural numbers = n(n+1)/2; sum of squares = n(n+1)(2n+1)/6; sum of cubes = [n(n+1)/2]².
- Look for squares, cubes, primes, or alternate patterns in mixed series.
- For "find the wrong term", locate the single value breaking the rule.
Number Patterns and Series — Flashcards (SSC CGL)
Cover the answer, recall, then check. 7 cards on the must-know Number Patterns and Series facts for SSC CGL.
Q1. Formula for the nth term of an AP.
A1. a + (n − 1)d, where a is the first term and d the common difference.
Q2. Sum of the first n natural numbers?
A2. n(n + 1)/2.
Q3. nth term of a GP with first term a and ratio r?
A3. a · r^(n−1).
Q4. Sum of the squares of the first n natural numbers?
A4. n(n + 1)(2n + 1)/6.
Q5. Find the next term: 2, 4, 8, 16, ?
A5. 32 — a GP with ratio 2.
Q6. Sum of the first 10 natural numbers?
A6. 55 — 10×11/2 = 55.
Q7. In a series 5, 8, 11, 14, what is the common difference?
A7. 3 (an arithmetic progression).