📚 IBPS PO – Reasoning Ability: Series
Series problems are staple in the IBPS PO exam.
They test pattern recognition, arithmetic/geometry, and quick mental math.
1. Simple Explanation
A series is a list of numbers written in a particular order.
In exam questions you will usually be:
- Asked to find the next term(s)
- Asked to fill in a missing term
- Asked to determine the kind of series (AP, GP, Fibonacci, etc.)
Key Idea:
Identify the rule that generates the sequence and apply it to the required position.
2. Key Points – Types of Series & Tricks
| Type | Typical Pattern | Formula / Trick | Quick Check |
|---|---|---|---|
| Arithmetic Progression (AP) | Constant difference d |
a_n = a_1 + (n-1)d |
a₂ - a₁ = d |
| Geometric Progression (GP) | Constant ratio r |
a_n = a_1 * r^(n-1) |
a₂ / a₁ = r |
| Fibonacci (or similar) | Each term = sum of previous two | F_n = F_{n-1} + F_{n-2} |
Check two preceding terms |
| Alternating Series | Two sub‑sequences interleaved | Treat as two separate AP/GP | Separate odd & even indices |
| Prime/Composite Pattern | Uses prime numbers / squares | Recognize sequence of primes / squares | Write next prime / square |
| Digit/Letter Pattern | Use digits of numbers or letters | Convert to numbers (A=1, B=2…) | Check incremental change |
| Sum/Multiple Pattern | Each term = sum/multiple of previous | a_n = a_{n-1} + k or = k*a_{n-1} |
Check ratio or difference |
General Tricks
- Look at the first 3–4 terms – they usually reveal the rule.
- Check differences – if constant → AP; if varying → maybe GP or Fibonacci.
- Check ratios – constant → GP.
- Look for square/cube patterns – e.g., 1, 4, 9, 16 → (n^2).
- Check for alternating patterns – split into two subsequences.
- Use “sum of digits” or “product of digits” for digit‑based series.
- Remember special sequences:
1, 2, 4, 8, 16…→ powers of 2 (GP, r=2)1, 1, 2, 3, 5, 8…→ Fibonacci2, 4, 8, 16…→ powers of 2 (GP, r=2)1, 2, 3, 5, 8, 13…→ Fibonacci with first 2 terms 1,2
3. Worked Examples
Example 1 – Arithmetic Progression
Series: 3, 7, 11, 15, …
Find: 8th term.
| Step | Calculation |
|---|---|
| 1 | First term (a_1 = 3) |
| 2 | Common difference (d = 7-3 = 4) |
| 3 | (a_8 = a_1 + (8-1)d = 3 + 7*4 = 31) |
Answer: 31
Example 2 – Geometric Progression
Series: 5, 15, 45, 135, …
Find: 6th term.
| Step | Calculation |
|---|---|
| 1 | First term (a_1 = 5) |
| 2 | Common ratio (r = 15/5 = 3) |
| 3 | (a_6 = a_1 * r^(6-1) = 5 * 3^5 = 5 * 243 = 1215) |
Answer: 1215
Example 3 – Fibonacci‑like
Series: 2, 5, 7, 12, 19, …
Find: 8th term.
| Step | Calculation |
|---|---|
| 1 | Recognize a_n = a_{n-1} + a_{n-2} |
| 2 | Write terms: |
| (a_1=2, a_2=5, a_3=7, a_4=12, a_5=19) | |
| 3 | Compute next terms: |
| (a_6 = 19+12 = 31) | |
| (a_7 = 31+19 = 50) | |
| (a_8 = 50+31 = 81) |
Answer: 81
Example 4 – Alternating Series
Series: 4, 8, 12, 20, 24, 32, …
Find: 7th term.
| Step | Observation |
|---|---|
| 1 | Odd positions: 4, 12, 24 → AP with (d = 8) |
| 2 | Even positions: 8, 20, 32 → AP with (d = 12) |
| 3 | 7th term is odd position → use odd sequence formula: |
| (a_{odd, n} = 4 + (n-1)*8) where (n=\frac{7+1}{2}=4). | |
| 4 | (a_7 = 4 + (4-1)*8 = 4 + 24 = 28) |
Answer: 28
Example 5 – Digit Sum Pattern
Series: 5, 10, 19, 28, …
Find: 5th term.
| Step | Observation |
|---|---|
| 1 | Notice difference sequence |