📚 Reasoning Ability – Series (Uttarakhand Police SI Exam)
Why series?
A “Series” question tests your ability to spot patterns, apply algebraic identities, and perform quick mental calculations. In competitive exams (IIT‑JEE, UPSC, Police SI, etc.) they are the most frequent and often the hardest type of reasoning problem.
1. Quick Reference: Common Series Types
| Type | General Form | Key Characteristic | Quick Test |
|---|---|---|---|
| Arithmetic | (a, a+d, a+2d, \ldots) | Constant difference (d) | Find two consecutive terms → difference |
| Geometric | (a, ar, ar^2, \ldots) | Constant ratio (r) | Multiply/divide two consecutive terms |
| Fibonacci‑type | (a, b, a+b, a+2b, 2a+3b, \ldots) | Each term = sum of previous two | Add previous two terms |
| Alternating | (a, -a, a, -a, \ldots) | Sign alternates | Check sign pattern |
| Mixed | Combination of above (e.g., (a, a+d, a+2d, a+3d), then start geometric) | Two or more subsequences | Split into subsequences |
| Square/Cube | (1, 4, 9, 16, 25,\ldots) | (n^2) or (n^3) | Recognise perfect powers |
| Prime‑based | (2, 3, 5, 7, 11,\ldots) | Prime numbers | Verify primality |
| Other | e.g., (1, 2, 4, 7, 11,\ldots) (triangular + 1) | Combination of arithmetic & other pattern | Decompose into simple sequences |
Tip: Always start by looking for differences or ratios; if none, try splitting into 2‑term groups.
2. Detailed Study
2.1 Arithmetic Series
| Property | Formula / Derivation |
|---|---|
| (n^{th}) term: (a_n = a + (n-1)d) | By definition: first term (a), add (d) each time |
| Sum of first (n) terms: (S_n = \frac{n}{2}(2a + (n-1)d)) | Derived by pairing terms: (S_n = a_n + a_1 = 2a + (n-1)d), etc. |
| Sum to infinity | Does not exist if (d \neq 0). Only for (d=0) (constant series) → (S_{\infty}=na). |
Tricky point:
- Negative difference → sequence decreases but still arithmetic.
- Large (n) → mental compute: use formula, avoid writing all terms.
2.2 Geometric Series
| Property | Formula / Derivation |
|---|---|
| (n^{th}) term: (a_n = ar^{,n-1}) | Multiply by (r) each step. |
| Sum of first (n) terms: (S_n = a \frac{1-r^n}{1-r}) for (r \neq 1) | Multiply series by (r) and subtract. |
| Sum to infinity: (S_\infty = \frac{a}{1-r}) if ( | r |
Tricky point:
- Negative ratio → signs alternate.
- Fractional ratio → mental power calculation using exponent rules.
2.3 Fibonacci‑type (Recursive) Series
| Property | Formula / Derivation |
|---|---|
| Recurrence: (F_n = F_{n-1} + F_{n-2}) | Given two initial terms. |
| Closed form (Binet's formula): (F_n = \frac{\phi^n - \psi^n}{\sqrt5}) where (\phi=\frac{1+\sqrt5}{2}), (\psi=\frac{1-\sqrt5}{2}) | Derived from solving the characteristic equation. |
| Sum of first (n) terms: (S_n = F_{n+2}-1) | Because (F_1+F_2+\dots+F_n = F_{n+2}-1). |
Tricky point:
- Non‑standard initial terms: e.g., (2, 3, 5, 8,\ldots). Subtract the first two terms from the standard Fibonacci to adjust.
- Large indices → use Binet’s formula or modular arithmetic for quick mental calculation.
2.4 Alternating Series
Often of the form ((-1)^{n+1} a_n).
- If (|