Official Syllabus Item: 39, 41 Data & Number Patterns

Data Handling, Pictographs, Tally Charts & Number Patterns

Master recording and organization of data, frequency tables, tally marks, pictographs with scale keys, bar graphs, arithmetic/geometric number patterns, triangular and square numbers for RIMC Maths 2026.

Overview & Fundamental Concept

Data Handling, Pictographs, and Mathematical Patterns form an essential analytical component of the RIMC Mathematics Paper (Paper I, 200 Marks, 90 Minutes). Prescribed under Serial Numbers 39 ('Recording & Organization of Data Pictograph & its Interpretation') and 41 ('Patterns in Mathematics') of Chapter 8 in the official brochure, these modules test a candidate's organizational precision and inductive reasoning across Part A (5 Marks) and Part B (10 Marks). RIMC examiners look for strict descriptive rigor: neatly ruled frequency tables featuring 5-bar bundled tally marks (four vertical strokes crossed diagonally by the fifth), explicit statements of pictographic scale keys (e.g., 'Key: 1 symbol represents 20 items'), drawing and interpreting fractional half-symbols, and deducing algebraic $n$-th term formulas for linear, quadratic, triangular ($T_n = \frac{n(n+1)}{2}$), and square number sequences.

Official Syllabus Coverage Breakdown

Syllabus Item	Prescribed Scope	Exam Relevance & Pedagogical Objective
Item 39	Recording & Organization of Data: Raw data tabulation, tally charts, frequency distributions, and pictograph construction/interpretation.	Tests systematic data synthesis, scale selection, and tabular presentation in 5-mark and 10-mark descriptive formats.
Item 41	Patterns in Mathematics: Arithmetic/geometric sequences, geometric matchstick patterns, square number identities, and triangular numbers.	Directly specified in syllabus item 41. Assesses inductive pattern discovery, difference arrays, and $n$-th term formula derivation.

1. Data Collection, Organization & Frequency Tables

In statistics, data is a collection of numerical facts, measurements, or observations gathered for a specific purpose:

• Raw Data: The initial collection of observations recorded in haphazard, unorganized sequence.
• Array: Raw numerical data arranged in ascending or descending order of magnitude.
• Frequency ($f$): The number of times a particular observation occurs in the dataset.
• Range: The difference between the highest (maximum) and lowest (minimum) observation in the dataset: $\text{Range} = X_{\max} - X_{\min}$.

The Standard Tally Mark Convention:
To record frequencies systematically from raw data, tally marks are grouped in bundles of five:
• Numbers $1$ to $4$ are drawn as vertical strokes: $|, ||, |||, ||||$.
• The fifth count is drawn as a diagonal slash crossing the first four strokes from top-right to bottom-left: $\cancel{||||}$ (denoted as a complete bundle of $5$).
• Count $7$ is written as $\cancel{||||} \; ||$; count $12$ is written as $\cancel{||||} \; \cancel{||||} \; ||$.

2. Pictographs: Construction & Interpretation

A pictograph represents statistical data through standardized pictures, icons, or symbols of objects. It conveys qualitative and quantitative relationships at a glance.

Rules for Constructing a Descriptive RIMC Pictograph:
1. Choosing an Optimal Scale Key: Find the Highest Common Factor (HCF) or a convenient round divisor of the frequencies to serve as the unit value of one symbol (e.g., '$\text{Let } 1 \odot = 10 \text{ cycles}$'). State this scale key prominently above the table.
2. Handling Fractional Values: When an observation is not an exact multiple of the scale key, draw fractional symbols. For instance, if $1 \odot = 20\text{ items}$, then half a symbol represents $10\text{ items}$, and a quarter symbol represents $5\text{ items}$.
3. Ruled Table Layout: Each category must have its own horizontal row with uniformly spaced icons of identical size.

3. Bar Graphs & Comparative Data Representations

A bar graph (or column chart) represents numerical data using rectangular bars of uniform width drawn vertically or horizontally with equal spacing between them:

• Key Principles:
1. The width of all bars must be strictly equal (width has no numerical significance).
2. The gap between consecutive bars must remain uniform throughout.
3. The height (or length) of each bar is directly proportional to the corresponding numerical frequency.
4. Both axes must be labeled with clear titles and a declared scale along the frequency axis (e.g., '$1\text{ cm} = 5\text{ units}$').

4. Mathematical Patterns & Number Sequences

Inductive reasoning in RIMC focuses on discovering mathematical rules governing sequences and spatial figures:

1. Arithmetic Sequences (Linear Patterns):
A sequence where the difference between consecutive terms is constant ($d$):
$$T_n = a + (n - 1)d$$where $a$ is the first term, $d$ is the common difference, and $n$ is the term index.
$$\text{Sum of first } n \text{ terms: } S_n = \frac{n}{2}[2a + (n - 1)d] = \frac{n}{2}(a + l)$$

2. Triangular Numbers:
Numbers that can be arranged in an equilateral triangular dot pattern: $1, 3, 6, 10, 15, 21, 28...$
$$T_n = \frac{n(n + 1)}{2} = \sum_{k=1}^n k = 1 + 2 + 3 + \dots + n$$Fundamental Theorem: The sum of any two consecutive triangular numbers is always a perfect square number:
$$T_{n-1} + T_n = \frac{(n-1)n}{2} + \frac{n(n+1)}{2} = \frac{n(n-1 + n+1)}{2} = \frac{n(2n)}{2} = n^2$$

3. Square Numbers & Odd Number Sums:
The $n$-th square number equals the sum of the first $n$ consecutive odd natural numbers:
$$n^2 = 1 + 3 + 5 + 7 + \dots + (2n - 1) = \sum_{k=1}^n (2k - 1)$$$$\text{Difference between consecutive squares: } (n + 1)^2 - n^2 = 2n + 1$$

4. Method of Finite Differences:
To find the rule of any unknown numerical pattern, write down the successive differences between terms:
• If the 1st differences are constant $\implies$ the rule is Linear: $T_n = an + b$.
• If the 2nd differences are constant $\implies$ the rule is Quadratic: $T_n = an^2 + bn + c$.

Illustrated Conceptual Diagrams & Visual Schematics

Descriptive RIMC Pictograph Chart with Standard Scale Key

Examiner Marking Standard: Always declare the scale key prominently at the top right. Draw full symbols for whole values and half-symbols for fractional values. Use neat rows with house/cadet labels.

Topper Strategy & Descriptive Step Breakdown

Mastering RIMC Mathematics Standard

In RIMC, clear step-by-step layout is essential. Write down all given values, declare formulas before substitution, and conclude with full sentences with proper units.

Official Descriptive Marking Scheme & Step Allocation

In RIMC Stage-I, answers without step justification receive only 20-30% of total marks even if the numeric answer is correct. Follow this strict 4-tier layout on every question:

Step 1: Systematic Organization & Labeled Tables 25% of marks

For data handling, construct a clean, ruled 3-column table: 'Observation / Category', 'Tally Marks', and 'Frequency ($f$)'. For sequences, write out the terms in tabular order with index $n$.

Step 2: Explicit Statement of Scale Keys / Pattern Differences 25% of marks

In pictographs, declare the scale key explicitly (e.g. 'Key: 1 ⚑ = 25 students'). In pattern problems, compute and exhibit the first and second difference arrays to justify the algebraic model.

Step 3: Algebraic Derivation of $n$-th Term 30% of marks

Derive the general formula $T_n$ algebraically (e.g., $T_n = a + (n-1)d$ or $T_n = n^2 + 1$) before substituting large values of $n$ (like $n = 50$ or $n = 100$).

Step 4: Analytical Interpretation & Boxed Conclusions 20% of marks

Answer each sub-question in a complete grammatical sentence with correct units. Box the final numbers and verify the total frequency sum $\sum f = N$.

Examiner Traps & Common Blunders (Marks Penalties)

Omitting the Pictograph Scale Key Loss of 2 to 3 marks out of 5 marks for uncalibrated representation.

Examiner Trap: Drawing icons without specifying the numerical value of one icon above the pictograph.

Incorrect Tally Mark Bundling 1 to 2 marks deducted for non-standard statistical convention.

Examiner Trap: Drawing five parallel vertical strokes without the fifth diagonal slash, or grouping in bundles other than 5.

Off-by-One Error in Pattern Terms Severe: invalidates all subsequent numerical predictions, losing 4 to 6 marks.

Examiner Trap: Writing $T_n = an + d$ instead of $a + (n-1)d$, causing all evaluated terms to be shifted by one index.

Confusing Term Index $n$ with Term Value $T_n$ Fatal algebraic error resulting in 0 marks for that sub-part.

Examiner Trap: Substituting a given large number as the index $n$ rather than setting $T_n = \text{given number}$ to solve for $n$.

Non-Uniform Bar Graph Spacing 2 marks deducted by RIMC examiners for sloppy graphical geometry.

Examiner Trap: Drawing bars with varying widths or unequal gaps on graph paper.

Cadet Topper Mental Verification Shortcuts

Difference Array Quadratic Rule Shortcut

If second difference is constant and equals $2A$, then the coefficient of $n^2$ in $T_n = An^2 + Bn + C$ is simply $A = \frac{\text{2nd difference}}{2}$. This deduces quadratic formulas in 10 seconds!

Consecutive Triangular Sum Property

Remember $T_{n-1} + T_n = n^2$. For example, $T_4 = 10$ and $T_5 = 15 \implies 10 + 15 = 25 = 5^2$. This bypasses tedious summations.

Odd Number Sum Shortcut

The sum of the first $n$ odd numbers is always $n^2$. Example: $1 + 3 + 5 + \dots + 39$: here last term is $2n - 1 = 39 \implies 2n = 40 \implies n = 20$. Sum is $20^2 = 400$ instantly!

Scale Factor HCF Rule for Pictographs

To choose the perfect pictograph key, compute $\gcd$ of all given data values. If frequencies are $40, 60, 90, 70$, $\gcd = 10$, so set $1 \text{ icon} = 10\text{ units}$.

Key Formulas & Conversion Cheat Sheet

Range of Dataset

$$\text{Range} = X_{\max} - X_{\min}$$

Difference between extreme observations.

Arithmetic Progression General Term

$$T_n = a + (n - 1)d$$

$a$ is first term, $d$ is common difference ($T_2 - T_1$), $n$ is term index.

Sum of First $n$ Terms of AP

$$S_n = \frac{n}{2}[2a + (n - 1)d] = \frac{n}{2}(a + l)$$

$l$ is the last term.

Triangular Number Formula

$$T_n = \frac{n(n + 1)}{2} = 1 + 2 + 3 + \dots + n$$

Sequence: $1, 3, 6, 10, 15, 21, 28...$

Consecutive Triangular Sum Identity

$$T_{n-1} + T_n = n^2$$

The sum of any two adjacent triangular numbers is a perfect square.

Sum of First $n$ Odd Numbers

$$\sum_{k=1}^n (2k - 1) = 1 + 3 + 5 + \dots + (2n - 1) = n^2$$

Every square number $n^2$ is the sum of the first $n$ odd integers.

Solved Model Examples (Descriptive Format with Step Mark Distribution)

Part A - Descriptive Frequency Table 5 Marks

The marks obtained by $30$ cadets in a Mathematics test (out of $10$) are recorded as follows:
$$6, 7, 5, 8, 9, 6, 7, 8, 4, 5, 8, 9, 7, 6, 8, 5, 7, 8, 9, 7, 6, 5, 8, 7, 6, 8, 9, 7, 8, 6$$
(i) Construct a frequency distribution table using tally marks.
(ii) Find the range of marks.
(iii) How many cadets scored $8$ or more marks?

Step-by-Step Model Solution:

Step 1: (i) Construct the Frequency Distribution Table
Organize the raw data into a 3-column table showing Marks ($x$), Tally Marks, and Frequency ($f$):
$$\begin{array}{|c|c|c|} \hline \textbf{Marks } (x) & \textbf{Tally Marks} & \textbf{Frequency } (f) \\ \hline 4 & | & 1 \\ 5 & |||| & 4 \\ 6 & \cancel{||||} \; | & 6 \\ 7 & \cancel{||||} \; || & 7 \\ 8 & \cancel{||||} \; ||| & 8 \\ 9 & |||| & 4 \\ \hline \textbf{Total} & & \mathbf{N = 30} \\ \hline \end{array}$$
Step 2: (ii) Calculate the Range of Marks
$$\text{Highest Mark } (X_{\max}) = 9$$ $$\text{Lowest Mark } (X_{\min}) = 4$$ $$\text{Range} = X_{\max} - X_{\min} = 9 - 4 = 5\text{ marks}$$
$$\mathbf{\text{Range of marks } = 5.}$$
Step 3: (iii) Count Cadets Scoring $8$ or More Marks
Cadets scoring $8$ or more marks are those who scored $8$ marks or $9$ marks:
$$\text{Number of Cadets} = f(8) + f(9) = 8 + 4 = 12\text{ cadets}$$
$$\mathbf{\text{Conclusion: (i) Table constructed; } (ii) \text{ Range } = 5; \; (iii) \; 12\text{ cadets scored } 8 \text{ or more marks.}}$$

Rough Column Working:
Marks count:
4: 1
5: 4
6: 6
7: 7
8: 8
9: 4
Sum = 1+4+6+7+8+4 = 30 (Verified!)
Range = 9 - 4 = 5
Score >= 8: 8 + 4 = 12.

Step Mark Allocation:

Drawing neat 3-column table with correct tally mark bundling: 2 Marks
Verifying total frequency sum N = 30: 1 Marks
Calculating range = 9 - 4 = 5: 1 Marks
Computing cadets scoring >= 8 (12 cadets) and boxed answer: 1 Marks

Examiner Scoring Key: Cross tally marks diagonally on the fifth count. Always sum the frequency column to verify N = 30.

Part A - Pictograph Construction 5 Marks

The number of bicycles sold by a dealership over five days of a week is recorded below:
$$\begin{array}{|c|c|c|c|c|} \hline \textbf{Monday} & \textbf{Tuesday} & \textbf{Wednesday} & \textbf{Thursday} & \textbf{Friday} \\ \hline 30 & 45 & 20 & 35 & 50 \\ \hline \end{array}$$
Construct a pictograph to represent this data. State your scale key explicitly and explain how fractional symbols are handled.

Step-by-Step Model Solution:

Step 1: Determine an Appropriate Scale Key
Observe the numbers: $30, 45, 20, 35, 50$.
The Highest Common Factor (HCF) of these values is $\gcd(30, 45, 20, 35, 50) = 5$, but choosing $10$ as the base scale provides an elegant, compact pictograph with half-symbols.
$$\mathbf{\text{Scale Key: } \text{Let } 1 \; \otimes = 10 \text{ bicycles, and } \frac{1}{2} \; \otimes = 5 \text{ bicycles.}}$$
Step 2: Determine Symbol Count for Each Day
• Monday ($30$ bicycles): $\frac{30}{10} = 3$ full symbols $\implies \otimes \; \otimes \; \otimes$
• Tuesday ($45$ bicycles): $\frac{45}{10} = 4.5 \implies 4$ full symbols and $1$ half-symbol $\implies \otimes \; \otimes \; \otimes \; \otimes \; (\frac{1}{2}\otimes)$
• Wednesday ($20$ bicycles): $\frac{20}{10} = 2$ full symbols $\implies \otimes \; \otimes$
• Thursday ($35$ bicycles): $\frac{35}{10} = 3.5 \implies 3$ full symbols and $1$ half-symbol $\implies \otimes \; \otimes \; \otimes \; (\frac{1}{2}\otimes)$
• Friday ($50$ bicycles): $\frac{50}{10} = 5$ full symbols $\implies \otimes \; \otimes \; \otimes \; \otimes \; \otimes$

Step 3: Pictograph Representation
$$\begin{array}{|l|l|c|} \hline \textbf{Day} & \textbf{Pictographic Representation} & \textbf{Sales} \\ \hline \text{Monday} & \otimes \; \otimes \; \otimes & 30 \\ \text{Tuesday} & \otimes \; \otimes \; \otimes \; \otimes \; [\text{Half}] & 45 \\ \text{Wednesday} & \otimes \; \otimes & 20 \\ \text{Thursday} & \otimes \; \otimes \; \otimes \; [\text{Half}] & 35 \\ \text{Friday} & \otimes \; \otimes \; \otimes \; \otimes \; \otimes & 50 \\ \hline \end{array}$$
$$\mathbf{\text{Total Bicycles Sold} = 30 + 45 + 20 + 35 + 50 = 180 \text{ bicycles.}}$$

Rough Column Working:
Scale: 1 symbol = 10 cycles
Mon: 3
Tue: 4.5
Wed: 2
Thu: 3.5
Fri: 5
Total = 18 symbols = 180 cycles.

Step Mark Allocation:

Explicit declaration of scale key (1 symbol = 10, half = 5): 1.5 Marks
Correct calculation of symbol counts for all 5 days: 1.5 Marks
Neat, ruled pictograph layout with uniform symbols: 1 Marks
Total sales verification (180 bicycles) and boxed statement: 1 Marks

Examiner Scoring Key: Stating the scale key at the top is worth 1.5 marks. Never draw a pictograph without an explicitly declared scale.

Part B - Matchstick Geometric Pattern 10 Marks

Cadets form regular hexagons using matchsticks in a connected chain where adjacent hexagons share a common matchstick:
• $1$ hexagon requires $6$ matchsticks.
• $2$ connected hexagons require $11$ matchsticks.
• $3$ connected hexagons require $16$ matchsticks.
(i) Write down the number of matchsticks for $1, 2, 3, 4,$ and $5$ connected hexagons in a table.
(ii) Find the general algebraic formula for the number of matchsticks $M$ required for $n$ connected hexagons.
(iii) How many matchsticks are required to construct a chain of $50$ connected hexagons?
(iv) If a box contains $241$ matchsticks, how many complete connected hexagons can be formed?

Step-by-Step Model Solution:

Step 1: (i) Analyze Pattern Differences and Tabulate
Observe the given terms:
$$T_1 = 6, \quad T_2 = 11, \quad T_3 = 16$$
Compute the first differences between consecutive terms:
$$T_2 - T_1 = 11 - 6 = 5$$ $$T_3 - T_2 = 16 - 11 = 5$$
Since the first difference is constant ($d = 5$), the sequence is an Arithmetic Progression with common difference $d = 5$.
Continuing the sequence:
$$T_4 = 16 + 5 = 21$$ $$T_5 = 21 + 5 = 26$$
$$\begin{array}{|l|c|c|c|c|c|} \hline \textbf{Hexagons } (n) & 1 & 2 & 3 & 4 & 5 \\ \hline \textbf{Matchsticks } (M) & 6 & 11 & 16 & 21 & 26 \\ \hline \end{array}$$
Step 2: (ii) Derive the General Algebraic Formula ($M$)
For an arithmetic sequence with first term $a = 6$ and common difference $d = 5$:
$$M = a + (n - 1)d$$ $$M = 6 + (n - 1) \times 5$$ $$M = 6 + 5n - 5$$ $$M = 5n + 1$$
Geometric Interpretation: The first hexagon requires $1 + 5 = 6$ matchsticks. Each subsequent connected hexagon shares one existing boundary side and adds $5$ new matchsticks. Hence, $M = 5n + 1$.
$$\mathbf{\text{General Formula: } M = 5n + 1.}$$
Step 3: (iii) Calculate Matchsticks for $n = 50$ Hexagons
Substituting $n = 50$ into the derived formula:
$$M_{50} = 5(50) + 1 = 250 + 1 = 251\text{ matchsticks}$$
$$\mathbf{\text{A chain of } 50 \text{ hexagons requires } 251\text{ matchsticks.}}$$
Step 4: (iv) Determine Hexagons Formed with $241$ Matchsticks
Given $M = 241$. Set up the linear equation:
$$5n + 1 = 241$$ $$5n = 241 - 1$$ $$5n = 240$$ $$n = \frac{240}{5} = 48$$
$$\mathbf{\text{Conclusion: (i) Table completed; } (ii) \; M = 5n + 1; \; (iii) \; 251\text{ matchsticks}; \; (iv) \; 48\text{ complete hexagons.}}$$

Rough Column Working:
Diff = 5
a = 6
M = 6 + 5(n-1) = 5n + 1
n = 50 => 5(50) + 1 = 251
5n + 1 = 241 => 5n = 240 => n = 48.

Step Mark Allocation:

Table of values for n = 1 to 5 showing constant difference d = 5: 2 Marks
Derivation of general formula M = 5n + 1 with geometric explanation: 3 Marks
Evaluating M for n = 50 to get 251 matchsticks: 2.5 Marks
Solving 5n + 1 = 241 to get n = 48 hexagons and boxed conclusion: 2.5 Marks

Examiner Scoring Key: Explaining the geometric intuition (1 shared side + 5 new sides per hexagon) demonstrates conceptual mastery and earns top marks.

Part B - Triangular & Square Number Proof 10 Marks

(a) Prove algebraically that the sum of any two consecutive triangular numbers is a perfect square number.
(b) Find the $12\text{th}$ and $13\text{th}$ triangular numbers, verify your algebraic identity with these values, and express their sum as a square number.
(c) Express $196$ as the sum of consecutive odd natural numbers.

Step-by-Step Model Solution:

Step 1: (a) Algebraic Proof of Consecutive Triangular Sum Theorem
By definition, the $n$-th triangular number $T_n$ is the sum of the first $n$ natural numbers:
$$T_n = \frac{n(n + 1)}{2}$$
The preceding triangular number $T_{n-1}$ is obtained by replacing $n$ with $(n - 1)$:
$$T_{n-1} = \frac{(n - 1)(n - 1 + 1)}{2} = \frac{(n - 1)n}{2}$$
Now, consider the sum of these two consecutive triangular numbers:
$$T_{n-1} + T_n = \frac{(n - 1)n}{2} + \frac{n(n + 1)}{2}$$
Factor out the common term $\frac{n}{2}$:
$$T_{n-1} + T_n = \frac{n}{2} \left[ (n - 1) + (n + 1) \right]$$ $$T_{n-1} + T_n = \frac{n}{2} [2n] = n \times n = n^2$$
$$\mathbf{\text{Hence Proved: } T_{n-1} + T_n = n^2 \text{ (a perfect square number).}}$$
Step 2: (b) Numerical Verification with $n = 13$
Calculate $T_{12}$ and $T_{13}$:
$$T_{12} = \frac{12 \times 13}{2} = 6 \times 13 = 78$$ $$T_{13} = \frac{13 \times 14}{2} = 13 \times 7 = 91$$
Summing the two consecutive triangular numbers:
$$T_{12} + T_{13} = 78 + 91 = 169$$
Since $13^2 = 169$:
$$T_{12} + T_{13} = 13^2 = 169 \quad [\text{Verified}]$$
$$\mathbf{T_{12} = 78, \; T_{13} = 91, \; \text{Sum } = 169 = 13^2.}$$
Step 3: (c) Express $196$ as the Sum of Consecutive Odd Numbers
We know that the square of any natural number $n$ equals the sum of the first $n$ consecutive odd natural numbers:
$$n^2 = 1 + 3 + 5 + 7 + \dots + (2n - 1)$$
Given number is $196$. Finding its square root:
$$\sqrt{196} = 14 \implies n = 14$$
The $14\text{th}$ odd number is $2(14) - 1 = 28 - 1 = 27$.
Therefore, $196$ can be expressed as the sum of the first $14$ odd natural numbers:
$$196 = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 + 25 + 27$$
$$\mathbf{\text{Conclusion: (a) Proved } T_{n-1} + T_n = n^2; \; (b) \; 78 + 91 = 169 = 13^2; \; (c) \; 196 = \sum_{k=1}^{14}(2k-1).}}$$

Rough Column Working:
T_n = n(n+1)/2
T_{n-1} = n(n-1)/2
Sum = n/2 * (n - 1 + n + 1) = n^2
T_12 = 78, T_13 = 91 => 78 + 91 = 169 = 13^2
196 = 14^2 = sum of first 14 odd numbers (1 to 27).

Step Mark Allocation:

Algebraic formulation of Tn and T_{n-1}: 2 Marks
Factoring to complete proof T_{n-1} + Tn = n^2 and stating Hence Proved: 2.5 Marks
Calculating T12 = 78 and T13 = 91 and verifying sum = 169 = 13^2: 2.5 Marks
Identifying n = 14 and writing out sum of first 14 odd numbers for 196: 3 Marks

Examiner Scoring Key: When asked to express a square as the sum of odd numbers, write out all individual terms explicitly.

Part A - Sequence Analysis via Differences 5 Marks

Find the next two terms and the $20\text{th}$ term of the sequence:
$$2, 5, 10, 17, 26, 37, \dots$$

Step-by-Step Model Solution:

Step 1: Examine the Difference Array
Write down the given terms:
$$T_1 = 2, \quad T_2 = 5, \quad T_3 = 10, \quad T_4 = 17, \quad T_5 = 26, \quad T_6 = 37$$
Compute the first differences between consecutive terms:
$$5 - 2 = 3$$ $$10 - 5 = 5$$ $$17 - 10 = 7$$ $$26 - 17 = 9$$ $$37 - 26 = 11$$
The first differences are consecutive odd numbers: $3, 5, 7, 9, 11, \dots$
Compute the second differences:
$$5 - 3 = 2, \quad 7 - 5 = 2, \quad 9 - 7 = 2, \quad 11 - 9 = 2$$
Since the second difference is constant ($2$), the sequence is quadratic in $n$.

Step 2: Calculate the Next Two Terms
• Next difference after $11$ is $13$:
$$T_7 = 37 + 13 = 50$$
• Next difference after $13$ is $15$:
$$T_8 = 50 + 15 = 65$$
Step 3: Deduce the General Rule ($T_n$)
Notice the relationship between each term and perfect squares:
$$T_1 = 1^2 + 1 = 1 + 1 = 2$$ $$T_2 = 2^2 + 1 = 4 + 1 = 5$$ $$T_3 = 3^2 + 1 = 9 + 1 = 10$$ $$T_4 = 4^2 + 1 = 16 + 1 = 17$$ $$T_n = n^2 + 1$$
Step 4: Calculate the $20\text{th}$ Term ($T_{20}$)
Substituting $n = 20$:
$$T_{20} = 20^2 + 1 = 400 + 1 = 401$$
$$\mathbf{\text{Conclusion: Next two terms are } 50 \text{ and } 65; \; \text{The } 20\text{th} \text{ term is } 401.}$$

Rough Column Working:
Terms: 2, 5, 10, 17, 26, 37
Squares: 1, 4, 9, 16, 25, 36
Pattern: n^2 + 1
T_7 = 7^2 + 1 = 50
T_8 = 8^2 + 1 = 65
T_20 = 20^2 + 1 = 401.

Step Mark Allocation:

Showing first and second difference arrays: 1.5 Marks
Finding next two terms (50 and 65): 1 Marks
Deducing formula Tn = n^2 + 1: 1.5 Marks
Computing T20 = 401 and boxed answer: 1 Marks

Examiner Scoring Key: Recognizing n^2 + 1 directly saves minutes of algebraic substitution.

Part B - Advanced Sequence Decomposition 10 Marks

Consider the sequence: $3, 8, 15, 24, 35, 48, \dots$
(i) Find the first and second differences of the sequence.
(ii) Derive the general formula for the $n$-th term $T_n$.
(iii) Find the $25\text{th}$ term.
(iv) Which term of this sequence equals $960$?

Step-by-Step Model Solution:

Step 1: (i) Compute First and Second Differences
Given terms:
$$T_1 = 3, \quad T_2 = 8, \quad T_3 = 15, \quad T_4 = 24, \quad T_5 = 35, \quad T_6 = 48$$
• First Differences ($\Delta_1$):
$$8 - 3 = 5$$ $$15 - 8 = 7$$ $$24 - 15 = 9$$ $$35 - 24 = 11$$ $$48 - 35 = 13$$
The first differences form the sequence of odd numbers: $5, 7, 9, 11, 13, \dots$

• Second Differences ($\Delta_2$):
$$7 - 5 = 2, \quad 9 - 7 = 2, \quad 11 - 9 = 2, \quad 13 - 11 = 2$$
$$\mathbf{\text{First differences are } 5, 7, 9, 11, 13; \; \text{Second difference is constant and equals } 2.}$$
Step 2: (ii) Derive the General Formula ($T_n$)
Observe the relation of each term to consecutive integers and square numbers:
$$T_1 = 1 \times 3 = (2^2 - 1) = 3$$ $$T_2 = 2 \times 4 = (3^2 - 1) = 8$$ $$T_3 = 3 \times 5 = (4^2 - 1) = 15$$ $$T_4 = 4 \times 6 = (5^2 - 1) = 24$$ $$T_5 = 5 \times 7 = (6^2 - 1) = 35$$
Each term is of the form $n(n + 2)$ or $(n + 1)^2 - 1$:
$$T_n = n(n + 2) = n^2 + 2n = (n + 1)^2 - 1$$
$$\mathbf{\text{General Formula: } T_n = n^2 + 2n \; [\text{or } (n + 1)^2 - 1].}$$
Step 3: (iii) Calculate the $25\text{th}$ Term ($T_{25}$)
Substituting $n = 25$ into the derived formula:
$$T_{25} = 25(25 + 2) = 25 \times 27 = 675$$
Alternatively: $(25 + 1)^2 - 1 = 26^2 - 1 = 676 - 1 = 675$.
$$\mathbf{\text{The } 25\text{th} \text{ term is } 675.}$$
Step 4: (iv) Determine Which Term Equals $960$
Given $T_n = 960$. Set up the quadratic equation:
$$(n + 1)^2 - 1 = 960$$ $$(n + 1)^2 = 960 + 1$$ $$(n + 1)^2 = 961$$
Taking the positive square root on both sides (as $n$ must be a natural number):
$$n + 1 = \sqrt{961} = 31$$ $$n = 31 - 1 = 30$$
$$\mathbf{\text{Conclusion: (i) Differences computed; } (ii) \; T_n = n(n+2); \; (iii) \; T_{25} = 675; \; (iv) \; \text{The } 30\text{th} \text{ term equals } 960.}$$

Rough Column Working:
Pattern: n(n+2) = (n+1)^2 - 1
Terms: 3, 8, 15, 24, 35, 48
T_25 = 26^2 - 1 = 676 - 1 = 675
(n+1)^2 - 1 = 960 => (n+1)^2 = 961 => n+1 = 31 => n = 30.

Step Mark Allocation:

Tabulating first and constant second differences: 2 Marks
Deriving general formula Tn = n(n+2) or (n+1)^2 - 1: 3 Marks
Evaluating T25 = 675: 2.5 Marks
Solving (n+1)^2 = 961 to deduce n = 30 and boxed answer: 2.5 Marks

Examiner Scoring Key: Expressing the quadratic as (n+1)^2 - 1 avoids factoring large quadratic trinomials and allows direct square root extraction.

RIMC Practice & Drill Exercises

Attempt these questions on paper following proper step-wise conventions, then click Reveal Solution to verify your answers.

Question 1 5 Marks

The following data shows the favorite sports of $25$ cadets: Football, Cricket, Hockey, Football, Basketball, Cricket, Cricket, Football, Tennis, Cricket, Football, Basketball, Cricket, Football, Cricket, Hockey, Football, Cricket, Tennis, Cricket, Football, Basketball, Football, Cricket, Football. Prepare a frequency table using tally marks.