Chapter 8: Predicting What Comes Next: Exploring Sequences and Progressions - Ultimate 100/100 Master Study Guide

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Topic 1: Introduction to Sequences & Pattern Recognition

Foundational Definitions

Sequence: An ordered list of numbers arranged according to a definite rule or pattern.
Each number in the sequence is called a term.
Notation: Terms are denoted as $a_1, a_2, a_3, \dots, a_n$, where $a_n$ is the $n^{\text{th}}$ term (or general term).
Finite Sequence: A sequence containing a limited number of terms (e.g., $2, 4, 6, 8, 10$).
Infinite Sequence: A sequence that continues indefinitely without end (e.g., $1, 3, 5, 7, \dots$).
General Term Formula ($a_n$): Expresses any term in terms of its position index $n$.
Example: If $a_n = 2n + 1$, then $a_1 = 3, a_2 = 5, a_3 = 7, \dots$.

Fig 8.1: Step-ladder Progression showing constant increment $+4$ across terms

Teacher's Exam Secret — Sequence vs Progression

All progressions are sequences, but NOT all sequences are progressions! A progression is a sequence whose terms strictly obey a specific mathematical rule (such as constant difference in AP).
Example: Prime numbers $2, 3, 5, 7, 11, \dots$ form a sequence, but NOT an Arithmetic Progression.

Immediate Solved Exam Applications

Example 1.1: General Term Evaluation NCERT

Question: Find the first three terms of the sequence defined by $a_n = \frac{n-3}{4}$.

Solution:

For $n = 1 \implies a_1 = \frac{1-3}{4} = \frac{-2}{4} = \mathbf{-\frac{1}{2}}$.

For $n = 2 \implies a_2 = \frac{2-3}{4} = \mathbf{-\frac{1}{4}}$.

For $n = 3 \implies a_3 = \frac{3-3}{4} = \mathbf{0}$.

The first three terms are $-\frac{1}{2}, -\frac{1}{4}, 0$.

Topic 1: School Exam Practice Problem Kit

Problem 1.1 (Finding Term by Index) 2 MARKS

Find the $15^{\text{th}}$ term of the sequence whose $n^{\text{th}}$ term is $a_n = (n-1)(2-n)(3+n)$.

Step-by-Step Solution:

Substitute $n = 15$: $$a_{15} = (15-1)(2-15)(3+15) = (14)(-13)(18)$$ $$a_{15} = -182 \times 18 = \mathbf{-3276}$$

Fig 8.1A: Term Indexing

Problem 1.2 (Fibonacci Sequence Pattern) 1 MARK

The Fibonacci sequence is defined by $a_1 = 1 = a_2$ and $a_n = a_{n-1} + a_{n-2}$ for $n > 2$. Find $a_5$.

Step-by-Step Solution:

$a_1 = 1, a_2 = 1$.
$a_3 = 1+1 = 2$. $a_4 = 2+1 = 3$. $a_5 = 3+2 = \mathbf{5}$.

Topic 2: Arithmetic Progression (AP) Fundamentals

Core AP Properties

Arithmetic Progression (AP): A sequence in which each term after the first is obtained by adding a fixed constant $d$ to the preceding term.
First Term ($a$ or $a_1$): The starting value of the AP.
Common Difference ($d$): The constant difference between any two consecutive terms: $$\mathbf{d = a_{k+1} - a_k = a_2 - a_1 = a_3 - a_2}$$
Nature of $d$:
If $d > 0$, the AP is Increasing (e.g. $3, 7, 11, 15, \dots$).
If $d < 0$, the AP is Decreasing (e.g. $20, 15, 10, 5, \dots$).
If $d = 0$, the AP is Constant (e.g. $5, 5, 5, 5, \dots$).
Standard Form of an AP: $$\mathbf{a, \; a+d, \; a+2d, \; a+3d, \; \dots, \; a+(n-1)d}$$

Fig 8.2: Linear Number Line Representation of AP with Uniform Step $+d$

Immediate Solved Exam Applications

Example 2.1: Identifying Common Difference NCERT

Question: For the AP: $\frac{3}{2}, \frac{1}{2}, -\frac{1}{2}, -\frac{3}{2}, \dots$, write the first term $a$ and common difference $d$.

Solution:

First term $a = \mathbf{\frac{3}{2}}$.

Common difference $d = a_2 - a_1 = \frac{1}{2} - \frac{3}{2} = \frac{-2}{2} = \mathbf{-1}$.

Topic 2: School Exam Practice Problem Kit

Problem 2.1 (Testing if Sequence is AP) 2 MARKS / NCERT

Which of the following forms an AP? If they form an AP, find $d$ and write the next two terms:
(i) $2, 4, 8, 16, \dots$ (ii) $2, \frac{5}{2}, 3, \frac{7}{2}, \dots$

Step-by-Step Solution:

(i) $a_2 - a_1 = 4-2 = 2$; $a_3 - a_2 = 8-4 = 4$. Since $2 \neq 4$, Not an AP!
(ii) $a_2 - a_1 = \frac{5}{2} - 2 = \frac{1}{2}$; $a_3 - a_2 = 3 - \frac{5}{2} = \frac{1}{2}$.
Since difference is constant $d = \mathbf{\frac{1}{2}}$, it Forms an AP!
Next terms: $a_5 = \frac{7}{2} + \frac{1}{2} = \mathbf{4}$, $a_6 = 4 + \frac{1}{2} = \mathbf{\frac{9}{2}}$.

Fig 8.2A: AP Condition Test

Topic 3: The General Term ($n^{\text{th}}$ Term) of an AP

General Term Formulas

The $n^{\text{th}}$ term $a_n$ of an AP with first term $a$ and common difference $d$ is given by:

$$\mathbf{a_n = a + (n - 1)d}$$

If there are $m$ terms in an AP, then $a_m$ represents the last term, often denoted as $l$.
$n^{\text{th}}$ Term from the END of an AP: If an AP has last term $l$ and common difference $d$, the $n^{\text{th}}$ term from the end is: $$\mathbf{a_n' = l - (n - 1)d}$$

Fig 8.3: Linear Relationship between Position Index $n$ and Term Value $a_n$

Immediate Solved Exam Applications

Example 3.1: Finding 10th Term NCERT

Question: Find the $10^{\text{th}}$ term of the AP: $2, 7, 12, \dots$.

Solution:

Here $a = 2$, $d = 7 - 2 = 5$, and $n = 10$.

$$a_{10} = a + (10-1)d = 2 + 9(5) = 2 + 45 = \mathbf{47}$$

Topic 3: School Exam Practice Problem Kit

Problem 3.1 (Which Term is Given Number) 3 MARKS / NCERT

Which term of the AP: $3, 8, 13, 18, \dots$ is $78$?

Step-by-Step Solution:

$a = 3$, $d = 8-3 = 5$. Let $a_n = 78$. $$3 + (n-1)5 = 78 \implies (n-1)5 = 75 \implies n-1 = 15 \implies n = \mathbf{16}$$ The $16^{\text{th}}$ term is $78$.

Fig 8.3A: Index Solving

Problem 3.2 (Term from the End) 3 MARKS

Find the $11^{\text{th}}$ term from the last term (towards the first term) of the AP: $10, 7, 4, \dots, -62$.

Step-by-Step Solution:

$l = -62$, $d = 7-10 = -3$, $n = 11$. $$a_{11}' = l - (11-1)d = -62 - 10(-3) = -62 + 30 = \mathbf{-32}$$

Topic 4: Finding Number of Terms ($n$) & Term Verification

Core Rule — Integer Condition for Term Index $n$

When checking if a number $X$ belongs to an AP, set $a_n = X$ and solve for $n$.
The index $n$ MUST be a positive integer ($1, 2, 3, \dots$). If $n$ comes out to be a fraction or negative, then $X$ is NOT a term of the AP!

Topic 4: School Exam Practice Problem Kit

Problem 4.1 (Verification of Membership) 3 MARKS / NCERT

Check whether $-150$ is a term of the AP: $11, 8, 5, 2, \dots$.

Step-by-Step Solution:

$a = 11$, $d = 8-11 = -3$. Let $a_n = -150$. $$11 + (n-1)(-3) = -150 \implies (n-1)(-3) = -161 \implies n-1 = \frac{161}{3} \implies n = \frac{164}{3} = \mathbf{54.66}$$ Since $n$ is NOT a positive integer, $-150$ is NOT a term of this AP!

Topic 5: Sum of First $n$ Terms of an AP ($S_n$)

Sum Formulas

The sum $S_n$ of the first $n$ terms of an AP with first term $a$ and common difference $d$ is:

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

If the last term $l = a_n$ is known, the alternative compact formula is:

$$\mathbf{S_n = \frac{n}{2} (a + l)}$$

Sum of First $n$ Positive Integers ($1 + 2 + 3 + \dots + n$): $$\mathbf{S_n = \frac{n(n + 1)}{2}}$$
Crucial Relation between $S_n$ and $a_n$: $$\mathbf{a_n = S_n - S_{n-1}}$$

Fig 8.4: Gauss Geometric Pair Addition Method for Sum of First 100 Positive Integers

Immediate Solved Exam Applications

Example 5.1: Finding AP Sum NCERT

Question: Find the sum of first 22 terms of the AP: $8, 3, -2, \dots$.

Solution:

$a = 8$, $d = 3-8 = -5$, $n = 22$.

$$S_{22} = \frac{22}{2} \left[ 2(8) + (22-1)(-5) \right] = 11 \left[ 16 + 21(-5) \right] = 11 [16 - 105] = 11 \times (-89) = \mathbf{-979}$$

Topic 5: School Exam Practice Problem Kit

Problem 5.1 (How Many Terms Needed for Sum) 4 MARKS / NCERT

How many terms of the AP: $24, 21, 18, \dots$ must be taken so that their sum is $78$?

Step-by-Step Solution:

$a = 24$, $d = -3$, $S_n = 78$. $$\frac{n}{2} [ 2(24) + (n-1)(-3) ] = 78 \implies n [ 48 - 3n + 3 ] = 156 \implies n(51 - 3n) = 156$$ $$3n^2 - 51n + 156 = 0 \implies n^2 - 17n + 52 = 0 \implies (n-4)(n-13) = 0$$ $$n = \mathbf{4} \quad \text{or} \quad n = \mathbf{13}$$
Both values are valid! Sum of terms from $5^{\text{th}}$ to $13^{\text{th}}$ equals zero.

Topic 6: Arithmetic Mean (AM) & Three Numbers in AP

Arithmetic Mean Principles

Arithmetic Mean (AM) of two numbers $A$ and $B$: $$\text{AM} = \frac{A + B}{2}$$
Three terms $a, b, c$ in AP Condition: $$b - a = c - b \implies \mathbf{2b = a + c} \quad \text{or} \quad \mathbf{b = \frac{a + c}{2}}$$
Convenient Term Selections for Problem Solving:
3 terms in AP: $(a - d), \; a, \; (a + d)$ (Sum $= 3a$).
4 terms in AP: $(a - 3d), \; (a - d), \; (a + d), \; (a + 3d)$ (Sum $= 4a$, difference $= 2d$).

Topic 6: School Exam Practice Problem Kit

Problem 6.1 (Finding Variable in AP Terms) 2 MARKS

If $2k+1, 3k+3, 5k-1$ are three consecutive terms of an AP, find the value of $k$.

Step-by-Step Solution:

Use condition $2b = a + c$: $$2(3k + 3) = (2k + 1) + (5k - 1)$$ $$6k + 6 = 7k \implies 7k - 6k = 6 \implies \mathbf{k = 6}$$

Topic 7: Real-World Applications of Progressions

Example 7.1: Factory TV Production Growth NCERT

A manufacturer of TV sets produced 600 sets in the 3rd year and 700 sets in the 7th year. Assuming that the production increases uniformly by a fixed number every year, find: (i) Production in 1st year, (ii) Production in 10th year, (iii) Total production in first 7 years.

Solution:

1. $a_3 = a + 2d = 600$ and $a_7 = a + 6d = 700$.

2. Subtracting: $(a + 6d) - (a + 2d) = 700 - 600 \implies 4d = 100 \implies \mathbf{d = 25}$.

3. $a + 2(25) = 600 \implies a = 600 - 50 = \mathbf{550\text{ sets}}$.

(i) Production in 1st year $= \mathbf{550\text{ sets}}$.

(ii) Production in 10th year $a_{10} = 550 + 9(25) = 550 + 225 = \mathbf{775\text{ sets}}$.

(iii) Total production in 7 years $S_7 = \frac{7}{2} [ 550 + 700 ] = \frac{7}{2} \times 1250 = 7 \times 625 = \mathbf{4375\text{ sets}}$.

Topic 8: Comprehensive Exam HOTS & Case Study Kit

Case Study 1: Wooden Rungs Tapering Ladder CASE STUDY

A ladder has rungs $25\text{ cm}$ apart. The rungs decrease uniformly in length from $45\text{ cm}$ at the bottom to $25\text{ cm}$ at the top. If the top and bottom rungs are $2.5\text{ m}$ apart, what is the total length of wood required for the rungs?

Solution:

1. Distance between top and bottom rungs $= 2.5\text{ m} = 250\text{ cm}$.

$$\text{Number of rungs } n = \frac{250}{25} + 1 = 10 + 1 = \mathbf{11\text{ rungs}}$$

2. First rung $a = 45\text{ cm}$, Last rung $l = 25\text{ cm}$.

$$\text{Total wood length } S_{11} = \frac{11}{2} (a + l) = \frac{11}{2} (45 + 25) = \frac{11}{2} \times 70 = 11 \times 35 = \mathbf{385\text{ cm} = 3.85\text{ m}}$$

HOTS 1: m-th and n-th Term Proof HOTS FAVORITE

Question: If the $m^{\text{th}}$ term of an AP is $\frac{1}{n}$ and the $n^{\text{th}}$ term is $\frac{1}{m}$, prove that its $(mn)^{\text{th}}$ term is $1$.

Solution:

1. $a_m = a + (m-1)d = \frac{1}{n}$ --- (Eq 1)

2. $a_n = a + (n-1)d = \frac{1}{m}$ --- (Eq 2)

3. Subtract Eq 2 from Eq 1:

$$(m - n)d = \frac{1}{n} - \frac{1}{m} = \frac{m - n}{mn} \implies \mathbf{d = \frac{1}{mn}}$$

4. Substitute $d = \frac{1}{mn}$ into Eq 1:

$$a + (m-1)\frac{1}{mn} = \frac{1}{n} \implies a + \frac{m}{mn} - \frac{1}{mn} = \frac{1}{n} \implies a - \frac{1}{mn} = 0 \implies \mathbf{a = \frac{1}{mn}}$$

5. Compute $(mn)^{\text{th}}$ term:

$$a_{mn} = a + (mn - 1)d = \frac{1}{mn} + (mn - 1)\frac{1}{mn} = \frac{1 + mn - 1}{mn} = \frac{mn}{mn} = \mathbf{1}$$

Hence Proved! ■

9. 100/100 School Examination Master Checklist

Master Exam Checklist

AP Definition: Constant difference $d = a_{k+1} - a_k$.
General Term: $a_n = a + (n-1)d$.
Term from End: $a_n' = l - (n-1)d$.
Sum Formulas: $S_n = \frac{n}{2}[2a + (n-1)d] = \frac{n}{2}(a+l)$.
Sum Relation: $a_n = S_n - S_{n-1}$.
3 Terms in AP: $2b = a + c$.