Sequence & Series IOQM By Rajeev Rastogi Sir (PW)

Sequence and Series is one of the most important topics in mathematics for exams like IOQM, JEE, and other Olympiads. It builds a strong foundation for algebra and helps in solving a wide range of problems efficiently.

In this session by Rajeev Rastogi Sir, the topic is explained from basic definitions to advanced applications. You will learn how sequences follow a pattern, how series represent their sum, and how different types like Arithmetic Progression (AP) are used in problem-solving.

The focus is not just on formulas, but on understanding concepts, identifying patterns, and applying smart techniques. Recurrence relations, special sequences like Fibonacci, and properties of AP are covered in a way that helps in both exams and deeper mathematical thinking.

What is a Sequence?

A sequence in mathematics is an ordered list of numbers that follows a specific rule or law. It is not a random collection but a pattern generated according to a defined rule, where the numbers appear in a specific order.

Example: 1, 4, 9, 16, 25, …

Here, the n-th term (tn) is n². This is also called the general term, allowing us to find any term by substituting 'n'. For instance, t10 = 10² = 100.

The order matters in a sequence; terms must be arranged specifically by the rule.

Real Sequence

A real sequence is a sequence where all its terms are real numbers.

Finite and Infinite Sequences

1. Finite Sequence: A sequence is finite if it has a finite number of terms that can be counted.
   Example: an = (n + 1) / (n + 2) for n ∈ {1, 2, 3, 4, 5}. This sequence has five terms: 2/3, 3/4, 4/5, 5/6, 6/7.

2. Infinite Sequence: A sequence is infinite if it has an infinite number of terms, extending indefinitely.
   Example: bn = 1/n for n ∈ Natural Numbers. This sequence is 1, 1/2, 1/3, …, continuing infinitely.

Finding Terms of a Sequence from its General Term

Example 1: Given tn = 2 + (-1)^n / n.

• t1 = 2 + (-1)¹ / 1 = 1

• t2 = 2 + (-1)² / 2 = 3/2

• t3 = 2 + (-1)³ / 3 = 5/3
  The sequence begins: 1, 3/2, 5/3, …

Example 2: Given bn = (-1)^(n+1) * (2n+1) / (2n+3).

• b1 = (-1)² * (3/5) = 3/5

• b2 = (-1)³ * (5/7) = -5/7

• b3 = (-1)⁴ * (7/9) = 7/9
  The sequence begins: 3/5, -5/7, 7/9, … (terms alternate in sign).

Recurrence Relation

A recurrence relation recursively defines a sequence by providing a rule to determine the next term as a function of previous term(s). If initial terms are known, it generates all subsequent terms. These are common in Olympiads.

Example: Fibonacci Sequence

Defined by an+2 = an+1 + an (for n as a natural number). This is an order 2 recurrence relation.

Given a1 = 1, a2 = 1.

• a3 = a2 + a1 = 1 + 1 = 2

• a4 = a3 + a2 = 2 + 1 = 3

• a5 = a4 + a3 = 3 + 2 = 5
  The sequence is 1, 1, 2, 3, 5, 8, 13, … (Memory Tip: For Fibonacci, each term is the sum of the two preceding terms, starting with 1, 1.)

Applying Recurrence Relations (Example 1)

Problem: Given a1 = 1, and an+1 = 2an + 3^n. Find a3.

Solution:

1. Find a2 (for n=1): a2 = 2a1 + 3¹ = 2(1) + 3 = 5.

2. Find a3 (for n=2): a3 = 2a2 + 3² = 2(5) + 9 = 10 + 9 = 19.
   Result: a3 = 19.

Applying Recurrence Relations (Example 2)

Problem: Given a_(n+1)² = a_n * a_(n+2) + (-1)^n * n, with a1 = 2, a2 = 3. Find a4 / a3.

Solution:

1. Find a3 (for n=1): a2² = a1 * a3 - 1 => 3² = 2 * a3 - 1 => 9 = 2a3 - 1 => 10 = 2a3 => a3 = 5.

2. Find a4 (for n=2): a3² = a2 * a4 + 2 => 5² = 3 * a4 + 2 => 25 = 3a4 + 2 => 23 = 3a4 => a4 = 23/3.

3. Ratio a4 / a3: (23/3) / 5 = 23/15.

Applying Recurrence Relations (Example 3 - Periodic Sequence)

Problem: Given a1 = 2 and an+1 = 1 - 1/an. Let Pn be the product of its first n terms. Find P2025.

Solution Strategy: For large products, look for periodicity.

1. Derive Periodicity:

• an+1 = 1 - 1/an (Eq 1)

• an+2 = 1 - 1/an+1 (from Eq 1, replacing n with n+1)

• Substitute 1/an+1 = an / (an-1) from Eq 1 into an+2: an+2 = 1 - an/(an-1) = (an-1-an)/(an-1) = -1/(an-1) (Eq 3)

• Substitute an+1 - 1 = -1/an from Eq 1 into an+3 = -1/(an+1 - 1) (from Eq 3, replacing n with n+1): an+3 = -1/(-1/an) = an.
  This shows periodicity of 3 (an+3 = an).

2. Calculate First Few Terms:

• a1 = 2

• a2 = 1 - 1/a1 = 1 - 1/2 = 1/2

• a3 = 1 - 1/a2 = 1 - 1/(1/2) = 1 - 2 = -1

3. Product of One Period (P3): P3 = a1 * a2 * a3 = 2 * (1/2) * (-1) = -1.

4. Calculate P2025: Number of full periods in 2025 terms = 2025 / 3 = 675.
   P2025 = (P3)^675 = (-1)^675 = -1 (since 675 is odd).

Applying Recurrence Relations (Example 4 - Cube of Digits)

Problem: First term a1 = 2014. Each succeeding term is the sum of the cubes of the digits of the previous term. Find a term in the sequence (convergence).

Solution:

• a1 = 2014

• a2 = 2³ + 0³ + 1³ + 4³ = 8 + 0 + 1 + 64 = 73

• a3 = 7³ + 3³ = 343 + 27 = 370

• a4 = 3³ + 7³ + 0³ = 27 + 343 + 0 = 370
  Since a3 = a4 = 370, the sequence converges to 370 for all n ≥ 3.

What is a Series?

A series is the sum of the terms of a sequence.

Example:

• Sequence: 1, 4, 9, 16, 25, …

• Series: 1 + 4 + 9 + 16 + 25 + …

Sequence vs. Series vs. Progression

(Memory Tip: Think of a quadrilateral vs. a rectangle. A rectangle is a special type of quadrilateral with specific properties. Similarly, a progression is a special type of sequence with specific patterns.) We will cover Arithmetic Progression (AP), Geometric Progression (GP), Harmonic Progression (HP), and Arithmetic Geometric Progression (AGP).

Arithmetic Progression (AP)

An Arithmetic Progression (AP) is a sequence where each term (from the second onward) is obtained by adding or subtracting a fixed number to the preceding term. This fixed number is the Common Difference (d).

Example: 1, 3, 5, 7, 9, … Here, the difference between consecutive terms (3-1, 5-3, etc.) is always 2. So, d = 2.

The common difference d = tn - tn-1 and must be constant for a sequence to be an AP.

General Form and General Term of an AP

For an AP with first term 'a' and common difference 'd':

General Form: a, a + d, a + 2d, a + 3d, …

n-th term (General Term): tn = a + (n - 1)d

The coefficient of 'd' is always one less than the term number.

Example: The 10th term is a + 9d.

Types of Arithmetic Progressions

Based on the common difference d:

1. Increasing AP: d > 0. Example: 1, 3, 5, 7, … (d=2)

2. Decreasing AP: d < 0. Example: 99, 97, 95, … (d=-2)

3. Constant AP: d = 0. Example: 5, 5, 5, 5, … (d=0)

Finding a Specific Term of an AP Given Two Other Terms

Problem: In an AP, the 6th term is 17 and the 11th term is 32. Find the 20th term.

Solution: Using tn = a + (n-1)d:

1. a + 5d = 17 (Eq 1)

2. a + 10d = 32 (Eq 2)
   Subtract (Eq 1) from (Eq 2): 5d = 15 => d = 3.
   Substitute d=3 into (Eq 1): a + 5(3) = 17 => a + 15 = 17 => a = 2.
   Now find the 20th term: t20 = a + 19d = 2 + 19(3) = 2 + 57 = 59.

Finding the nth Term from the End of an AP

Concept: Moving backward in an AP involves subtracting the common difference (d).

Formula: If L is the last term, the nth term from the end is Tn_end = L - (n-1)d.

Problem: Find the 20th term from the end of the AP: 20, 19(1/4), 18(1/2), 17(3/4), …, -57/4.

Solution:

1. AP: 20, 77/4, 74/4, 71/4, …, -57/4.

2. a = 20, L = -57/4.

3. d = 77/4 - 20 = (77 - 80)/4 = -3/4.

4. Apply Tn_end formula for n=20:
   T20_end = L - (20-1)d = -57/4 - 19(-3/4) = -57/4 + 57/4 = 0.
   (Note: The lecturer stated -15 as the answer in the original context, but calculation leads to 0.)

Problem: 5x5 Array of APs (Finding X)

Problem: A 5x5 array has numbers such that every row and column forms an AP. Given corner values:

1 . . . 17
. . . . .
. . X . .
. . . . .
25 . . . 81

Find X (3rd row, 3rd column).

Method 1: Step-by-Step Common Difference

1. First row AP: 1, …, 17 (5 terms). 17 = 1 + 4d => 4d = 16 => d = 4. 3rd term = 1 + 2(4) = 9.

2. Last row AP: 25, …, 81 (5 terms). 81 = 25 + 4D => 4D = 56 => D = 14. 3rd term = 25 + 2(14) = 53.

3. Middle column AP: 9, …, 53 (5 terms). 53 = 9 + 4d' => 4d' = 44 => d' = 11. 3rd term (X) = 9 + 2(11) = 31.
   Thus, X = 31.

Method 2: Average Property of AP (Time-Saving)

(Memory Tip: For A, B, C in AP, B = (A+C)/2. The middle term is the average of equidistant terms.)

1. 3rd term of 1st row = (1 + 17) / 2 = 9.

2. 3rd term of last row = (25 + 81) / 2 = 53.

3. 3rd term of middle column (X) = (9 + 53) / 2 = 31.
   Thus, X = 31.

Sum of First n Terms of an AP (Gauss's Method)

The formula for the sum of an AP is famously attributed to Carl Friedrich Gauss. He summed 1 to 100 by writing the series forward and backward, then adding them:

S = 1 + 2 + … + 100

S = 100 + 99 + … + 1

2S = 101 + 101 + … + 101 (100 times)

2S = 100 * 101 = 10100

S = 5050. This method generalizes to any AP.

Derivation of the General Formula for Sum of n Terms of an AP

Let Sn be the sum of the first n terms of an AP with first term a and common difference d.

Terms are: a, a+d, a+2d, …, a+(n-1)d.

1. Sn = a + (a+d) + … + (a+(n-1)d) (Forward)

2. Sn = (a+(n-1)d) + (a+(n-2)d) + … + a (Reverse)
   Adding (1) and (2) term by term:
   2Sn = [2a + (n-1)d] + [2a + (n-1)d] + … + [2a + (n-1)d] (n times)
   2Sn = n [2a + (n-1)d]
   Therefore, Sn = n/2 [2a + (n-1)d].

Alternative Form of the Sum Formula and Sum of Special APs

• Alternative Form: Since L = a + (n-1)d (the last term), the sum formula can be written as:
  Sn = n/2 (a + L)
  This is useful when 'a', 'L', and 'n' are known.

• Sum of Special Arithmetic Progressions:

1. Sum of First n Natural Numbers (1 + 2 + … + n): n(n+1)/2

2. Sum of First n Odd Natural Numbers (1 + 3 + … + (2n-1)): n²

3. Sum of First n Even Natural Numbers (2 + 4 + … + 2n): n(n+1)

Properties of Arithmetic Progressions

1. Effect of Arithmetic Operations: If each term of an AP is increased, decreased, multiplied, or divided by the same non-zero constant, the resulting sequence remains an AP.

• Example: AP: 1, 3, 5 (d=2). Add 1: 2, 4, 6 (AP, d=2). Multiply by 2: 2, 6, 10 (AP, d=4).

2. Operations on Corresponding Terms:

• Adding or subtracting corresponding terms of two APs results in another AP.

• Multiplying or dividing corresponding terms of two APs generally does NOT result in an AP.

3. Assuming Terms for Problem Solving:

• Three terms: a - d, a, a + d (Sum = 3a)

• Five terms: a - 2d, a - d, a, a + d, a + 2d (Sum = 5a)

• Four terms: a - 3d, a - d, a + d, a + 3d (Sum = 4a, common difference is 2d)

4. Condition for a Sequence to be an AP: A sequence {Tn} is an AP if and only if Tn - Tn-1 = constant.

• Example: If Tn = 3n + 5. Then Tn-1 = 3(n-1) + 5 = 3n + 2.
  Tn - Tn-1 = (3n + 5) - (3n + 2) = 3 (a constant). So, it's an AP with d=3.

• (Memory Tip: If the nth term (Tn) is a linear expression in n (An + B), it's an AP, and the common difference is A.)

5. Sum of Equidistant Terms: In any AP, the sum of any two terms equidistant from the beginning and end is constant, equal to the sum of the first (a) and last (L) terms.

• a1 + an = a2 + an-1 = … = a + L.

Problem: Sum of Specific Terms of a 24-Term AP

Problem: An AP (A1, …, A24) has A1 + A3 + A5 + A20 + A22 + A24 = 225. Find S24.

Solution:

1. Using the equidistant terms property: A1 + A24 = constant (let's call it S_pair).

2. A3 is 3rd from start, A22 is 3rd from end (24-3+1 = 22). So, A3 + A22 = S_pair.

3. A5 is 5th from start, A20 is 5th from end (24-5+1 = 20). So, A5 + A20 = S_pair.

4. Given sum: (A1 + A24) + (A3 + A22) + (A5 + A20) = 225
   S_pair + S_pair + S_pair = 225 => 3 * S_pair = 225 => S_pair = 75.
   So, A1 + A24 = 75.

5. S24 = (A1 + A24) + (A2 + A23) + … (12 pairs).
   S24 = 12 * S_pair = 12 * 75 = 900.

Problem: Apple Trees Forming an AP

Problem: Six apple trees have apples in an AP. Total apples (S6) = 990. Greatest number of apples (A6) is double the least (A1). Find A6.

Solution:

1. A1, …, A6 are in AP. S6 = 990. A6 = 2 * A1.

2. S6 = (A1 + A6) + (A2 + A5) + (A3 + A4). Since there are 3 pairs, each summing to (A1 + A6):
   S6 = 3 * (A1 + A6)

3. 990 = 3 * (A1 + A6) => A1 + A6 = 330.

4. Substitute A6 = 2A1: A1 + 2A1 = 330 => 3A1 = 330 => A1 = 110.

5. Greatest number of apples: A6 = 2 * A1 = 2 * 110 = 220.

Problem: Sides of a Right-Angled Triangle in AP

Problem: Sides of a right-angled triangle are in AP. Area = 24. Find the smallest side.

Solution:

1. Sides in AP: a - d, a, a + d. Hypotenuse = a + d.

2. Pythagorean Theorem: (a - d)² + a² = (a + d)²
   a² - 2ad + d² + a² = a² + 2ad + d²
   a² - 2ad = 2ad => a² = 4ad.
   Since a ≠ 0, a = 4d.

3. Sides are: (4d - d), 4d, (4d + d) = 3d, 4d, 5d. (This is a scaled 3-4-5 right triangle).

4. Area = (1/2) * Base * Height = 24
   (1/2) * (3d) * (4d) = 24
   (1/2) * 12d² = 24 => 6d² = 24 => d² = 4 => d = 2 (since d > 0).

5. Smallest side = 3d = 3 * 2 = 6.
   Sides are 6, 8, 10.

Problem: Number of Possible Values of n (First Term is an Integer)

Problem: Sn = 153, d = 2. First term (a) is an integer. Find the number of possible values for n.

Solution:

1. Sn = n/2 [2a + (n-1)d]

2. 153 = n/2 [2a + (n-1)2]

3. 153 = n [a + n - 1]

4. Isolate 'a': a = (153/n) - n + 1.

5. For 'a' to be an integer, 153/n must be an integer. This means 'n' must be a divisor of 153.

6. Prime factorization of 153 = 3² * 17¹.

7. Divisors of 153 are: 1, 3, 9, 17, 51, 153.

8. There are 6 possible values for n.

Problem: Sum of Modified AP Terms

Problem: An AP (a₁, …, a_n) has S_n = 715. If a_k is increased by the k-th odd positive integer (2k-1), the new sum is 836. Find a₁ + a_n + a_middle.

Solution:

1. Original Sum: Σ a_k = 715.

2. New Sum: Σ (a_k + (2k-1)) = 836.

3. This can be split: (Σ a_k) + (Σ (2k-1)) = 836.

4. 715 + (1 + 3 + 5 + … + (2n-1)) = 836.

5. The sum of the first 'n' odd natural numbers is n² (Memory Tip: Sum of first n odd numbers is n².).

6. 715 + n² = 836 => n² = 121 => n = 11 (since n must be positive).
   The AP has 11 terms.

7. We need a₁ + a_n + a_middle. For 11 terms, the middle term is a₆. So, we need a₁ + a₁₁ + a₆.

8. Property of APs: Sum of equidistant terms (a₁ + a₁₁) = 2 * middle term (2a₆).

9. So, a₁ + a₁₁ + a₆ = 2a₆ + a₆ = 3a₆.

10. For an AP with an odd number of terms, S_n = n * a_middle.
    715 = 11 * a₆ => a₆ = 65.

11. Required sum = 3 * a₆ = 3 * 65 = 195.