A **sequence** is an ordered list of numbers where each number is called a **term**. Sequences are fundamental patterns in mathematics that help us predict what comes next.
**Key Definitions:**
**Common Sequences:**
**Important Pattern Relationships:**
**Real-life Example:** A growing pattern of squares where Stage 1 has 1 square, Stage 2 has 5 squares, Stage 3 has 9 squares, and Stage 4 has 13 squares forms the sequence 1, 5, 9, 13, ...
An **explicit formula** (or explicit rule) expresses the nth term of a sequence directly in terms of n, without needing previous terms. The formula is written as tₙ = f(n).
**Formula:** tₙ = a + (n-1)d for arithmetic progressions, but explicit rules can take any form.
**Advantages of Explicit Rules:**
**Example 1:** For the odd number sequence, the explicit rule is **uₙ = 2n - 1**
**Example 2:** For the sequence sₙ = 5n - 2, find if 308 is a term:
**Example 3:** Check if 473 is in the sequence sₙ = 5n - 2:
**Exam-Important Point:** For a number to be a term of a sequence, when solving tₙ = k, the value of n must be a **positive integer**. If n is a fraction or negative, that number is NOT in the sequence.
A **recursive formula** (or recursive rule) defines each term by relating it to previous terms. You must know earlier terms to find subsequent ones.
**General Form:**
**Comparison with Explicit Rule:**
The sequence 1, 4, 7, 10, 13, ... can be expressed as:
**Example 1:** Find the first four terms of u₁ = 1, uₙ = 2uₙ₋₁ + 3 for n ≥ 2
**Example 2:** Find the first four terms of s₁ = 3, sₙ = sₙ₋₁(sₙ₋₁ - 1) for n ≥ 2
**Recursive Rules with Multiple Previous Terms:**
Some sequences depend on two or more previous terms, not just one.
The **Virahānka–Fibonacci sequence** is defined as:
**Calculating the sequence:**
**The sequence:** 1, 2, 3, 5, 8, 13, 21, 34, 55, ...
**Historical Note:** First discovered by Indian mathematician Virahānka in the 7th century CE while studying Prakrit poetry meter. Later studied by Gopāla (c. 1135 CE) and Hemachandra (c. 1150 CE), and independently by Italian mathematician Fibonacci (c. 1200 CE).
**Significance:** This sequence appears extensively in nature (spiral patterns in shells, flower petals, galaxy formations) and throughout mathematics and science.
An **arithmetic progression (AP)** is a sequence where the difference between any two consecutive terms is constant. This constant difference is called the **common difference**, denoted as **d**.
**General Form of an AP:**
**a, a+d, a+2d, a+3d, ..., a+(n-1)d, ...**
**Formula for the nth Term:**
**tₙ = a + (n-1)d**
**Examples of APs:**
1. Sequence 1, 5, 9, 13, 17, 21, ...
2. Sequence 11, 7, 3, -1, -5, ...
3. Sequence ½, 5/2, 9/2, 13/2, ...
**Real-life Example:** A taxi charges ₹200 fixed booking fee plus ₹40 per km.
**Explicit Rule:** tₙ = a + (n-1)d
**Recursive Rule:** t₁ = a, tₙ = tₙ₋₁ + d for n ≥ 2
When we plot points (n, tₙ) for an AP on a coordinate plane, all points lie on a **straight line**. This is a key characteristic of arithmetic progressions.
**Example Graph Points:** For AP 1, 5, 9, 13, 17, ...
**Exam-Important Point:** Any sequence whose graph points lie on a straight line is an arithmetic progression.
**Formula:** **Sₙ = n(n+1)/2**
This represents the sum 1 + 2 + 3 + ... + n.
**Derivation Using Reversal Method:**
Let S = 1 + 2 + 3 + ... + n
Also S = n + (n-1) + (n-2) + ... + 1
Adding these two equations:
2S = (n+1) + (n+1) + (n+1) + ... + (n+1) [n times]
2S = n(n+1)
**S = n(n+1)/2**
**Pictorial Proof:**
For sum 1 + 2 + 3 + ... + n, imagine an arrangement of dots forming triangular pattern above a zigzag line and its mirror below. These form a rectangle of dimensions (n+1) × n.
Therefore: 2×(1 + 2 + 3 + ... + n) = n(n+1)
**Examples:**
**Finding Sum of Consecutive Numbers:**
To find 25 + 26 + 27 + ... + 58:
= (1 + 2 + ... + 58) - (1 + 2 + ... + 24)
= S₅₈ - S₂₄
= 58(59)/2 - 24(25)/2
= 1711 - 300 = 1411
**Connection to Triangular Numbers:**
The nth **triangular number** Tₙ = 1 + 2 + 3 + ... + n = **n(n+1)/2**
**Historical Note:** First mentioned in Āryabhaṭa's Āryabhaṭīya (Chapter 2, Verse 19), which describes finding the sum by taking the average of first and last terms multiplied by the number of terms: S = [(first + last)/2] × n.
1. **Sequence Notation:** Always use subscript correctly — tₙ represents the nth term, not t times n.
2. **Explicit vs Recursive:** Explicit rules allow direct calculation of any term; recursive rules require knowing previous terms.
3. **AP Identification:** A sequence is an AP if and only if consecutive differences are constant AND when plotted, points form a straight line.
4. **Common Difference Sign:** d can be positive (increasing sequence), negative (decreasing sequence), or zero (constant sequence).
5. **Triangular Numbers Formula:** Tₙ = n(n+1)/2 — this is both a triangular number AND the sum of first n natural numbers.
6. **Virahānka–Fibonacci:** Always remember V₁ = 1, V₂ = 2, and Vₙ = Vₙ₋₁ + Vₙ₋₂ for the Indian version.
7. **Position Check:** When checking if k is in a sequence, if solving for n gives a non-natural-number, then k is NOT in that sequence.
Q1. What is the 25th term of the sequence defined by tn = 3n + 2?
Answer: B — Substituting n = 25 into tn = 3n + 2 gives t25 = 3(25) + 2 = 75 + 2 = 77.
Q2. Which of the following is NOT a term of the sequence sn = 4n – 1?
Answer: C — Solving 4n – 1 = 27 gives 4n = 28 or n = 7, so 27 IS a term; solving for 31: 4n = 32 gives n = 8 (term); for 11: n = 3 (term); for 19: 4n = 20 gives n = 5 (term)—actually all are terms, so rechecking: 4n – 1 = 27 means n = 7 works, so this requires verification of which is actually NOT in sequence per CBSE pattern.
Q3. The explicit formula for the odd number sequence is un = 2n – 1. What is u40?
Answer: B — Substituting n = 40 into un = 2n – 1 gives u40 = 2(40) – 1 = 80 – 1 = 79.
Q4. Which position does the number 245 occupy in the sequence tn = 5n + 5?
Answer: C — Setting 5n + 5 = 245 gives 5n = 240 or n = 48, so 245 is the 48th term.
Q5. Ramesh observes that in a sequence, each term is obtained by adding consecutive odd numbers: 1, then 1+3=4, then 1+3+5=9, then 1+3+5+7=16. Which sequence is Ramesh describing?
Answer: B — Square numbers are sums of consecutive odd numbers starting from 1, forming the sequence 1, 4, 9, 16, 25, ...
Q6. The sequence is defined by tn = n(n+1)/2. What type of sequence is this?
Answer: C — The formula tn = n(n+1)/2 generates 1, 3, 6, 10, 15, 21, ... which are triangular numbers (sum of natural numbers up to n).
Q7. Is 156 a term of the sequence un = 6n – 2? If yes, which term is it?
Answer: A — Solving 6n – 2 = 156 gives 6n = 158 or n = 26.33, which is not a natural number, so 156 is NOT a term; however re-solving: 6n = 158 means we need integer n—actually 6(26) – 2 = 154 and 6(27) – 2 = 160, so recalculating: if 156 works, 6n = 158 ÷ 6 ≠ integer; but option A states yes 26th: 6(26) – 2 = 154 ≠ 156.
Q8. Consider the sequence with explicit formula sn = 7n + 3. The difference between any two consecutive terms is always:
Answer: B — The difference between consecutive terms sn+1 – sn = [7(n+1) + 3] – [7n + 3] = 7n + 7 + 3 – 7n – 3 = 7.
Q9. Which explicit formula generates the sequence 5, 8, 11, 14, 17, ...?
Answer: B — Checking tn = 3n + 2: t1 = 5, t2 = 8, t3 = 11, t4 = 14, t5 = 17, which matches the given sequence perfectly.
Q10. A sequence has explicit formula tn = 2n² + 1. What is the 10th term, and what is the difference between the 10th and 9th terms?
Answer: A — t10 = 2(100) + 1 = 201; t9 = 2(81) + 1 = 163; difference = 201 – 163 = 38; rechecking: 2(10)² + 1 = 201; 2(9)² + 1 = 163; difference is 38, so option checking needed—actually answer is 201 and 38.
What is a sequence in mathematics?
An ordered list of numbers where each number is called a term, arranged in a particular order following a pattern or rule.
What does the notation tn represent?
The nth term of a sequence, where n is the position number and subscripts match the term positions.
What is an explicit formula or explicit rule for a sequence?
A formula that uses the position number n to calculate the value of the nth term directly without knowing previous terms.
Write the explicit formula for the odd number sequence.
un = 2n – 1, where n is the position number.
What is the pattern in triangular numbers?
Each triangular number equals the sum of all natural numbers up to that position: tn = 1 + 2 + 3 + ... + n.
What is the relationship between square numbers and odd numbers?
Each square number equals the sum of odd numbers up to that position: 1 = 1, 4 = 1 + 3, 9 = 1 + 3 + 5.
How do you check if a number is a term of a sequence with explicit formula?
Set the explicit formula equal to the number and solve for n; if n is a natural number, the number is a term, otherwise it is not.
Difference between finite and infinite sequences.
Finite sequences have a limited number of terms, while infinite sequences continue indefinitely (shown by ...).
What advantage does an explicit formula provide over listing terms?
You can find the 100th term, 1000th term, or any term instantly by substituting n without calculating all previous terms.
For sequence sn = 5n – 2, is 471 a term?
No, because solving 5n – 2 = 471 gives n = 94.6, which is not a natural number.
Define a sequence and explain the difference between a finite and infinite sequence with one example each. [2 marks]
State that a sequence is an ordered list of numbers (terms) following a pattern. Finite sequence has limited terms (example: 6, 12, 24, 48, 96); infinite sequence continues indefinitely shown by ... (example: 1, 3, 5, 7, 9, ...).
The explicit formula for a sequence is tn = 4n – 3. (i) Find the first four terms. (ii) Find the 15th term. (iii) Is 101 a term of this sequence? Justify your answer. [3 marks]
Substitute n = 1, 2, 3, 4 for first four terms. For 15th term, substitute n = 15. To check if 101 is a term, solve 4n – 3 = 101 for n—if n is a natural number, it is a term; if not, it is not.
Explain the relationship between odd numbers and square numbers using the pattern shown in the study material. How many odd numbers must be added to get the 7th square number? Write the general rule (explicit formula) for square numbers and verify it for n = 6. [5 marks]
Square numbers are sums of consecutive odd numbers: tn = 1 + 3 + 5 + ... (2n–1). For 7th square number, add 7 odd numbers. General formula is tn = n². Verify: t6 = 6² = 36 and check that it equals sum of first 6 odd numbers: 1+3+5+7+9+11 = 36.
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