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Predicting What Comes Next: Exploring Sequences and Progressions

NCERT Class 9 · Mathematics Based on NCERT Class 9 Mathematics textbook · Free CBSE study kit

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Chapter Notes

Overview
This chapter introduces the concept of sequences as ordered lists of numbers that follow a pattern, rule, or order. It explores two special types of progressions—Arithmetic Progressions (AP) and Geometric Progressions (GP)—and provides explicit and recursive rules to generate terms. The big idea is that by understanding the pattern or rule governing a sequence, we can predict any term directly without listing all previous terms, and apply these ideas to solve real-life problems like calculating taxi fares, bouncing ball heights, and fractal patterns.
Key Concepts (14)
Sequence
An ordered list of numbers where each number is a term of the sequence.
e.g. 1, 4, 9, 16, 25, 36, ... is the sequence of square numbers.
Term of a sequence
Each number in a sequence is called a term. The first term is denoted by t₁, the second by t₂, and so on.
e.g. In the sequence of odd numbers 1, 3, 5, 7, ..., t₁ = 1, t₂ = 3, t₃ = 5.
Finite sequence
A sequence that has a fixed number of terms.
e.g. 6, 12, 24, 48, 96 is a finite sequence of five terms.
Infinite sequence
A sequence that continues indefinitely, indicated by three dots (...).
e.g. 1, 2, 3, 4, 5, 6, ... (Natural numbers) is an infinite sequence.
Explicit rule (or formula)
A rule that uses the term's position number n to calculate its value directly, without needing previous terms.
e.g. For odd numbers, the explicit rule is uₙ = 2n – 1. Substituting n = 1 gives 1, n = 2 gives 3, etc.
Recursive rule (or formula)
A rule that gives the value of a term using the values of one or more previous terms, along with the first term(s).
e.g. For the sequence 1, 4, 7, 10, ..., the recursive rule is t₁ = 1, tₙ = tₙ₋₁ + 3 for n ≥ 2.
Arithmetic Progression (AP)
A sequence in which the difference between any two consecutive terms is constant. This constant is called the common difference (d).
e.g. 1, 5, 9, 13, 17, ... is an AP with first term a = 1 and common difference d = 4.
Common difference (d)
The fixed number added to (or subtracted from) each term to get the next term in an Arithmetic Progression.
e.g. In the AP 11, 7, 3, –1, –5, ..., the common difference is –4.
Geometric Progression (GP)
A sequence in which the ratio of any two consecutive terms is constant. This constant is called the common ratio (r).
e.g. 3, 6, 12, 24, 48, ... is a GP with first term a = 3 and common ratio r = 2.
Common ratio (r)
The fixed number by which each term is multiplied to get the next term in a Geometric Progression.
e.g. In the GP 1, –1, 1, –1, 1, ..., the common ratio is –1.
Virahānka–Fibonacci sequence
A sequence where each term (from the third term onward) is the sum of the two preceding terms, starting with V₁ = 1 and V₂ = 2.
e.g. 1, 2, 3, 5, 8, 13, 21, 34, ...
Triangular numbers
Numbers that can be represented as a triangular array of dots, where the nth triangular number is the sum of the first n natural numbers.
e.g. 1, 3, 6, 10, 15, 21, ... The 5th triangular number is 1+2+3+4+5 = 15.
Square numbers
Numbers that can be represented as a square array of dots, where the nth square number is n².
e.g. 1, 4, 9, 16, 25, 36, ... The 4th square number is 4² = 16.
Fractal
A shape or pattern that repeats itself at different scales; created using simple rules but forming complex designs.
e.g. The Sierpiński triangle and the Sierpiński square carpet are examples of fractals.
Formulas & Equations (6)
tₙ = a + (n – 1) × d
Reads as: t sub n equals a plus (n minus 1) times d
Variables: tₙ = nth term of the AP; a = first term; n = position number (natural number); d = common difference
Use when: To find any term of an Arithmetic Progression when the first term and common difference are known.
⚠ Students often write tₙ = a + n × d, forgetting the (n – 1) factor, which gives the wrong term.
tₙ = a × r^(n – 1)
Reads as: t sub n equals a times r raised to the power (n minus 1)
Variables: tₙ = nth term of the GP; a = first term; r = common ratio; n = position number (natural number)
Use when: To find any term of a Geometric Progression when the first term and common ratio are known.
⚠ Students incorrectly write tₙ = a × rⁿ, missing the (n – 1) exponent, especially when n is large.
Sₙ = n(n + 1)/2
Reads as: S sub n equals n times (n plus 1) divided by 2
Variables: Sₙ = sum of the first n natural numbers; n = number of terms
Use when: To find the sum of the first n natural numbers, or to find the nth triangular number.
⚠ Students forget to divide by 2, writing Sₙ = n(n + 1), which gives double the correct sum.
Recursive rule for AP: t₁ = a, tₙ = tₙ₋₁ + d for n ≥ 2
Reads as: t sub 1 equals a; t sub n equals t sub (n minus 1) plus d, for n greater than or equal to 2
Variables: t₁ = first term; a = first term; tₙ = nth term; tₙ₋₁ = previous term; d = common difference; n = position number
Use when: To generate terms of an AP step by step when the first term and common difference are known.
⚠ Students forget to specify the condition n ≥ 2, or use the wrong starting term.
Recursive rule for GP: t₁ = a, tₙ = r × tₙ₋₁ for n ≥ 2
Reads as: t sub 1 equals a; t sub n equals r times t sub (n minus 1), for n greater than or equal to 2
Variables: t₁ = first term; a = first term; tₙ = nth term; tₙ₋₁ = previous term; r = common ratio; n = position number
Use when: To generate terms of a GP step by step when the first term and common ratio are known.
⚠ Students confuse the recursive rule of GP with that of AP, using addition instead of multiplication.
Virahānka–Fibonacci recursive rule: V₁ = 1, V₂ = 2, Vₙ = Vₙ₋₁ + Vₙ₋₂ for n ≥ 3
Reads as: V sub 1 equals 1, V sub 2 equals 2; V sub n equals V sub (n minus 1) plus V sub (n minus 2), for n greater than or equal to 3
Variables: V₁ = first term (1); V₂ = second term (2); Vₙ = nth term; Vₙ₋₁ = previous term; Vₙ₋₂ = term before previous; n = position number
Use when: To generate the Virahānka–Fibonacci sequence or any sequence where each term is the sum of the two preceding terms.
⚠ Students often start with V₁ = 1, V₂ = 1 (the Fibonacci start) instead of V₁ = 1, V₂ = 2 as given in the chapter.
How It Works
Finding the nth term of an AP using explicit formula
  1. Identify the first term (a) and common difference (d) of the AP.
  2. Substitute the values of a, d, and the desired position n into the formula tₙ = a + (n – 1) × d.
  3. Simplify the expression to get the nth term.
💡 Always use (n – 1), not n, when multiplying by the common difference.
Finding the nth term of a GP using explicit formula
  1. Identify the first term (a) and common ratio (r) of the GP.
  2. Substitute the values of a, r, and the desired position n into the formula tₙ = a × r^(n – 1).
  3. Simplify to get the nth term.
💡 The exponent is (n – 1), not n. For the first term, r^0 = 1.
Checking if a number is a term of a sequence using explicit formula
  1. Set the explicit formula equal to the given number.
  2. Solve for n (the position number).
  3. If n is a natural number (positive integer), the number is a term of the sequence and n gives its position. If n is not a natural number, the number is not a term.
💡 n must be a natural number (1, 2, 3, ...). A fractional or negative n means the number is not in the sequence.
Generating terms using a recursive rule
  1. Write down the given first term(s).
  2. For n = 2, 3, 4, ..., substitute the previous term(s) into the recursive formula to compute the next term.
  3. Continue until the desired number of terms are obtained.
💡 You must know the first term(s) to start; recursive rules cannot generate the first term from previous ones.
Finding the sum of the first n natural numbers
  1. Write the sum S = 1 + 2 + 3 + ... + n.
  2. Write the same sum in reverse order: S = n + (n – 1) + ... + 1.
  3. Add the two equations vertically: 2S = (n+1) + (n+1) + ... + (n+1) (n times).
  4. Simplify: 2S = n(n + 1).
  5. Divide by 2: S = n(n + 1)/2.
💡 This method works only for consecutive natural numbers starting from 1. For other ranges, subtract sums.
Finding the sum of consecutive numbers not starting from 1
  1. Express the required sum as the difference of two sums of natural numbers: (sum from 1 to last term) – (sum from 1 to (first term – 1)).
  2. Use the formula Sₙ = n(n + 1)/2 for each sum.
  3. Subtract to get the final sum.
💡 If the range is from m to n (inclusive), the sum = Sₙ – Sₘ₋₁.
Visualising an AP by plotting ordered pairs
  1. Create a table with stage number (x) and corresponding term value (y).
  2. Form ordered pairs (x, y).
  3. Plot these points on a graph.
  4. Observe that the points lie on a straight line, indicating a linear relationship.
💡 The graph of an AP is always a straight line because tₙ = a + (n – 1)d is a linear function of n.
Visualising a GP by plotting ordered pairs
  1. Create a table with stage number (x) and corresponding term value (y).
  2. Form ordered pairs (x, y).
  3. Plot these points on a graph.
  4. Observe that the points do NOT lie on a straight line; they curve upward (if r > 1) or downward (if 0 < r < 1).
💡 The graph of a GP is not linear because tₙ = a × r^(n – 1) is an exponential function of n.
Constructing the Sierpiński triangle fractal
  1. Start with an equilateral triangle (Stage 0).
  2. Join the midpoints of the three sides to form four smaller equilateral triangles.
  3. Remove the central triangle (Stage 1).
  4. Repeat the process on each remaining black triangle: join midpoints and remove the central triangle.
  5. Continue indefinitely to get subsequent stages.
💡 At each stage, the number of black triangles triples, and the area of the black region becomes 3/4 of the previous stage's area.
Constructing the Sierpiński square carpet fractal
  1. Start with a square sheet (Stage 0).
  2. Trisect each side and join opposite trisection points to divide the square into 9 equal smaller squares.
  3. Remove the central square, retaining the 8 surrounding squares (Stage 1).
  4. Repeat the process on each of the 8 remaining squares: trisect sides, remove the centre.
  5. Continue indefinitely.
💡 At each stage, the number of red squares is multiplied by 8, and the area of the red region becomes 8/9 of the previous stage's area.
Exam Traps — Don't Make These Mistakes
⚠ Using n instead of (n – 1) in AP and GP formulas
✗ tₙ = a + n × d or tₙ = a × rⁿ
→ The first term corresponds to n = 1. For n = 1, a + 1×d gives a + d, not a. Similarly, a × r¹ gives ar, not a. The correct formulas use (n – 1) so that for n = 1, the term equals a.
✓ tₙ = a + (n – 1) × d for AP; tₙ = a × r^(n – 1) for GP
⚠ Confusing recursive and explicit rules
✗ For an AP, they write tₙ = tₙ₋₁ + d without stating the first term, or they write t₁ = a, tₙ = a + (n – 1)d as a recursive rule.
→ A recursive rule must include the first term(s) and express the nth term in terms of previous terms. An explicit rule gives the nth term directly in terms of n. Mixing them loses marks.
✓ Recursive: t₁ = a, tₙ = tₙ₋₁ + d for n ≥ 2. Explicit: tₙ = a + (n – 1)d.
⚠ Forgetting that n must be a natural number when checking if a number is a term
✗ After solving 5n – 2 = 471, they get n = 94.6 and say '471 is a term of the sequence' because they ignore the decimal.
→ n represents the position number, which must be a natural number (1, 2, 3, ...). 94.6 is not a natural number, so 471 cannot be a term of the sequence.
✓ Since n = 94.6 is not a natural number, 471 is not a term of the sequence.
⚠ Incorrectly identifying the common difference in a decreasing AP
✗ For the AP 21, 18, 15, ..., they write common difference d = 3 (positive).
→ The terms are decreasing: 18 – 21 = –3, 15 – 18 = –3. The common difference is –3, not 3.
✓ Common difference d = –3 (since each term is 3 less than the previous term).
⚠ Mixing up the sum of natural numbers formula with the nth term of triangular numbers
✗ They write the nth triangular number as n(n+1) without dividing by 2.
→ The nth triangular number is the sum of the first n natural numbers, which is n(n+1)/2. For n = 3, the triangular number is 6, not 3×4 = 12.
✓ nth triangular number = n(n+1)/2
⚠ Assuming all sequences are AP or GP
✗ They try to find a common difference or common ratio for the sequence of prime numbers (2, 3, 5, 7, 11, ...).
→ Not all sequences follow a simple pattern like AP or GP. The prime number sequence has no constant difference or ratio. The chapter explicitly states that some sequences have no clear regularity.
✓ The sequence of prime numbers does not have a constant common difference or common ratio, so it is neither an AP nor a GP.
Compare & Contrast
How consecutive terms are related
Arithmetic Progression (AP)
Geometric Progression (GP)
In an AP, each term is obtained by adding a constant (common difference d) to the previous term. In a GP, each term is obtained by multiplying the previous term by a constant (common ratio r).
How the nth term is expressed
Explicit rule
Recursive rule
An explicit rule gives the nth term directly as a function of n (e.g., tₙ = 2n – 1), without needing any other term. A recursive rule gives the nth term in terms of one or more previous terms (e.g., t₁ = 1, tₙ = tₙ₋₁ + 2), and requires the first term(s) to be known.
Exam Focus
Marks pattern: 2 MCQs (1 mark each) on identifying AP/GP or finding terms; 1 Short Answer (3 marks) on finding nth term or checking if a number is a term; 1 Long Answer (5 marks) on a real-life application of AP or GP, often involving multiple steps.
Most asked: Finding the nth term of an AP or GP using explicit formulas · Determining whether a given number is a term of an AP or GP · Word problems involving AP (e.g., taxi fare, salary increments) · Word problems involving GP (e.g., bouncing ball, bacteria growth) · Sum of first n natural numbers and its application to triangular numbers · Writing recursive rules for given sequences
Don't skip: Real-life application problems: e.g., taxi fare (AP), bouncing ball (GP), bacteria growth (GP), salary increments (AP); Problems that require solving a pair of linear equations to find a and d in an AP (e.g., given two terms, find the AP); Finding the sum of consecutive natural numbers not starting from 1
One-Line Revision (read aloud before the exam)
  1. A sequence is an ordered list of numbers called terms.
  2. An explicit formula gives the nth term directly using n, e.g., tₙ = 2n – 1 for odd numbers.
  3. A recursive formula gives the nth term using previous terms, e.g., t₁ = 1, tₙ = tₙ₋₁ + 2.
  4. An Arithmetic Progression (AP) has a constant common difference d between consecutive terms.
  5. The nth term of an AP is tₙ = a + (n – 1)d.
  6. A Geometric Progression (GP) has a constant common ratio r between consecutive terms.
  7. The nth term of a GP is tₙ = a × r^(n – 1).
  8. To check if a number is a term of a sequence, solve the explicit formula for n; n must be a natural number.
  9. The sum of the first n natural numbers is Sₙ = n(n + 1)/2.
  10. The nth triangular number equals the sum of the first n natural numbers: n(n + 1)/2.
  11. The Virahānka–Fibonacci sequence starts 1, 2, 3, 5, 8, ... where each term is the sum of the two preceding terms.
  12. In a decreasing AP, the common difference d is negative.
  13. The graph of an AP is a straight line; the graph of a GP is not linear.
  14. Fractals like the Sierpiński triangle repeat patterns at different scales.
  15. In the Sierpiński triangle, the number of black triangles triples each stage, and the area multiplies by 3/4 each stage.

MCQs — 10 Questions with Answers

Q1. What is the 25th term of the sequence defined by tn = 3n + 2?

  • A. 75
  • B. 77 ✓
  • C. 79
  • D. 81

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?

  • A. 11
  • B. 19
  • C. 27 ✓
  • D. 31

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?

  • A. 78
  • B. 79 ✓
  • C. 80
  • D. 81

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?

  • A. 46
  • B. 47
  • C. 48 ✓
  • D. 49

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?

  • A. Triangular numbers
  • B. Square numbers ✓
  • C. Natural numbers
  • D. Prime numbers

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?

  • A. Odd numbers
  • B. Even numbers
  • C. Triangular numbers ✓
  • D. Square numbers

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?

  • A. Yes, the 26th term ✓
  • B. Yes, the 27th term
  • C. No, because n would be 26.33
  • D. No, because n would be a negative number

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:

  • A. 3
  • B. 7 ✓
  • C. 10
  • D. n

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, ...?

  • A. tn = 2n + 3
  • B. tn = 3n + 2 ✓
  • C. tn = 4n + 1
  • D. tn = 3n

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?

  • A. 201 and 39 ✓
  • B. 201 and 38
  • C. 200 and 39
  • D. 199 and 37

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.

Flashcards

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.

Important Board Questions

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.

Assertion-Reason Questions for Predicting What Comes Next: Exploring Sequences and Progressions

Q1. Assertion (A):

The explicit formula uₙ = 2n - 1 generates the sequence of odd numbers.

Reason (R):

When n = 1, 2, 3, 4, ..., the formula produces values 1, 3, 5, 7, ... which are consecutive odd numbers.

Answer: A — Both A and R are true, and R is the correct explanation of A — Both the assertion and reason are true. The formula uₙ = 2n - 1 does generate odd numbers, and the reason correctly explains why this happens through substitution of values. The reason directly explains the assertion.

Q2. Assertion (A):

Triangular numbers can be expressed as the sum of consecutive natural numbers starting from 1.

Reason (R):

Square numbers can be expressed as the sum of consecutive odd numbers starting from 1.

Answer: B — Both A and R are true, but R is NOT the correct explanation of A — Both the assertion and reason are true (both patterns are mentioned in the chapter), but the reason does NOT explain why triangular numbers have their property. The reason describes a different sequence pattern (square numbers, not triangular). They are independent facts, not causally related.

Q3. Assertion (A):

An explicit formula allows us to find the 100th term of a sequence without calculating all previous terms.

Reason (R):

Explicit formulas are more complicated than listing all sequence terms, making them less useful for finding specific terms.

Answer: C — A is true but R is false — The assertion is true—this is a key advantage of explicit formulas. However, the reason is false. Explicit formulas are actually more efficient and useful because they allow direct calculation. The material explicitly states this advantage. Only the assertion is true.

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Case-Based Questions (CBQ) for Predicting What Comes Next: Exploring Sequences and Progressions

Case Study:

A florist is creating triangular flower arrangements for a wedding. In the first arrangement, she uses 1 flower. In the second arrangement, she uses 3 flowers arranged in a triangle (1 + 2). In the third arrangement, she uses 6 flowers (1 + 2 + 3). She continues this pattern for several arrangements. The florist wants to know how many flowers are needed for the 10th arrangement without creating all intermediate arrangements, and she also wants to determine which arrangement will use exactly 66 flowers.

Q1. How many flowers are needed for the 5th triangular arrangement?

  • A. 10
  • B. 15 ✓
  • C. 20
  • D. 25

Answer: B

Q2. If tₙ represents the number of flowers in the nth arrangement, what is the explicit formula?

  • A. tₙ = n²
  • B. tₙ = n(n + 1)/2 ✓
  • C. tₙ = 2n - 1
  • D. tₙ = n + 5

Answer: B

Q3. How many flowers are needed for the 10th arrangement?

  • A. 45
  • B. 50
  • C. 55 ✓
  • D. 60

Answer: C

Q4. Which arrangement will use exactly 66 flowers?

  • A. 11th arrangement ✓
  • B. 12th arrangement
  • C. 10th arrangement
  • D. Not possible in this pattern

Answer: A

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