HOTS Questions on Class 9 Maths Chapter 8 Predicting What Comes Next: Exploring Sequences and Progressions

HOTS Questions on Class 9 Maths Chapter 8 Predicting What Comes Next: Exploring Sequences and Progressions help students understand patterns, number sequences, and progressions in a simple way. They improve knowledge of arithmetic progression, common difference, nth term, and sequence rules. This chapter builds a strong base for solving pattern-based problems and improving math skills. A downloadable PDF is also available for easy revision and practice. 

HOTS Questions Based on Number Patterns

Q1. Look at the sequence: 1, 8, 27, 64, 125… Identify the pattern and write the 10th term.

Answer: Each term is the cube of its position number: 1³, 2³, 3³, 4³, 5³…

The nth term = n³

10th term = 10³ = 1000

Q2. A sequence is: 2, 3, 5, 8, 13, 21… What rule connects consecutive terms? Is this an arithmetic progression?

Answer: Observe the differences: 3 − 2 = 1, 5 − 3 = 2, 8 − 5 = 3, 13 − 8 = 5, 21 − 13 = 8.

The differences are not constant, so it is not an AP.

The rule is: each term is the sum of the two terms before it (this is the Fibonacci sequence: aₙ = aₙ₋₁ + aₙ₋₂).

Next term = 21 + 13 = 34

Q3. The sequence 3, 7, 11, 15, 19… follows a pattern. Without using a formula, explain why the 50th term must be odd.

Answer: The sequence is an AP with a = 3 (odd) and d = 4 (even).

Every new term is obtained by adding 4 (even) to the previous term.

Odd + Even = Odd always.

Since the first term is odd and every step adds an even number, every term in this sequence remains odd.

Therefore the 50th term is odd no calculation needed.

Q4. Find the missing terms in the sequence: 5, ?, 17, ?, 29.

Answer: Check if this is an AP. If it is, the common difference d must be constant.

From 5 to 17 (skipping one term): two steps of d: 5 + 2d = 17, 2d = 12, d = 6

Missing term 2 = 5 + 6 = 11

Missing term 4 = 17 + 6 = 23

Sequence: 5, 11, 17, 23, 29 (all differences = 6)

Q5. Three consecutive terms of an AP are written as (x − 2), (2x + 1), and (4x − 5). Find x and write the three terms.

Answer: In an AP, the middle term is the average of the other two:

2(2x + 1) = (x − 2) + (4x − 5)

4x + 2 = 5x − 7

x = 9

Three terms: (9 − 2), (2 × 9 + 1), (4 × 9 − 5) = 7, 19, 31

Check: 19 − 7 = 12, 31 − 19 = 12 (d = 12)

Q6. In a sequence, the 5th term is 22 and the 9th term is 38. Find the first term and the common difference, assuming it is an AP.

Answer: Using aₙ = a + (n − 1)d:

a₅ = a + 4d = 22 … (i) a₉ = a + 8d = 38 … (ii)

Subtract (i) from (ii): 4d = 16, d = 4

From (i): a + 16 = 22, a = 6

First term = 6, common difference = 4

Verify: a₅ = 6 + 4(4) = 22 ✓, a₉ = 6 + 8(4) = 38

HOTS Questions Based on Arithmetic Progressions

Q1. The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is 32. Find the first term and common difference.

Answer:

a₄ + a₈ = (a + 3d) + (a + 7d) = 2a + 10d = 24, a + 5d = 12 … (i)

a₆ + a₁₀ = (a + 5d) + (a + 9d) = 2a + 14d = 32, a + 7d = 16 … (ii)

Subtract (i) from (ii): 2d = 4, d = 2

From (i): a + 10 = 12, a = 2

Q2. In an AP, the 3rd term is 7 and the 7th term is 3. Find the 10th term. Is there a term equal to zero in this sequence?

Answer:

a₃ = a + 2d = 7 … (i) a₇ = a + 6d = 3 … (ii)

Subtract (i) from (ii): 4d = − 4, d = − 1

From (i): a − 2 = 7, a = 9

a₁₀ = 9 + 9(−1) = 9 − 9 = 0

Yes the 10th term is zero. This confirms a term equal to zero exists in the sequence.

Q3. An AP has first term 5 and common difference 3. Which term of the AP is 74? If no such term exists, explain why.

Answer: Set aₙ = 74:

5 + (n − 1)(3) = 74

(n − 1)(3) = 69 

n − 1 = 23

n = 24

Since n = 24 is a positive whole number, 74 is the 24th term of the AP.

Q4. A student says that the number 301 is a term of the AP: 5, 11, 17, 23… Is the student correct? Justify.

Answer: Here a = 5, d = 6. Set aₙ = 301:

5 + (n − 1)(6) = 301

(n − 1)(6) = 296

n − 1 = 296/6 = 49.33…

Since n is not a whole number, 301 is not a term of this AP. The student is incorrect.

Q5. The nth term of a sequence is given by aₙ = 3n + 2. (i) Is this sequence an AP? (ii) Find a₁, a₅, and a₁₀. (iii) What is the first term greater than 50?

Answer:

(i) aₙ = 3n + 2. aₙ₊₁ − aₙ = [3(n+1)+2] − [3n+2] = 3 = constant.

Yes, this is an AP with d = 3.

(ii) a₁ = 3(1)+2 = 5, a₅ = 3(5)+2 = 17, a₁₀ = 3(10)+2 = 32

(iii) 3n + 2 > 50, 3n > 48, n > 16.

So n = 17: a₁₇ = 3(17)+2 = 53

Q6. If the kth term of an AP is (2k + 1) and the sum of the first n terms is expressed using the formula for AP, find d and a.

Answer: kth term = 2k + 1

1st term: a₁ = 2(1) + 1 = 3

2nd term: a₂ = 2(2) + 1 = 5

Common difference: d = 5 − 3 = 2

a = 3, d = 2

This AP is 3, 5, 7, 9…the sequence of odd numbers starting from 3.

HOTS Questions Based on Finding Rules in Sequences

Q1. A sequence starts: 1, 1, 2, 3, 5, 8, 13… Two students claim two different rules. Student A: "Each term is double the previous one." Student B: "Each term is the sum of the two before it." Who is correct? Justify with at least three checks.

Answer: Check Student A (double the previous):

1×2 = 2 ≠ 1 (second term). Student A is wrong at the very first step.

Check Student B (sum of two before):

1 + 1 = 2, 1 + 2 = 3, 2 + 3 = 5, 3 + 5 = 8, 5 + 8 = 13

Student B is correct. The rule is aₙ = aₙ₋₁ + aₙ₋₂ (the Fibonacci sequence).

Q2. Find the rule for the sequence: 1, 4, 9, 16, 25, 36… and write two properties of its terms that always hold.

Answer: Rule: each term is the square of its position number. nth term = n²

Property 1: Every term is a perfect square you can always find an integer whose square equals each term.

Property 2: The difference between consecutive terms increases by 2 each time: 4 − 1 = 3, 9 − 4 = 5, 16 − 9 = 7, 25 − 16 = 9.

The differences form an AP with d = 2. This means n² is not itself an AP.

Q3. A sequence follows the rule: "multiply the position number by itself and subtract the position number." Write the first five terms and find the rule as an expression.

Answer: Rule: aₙ = n × n − n = n² − n = n(n − 1)

• a₁ = 1(0) = 0

• a₂ = 2(1) = 2

• a₃ = 3(2) = 6

• a₄ = 4(3) = 12

• a₅ = 5(4) = 20

Sequence: 0, 2, 6, 12, 20…

These are the products of consecutive integers (also called oblong numbers).

Q4. The sequence 2, 6, 12, 20, 30… does not have a constant difference. Find the rule and the 8th term.

Answer: Differences: 4, 6, 8, 10… (differences increase by 2 each time these are consecutive even numbers).

Look at the terms: 1×2, 2×3, 3×4, 4×5, 5×6…

Rule: aₙ = n(n + 1)

8th term = 8 × 9 = 72

Q5. The differences of a sequence are themselves an AP: 2, 5, 8, 11, 14… If the first term of the original sequence is 3, write the first six terms.

Answer: The differences are 2, 5, 8, 11, 14, 17…

Build the original sequence by adding each difference:

a₁ = 3 a₂ = 3 + 2 = 5 a₃ = 5 + 5 = 10 a₄ = 10 + 8 = 18 a₅ = 18 + 11 = 29 a₆ = 29 + 14 = 43

Original sequence: 3, 5, 10, 18, 29, 43…

This is not an AP (differences vary), but its differences form an AP — it is a quadratic sequence.

HOTS Questions Based on Real-Life Applications of Sequences

Q1. A new company has 10 employees in its first year. It hires 8 more employees each year. In which year will the company have exactly 90 employees?

Answer: AP: a = 10, d = 8. Set aₙ = 90:

10 + (n − 1)(8) = 90

(n − 1)(8) = 80 

n − 1 = 10 

n = 11

The company will have 90 employees in the 11th year.

Q2. A runner trains by running 500 m on Day 1 and increasing the distance by 200 m each day. On which day does the daily run first exceed 3 km (3000 m)?

Answer: AP: a = 500, d = 200. Set aₙ > 3000:

500 + (n − 1)(200) > 3000

(n − 1)(200) > 2500

n − 1 > 12.5

n > 13.5

So n = 14: a₁₄ = 500 + 13(200) = 500 + 2600 = 3100 m

The daily run first exceeds 3 km on Day 14.

Q3. A school library has 200 books in its first year. It receives 50 new books every year. A separate school has 500 books and receives 20 new books per year. In which year will both libraries have the same number of books?

Answer: Library A: 200 + 50(n − 1) books in year n

Library B: 500 + 20(n − 1) books in year n

Set equal: 200 + 50(n−1) = 500 + 20(n−1)

30(n−1) = 300, n − 1 = 10, n = 11

Both libraries will have the same number of books in the 11th year.

Verify: A = 200 + 500 = 700. B = 500 + 200 = 700

Q4. The monthly electricity bill (₹) for a household over six months is: 400, 450, 500, 550, 600, 650. Does this form an AP? Predict the 10th month's bill.

Answer: Differences: 50, 50, 50, 50, 50 all equal.

Yes, it is an AP with a = 400, d = 50.

a₁₀ = 400 + 9(50) = 400 + 450 = ₹850

Q5. Seats in a cinema hall are arranged so that Row 1 has 15 seats, Row 2 has 18, Row 3 has 21, and so on. There are 20 rows. How many seats does the last row have?

Answer: AP: a = 15, d = 3.

a₂₀ = 15 + 19(3) = 15 + 57 = 72 seats in Row 20.

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