Case Study on Class 9 Maths Chapter 8 ‘Predicting What Comes Next - Exploring Sequences and Progressions’

The Case Study Questions for Class 9 Maths Chapter 8 "Predicting What Comes Next - Exploring Sequences and Progressions" include short, real life problem situations that have clear answers and step by step solutions to help students gain confidence for exams. It covers important topics including identifying patterns in sequences, understanding arithmetic progressions (AP) with common difference, exploring geometric progressions (GP) with common ratio, finding the nth term of a sequence, calculating the sum of terms in AP and GP, distinguishing between finite and infinite sequences, applying recursive and explicit rules, recognizing the Virahanka-Fibonacci sequence, and solving word problems involving sequences in daily life situations like savings plans, population growth, and depreciation. These practice questions help the students in better understanding of the concepts, handling sequence problems smoothly and to be faster and accurate for their board exams. A free PDF is included for offline practice.

Introduction to Case Study Questions on Sequences and Progressions

Understanding Sequences

A sequence is an ordered list of numbers in which each term follows a rule. The terms are written as a₁, a₂, a₃… where the subscript shows the position. To understand a sequence, always find the rule connecting consecutive terms.

Understanding Progressions

An Arithmetic Progression (AP) is a sequence where the difference between any two consecutive terms is always the same. This fixed difference is the common difference d. If d is positive, the sequence increases. If d is negative, it decreases.

Key formula: aₙ = a + (n − 1)d

Where a = first term, d = common difference, n = position, aₙ = nth term.

Arithmetic Patterns and Rules

Before applying the formula, always verify the sequence is arithmetic by checking two or three differences. If a₂ − a₁ = a₃ − a₂ = a₄ − a₃, the sequence is an AP and the formula applies.

Finding Missing Terms

If a term in the middle of a sequence is unknown, use the fact that consecutive differences must be equal. Set up a simple equation and solve. For example, if the sequence is 5, ?, 17, the missing term x must satisfy x − 5 = 17 − x, giving 2x = 22, so x = 11.

Case Study 1: Growing Number Patterns

A mathematics teacher writes the following sequence on the board for her Class 9 students: 4, 9, 14, 19, 24… She tells them this pattern represents the number of students who have joined a school club over consecutive weeks, starting from Week 1. She asks the class to study the pattern and answer questions based on it.

Questions:

(i) What is the first term and common difference of this sequence?

(ii) Is this sequence an arithmetic progression? Justify.

(iii) Write the nth term formula for this sequence.

(iv) How many students will have joined by Week 10?

(v) In which week will the total reach 54?

Solution:

(i) First term a = 4. Common difference d = 9 − 4 = 5.

(ii) Yes. Check: 9 − 4 = 5, 14 − 9 = 5, 19 − 14 = 5, 24 − 19 = 5. The difference is constant throughout. This confirms it is an AP with d = 5.

(iii) aₙ = a + (n − 1)d = 4 + (n − 1) × 5 = 4 + 5n − 5 = 5n − 1.

(iv) a₁₀ = 5(10) − 1 = 50 − 1 = 49 students will join in Week 10.

(v) Set aₙ = 54: 5n − 1 = 54 → 5n = 55 → n = 11. The total reaches 54 in Week 11.

Case Study 2: Seating Arrangement in a School Auditorium

The principal of a school is arranging seats for the annual function. The first row has 20 seats, the second row has 24 seats, the third row has 28 seats, and this pattern continues throughout the auditorium. There are a total of 15 rows in the auditorium.

Questions:

(i) Identify the first term and common difference of this AP.

(ii) Write the general term (nth term) for the number of seats in the nth row.

(iii) How many seats are in the 15th row?

(iv) Which row has exactly 48 seats?

(v) The school wants to add a 16th row. How many seats should it have to continue the pattern?

Solution:

(i) First term a = 20 (Row 1). Common difference d = 24 − 20 = 4.

(ii) aₙ = 20 + (n − 1) × 4 = 20 + 4n − 4 = 4n + 16.

(iii) a₁₅ = 4(15) + 16 = 60 + 16 = 76 seats in the 15th row.

(iv) Set aₙ = 48: 4n + 16 = 48 → 4n = 32 → n = 8. Row 8 has exactly 48 seats.

(v) a₁₆ = 4(16) + 16 = 64 + 16 = 80 seats. The 16th row should have 80 seats.

Case Study 3: Matchstick Pattern Formation

During a craft class, students are asked to make a row of pentagons using matchsticks. The first pentagon uses 5 matchsticks. Each additional pentagon added to the row shares one side with the previous one and uses 4 more matchsticks. The pattern continues as more pentagons are added.

Questions:

(i) Write the number of matchsticks used for 1, 2, 3, and 4 pentagons.

(ii) Is the number of matchsticks an arithmetic progression? What is d?

(iii) Write the nth term formula for the number of matchsticks needed for n pentagons.

(iv) How many matchsticks are needed to make a row of 12 pentagons?

(v). A student has 45 matchsticks. What is the maximum number of pentagons she can make?

Solution:

(i) Pentagon 1: 5 sticks. Pentagon 2: 5 + 4 = 9 sticks. Pentagon 3: 9 + 4 = 13 sticks. Pentagon 4: 13 + 4 = 17 sticks. Sequence: 5, 9, 13, 17…

(ii) Yes. Check: 9 − 5 = 4, 13 − 9 = 4, 17 − 13 = 4. Constant difference. AP with d = 4.

(iii) a = 5, d = 4. aₙ = 5 + (n − 1) × 4 = 5 + 4n − 4 = 4n + 1.

(iv) a₁₂ = 4(12) + 1 = 48 + 1 = 49 matchsticks.

(v) Set aₙ ≤ 45: 4n + 1 ≤ 45, 4n ≤ 44, n ≤ 11. She can make a maximum of 11 pentagons (which needs 4(11) + 1 = 45 matchsticks exactly).

Case Study 4: Saving Money Every Month

Kavya decides to start saving money from January. In January, she saves ₹200. Each month after that, she saves ₹50 more than the previous month. So in February she saves ₹250, in March ₹300, and so on. She follows this plan for a full year (12 months).

Questions:

(i) List Kavya's savings for the first four months. Is it an AP?

(ii) What is the first term and common difference of this AP?

(iii) Write the nth term formula for her monthly savings.

(iv) How much does Kavya save in the 12th month (December)?

(v) In which month does her monthly saving first exceed ₹600?

Solution:

(i) January: ₹200, February: ₹250, March: ₹300, April: ₹350. Differences: 250 − 200 = 50, 300 − 250 = 50. Yes, it is an AP with constant difference ₹50.

(ii). First term a = ₹200. Common difference d = ₹50.

(iii) aₙ = 200 + (n − 1) × 50 = 200 + 50n − 50 = 50n + 150.

(iv) a₁₂ = 50(12) + 150 = 600 + 150 = ₹750 saved in December.

(v) Set aₙ > 600: 50n + 150 > 600, 50n > 450, n > 9. So in the 10th month (October), her saving first exceeds ₹600 (a₁₀ = 50(10) + 150 = ₹650).

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