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MathematicsClass 9Predicting What Comes Next: Exploring Sequences and

Predicting What Comes Next: Exploring Sequences and | Class 9 Mathematics Notes

By ConceptScroll Team · Published on 17 July 2026 · 3 min read

Predicting What Comes Next: Exploring Sequences and – this guide gives you a concise, exam-ready overview of Predicting What Comes Next: Exploring Sequences and from Class 9 Mathematics, written by ConceptScroll editors and reviewed against the latest NCERT textbook.

Applications of Sequences

Sequences have numerous applications in everyday life and various fields of study. Arithmetic progressions are used in calculating monthly savings, installment payments, and distributing resources evenly. Geometric progressions model population growth, radioactive decay, and compound interest in finance. Recursive sequences like the Fibonacci sequence appear in biology, such as the arrangement of leaves, flowers, and branching patterns in trees. Understanding sequences helps in solving problems involving patterns and predictions. They are also fundamental in computer science algorithms, physics, and economics. This section highlights the importance of sequences beyond mathematics and encourages students to appreciate their practical relevance.

📊 Diagram: Illustrations showing examples of sequences in nature (e.g., sunflower seeds) and finance (compound interest growth graph).

🧪 Activity: Activity 8.7: Students identify sequences in real-life contexts and explain the pattern and type of sequence involved.

🔗 Connection: This section connects to the chapter summary and review exercises reinforcing the concepts learned.

Frequently asked questions

Find the 12th term of a GP with common ratio 2, whose 8th term is 192.

Given: Common ratio r = 2, 8th term t_8 = 192.

Formula for nth term of GP: t_n = a * r^(n-1)

So, t_8 = a 2^(8-1) = a 2^7 = a * 128 = 192 => a = 192 / 128 = 1.5

Now, find 12th term: t_12 = a 2^(12-1) = 1.5 2^11 = 1.5 * 2048 = 3072

Answer: The 12th term is 3072.

Find the 10th and nth terms of the GP: 5, 25, 125, ...

Given GP: 5, 25, 125, ... First term a = 5 Common ratio r = 25 / 5 = 5

10th term: t_10 = a r^(10-1) = 5 5^9 = 5 * 1953125 = 9765625

nth term: t_n = a r^{n-1} = 5 5^{n-1} = 5^n

Answer: 10th term = 9765625 nth term = 5^n

A sequence is given by the recursive rule t_1 = 2, t_{n+1} = 3t_n - 2 for n ≥ 1. Which term of the sequence is 730?

Given: t_1 = 2, t_{n+1} = 3t_n - 2

We want to find n such that t_n = 730.

First, solve the recurrence: Assume t_n = A * 3^{n-1} + B (particular solution)

Substitute into recurrence: t_{n+1} = 3 t_n - 2 => A 3^n + B = 3 (A 3^{n-1} + B) - 2 => A 3^n + B = A 3^n + 3B - 2

Equate constants: B = 3B - 2 => 2B = 2 => B = 1

Use initial condition t_1 = 2: 2 = A * 3^{0} + 1 => A + 1 = 2 => A = 1

So, t_n = 3^{n-1} + 1

Set t_n = 730: 3^{n-1} + 1 = 730 => 3^{n-1} = 729

Since 729 = 3^6,

n - 1

Which term of the GP: 2, 6, 18, ... is 4374? Write the explicit formula as well as the recursive formula for the nth term.

Given GP: 2, 6, 18, ... First term a = 2 Common ratio r = 6 / 2 = 3

Find n such that t_n = 4374

Formula: t_n = a r^{n-1} = 2 3^{n-1}

Set equal: 2 * 3^{n-1} = 4374 => 3^{n-1} = 4374 / 2 = 2187

Note 2187 = 3^7

So, n - 1 = 7 => n = 8

Explicit formula: t_n = 2 3^{n-1} Recursive formula: t_1 = 2, t_{n} = 3 t_{n-1}

Answer: The 8th term is 4374. Explicit formula: t_n = 2 * 3^{n-1} Recursive formula: t_1 = 2, t_n = 3 t_{n-1}

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