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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 · 4 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.

Geometric Progression

A geometric progression (GP) is a sequence where each term after the first is obtained by multiplying the previous term by a fixed non-zero number called the common ratio (r). For example, the sequence 3, 6, 12, 24, ... is a GP with common ratio 2. The general form of a GP is a, ar, ar², ar³, ..., where 'a' is the first term. The nth term of a GP is given by Tₙ = a × r^(n - 1). Geometric progressions model many real-world phenomena such as population growth, compound interest, and radioactive decay. Understanding GP helps in solving problems involving exponential growth or decay. Unlike arithmetic progression where terms increase or decrease by addition or subtraction, GP involves multiplication or division. This section introduces the concept, formula, and examples to help students grasp the nature of geometric sequences.

📊 Diagram: Number line or bar diagram showing terms of a GP: 1, 2, 4, 8, 16 with increasing gaps representing multiplication by 2.

🧪 Activity: Activity 8.4: Students identify the common ratio in sequences and write the general term for each geometric progression.

🔗 Connection: This section prepares students for the next topic: sum of terms of a geometric progression.

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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