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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.

Arithmetic Progression

An arithmetic progression (AP) is a sequence in which the difference between any two consecutive terms is constant. This constant difference is called the common difference (d). For example, in the sequence 5, 8, 11, 14, ..., the common difference is 3 because each term increases by 3. The general form of an AP is a, a + d, a + 2d, a + 3d, ..., where 'a' is the first term. The nth term of an AP, denoted by Tn, can be found by the formula Tn = a + (n - 1)d. This formula allows us to find any term in the sequence directly without listing all previous terms. Arithmetic progressions are widely used in daily life, such as calculating monthly savings, distributing items equally, or planning events at regular intervals. Understanding AP helps in solving problems involving evenly spaced numbers and forms the basis for more advanced topics like series and sequences in mathematics.

📊 Diagram: A number line diagram showing terms of an AP: 2, 5, 8, 11, 14 with equal spacing representing the common difference of 3.

🧪 Activity: Activity 8.2: Students identify the common difference in given sequences and write the general term for each.

🔗 Connection: This section prepares students for the next part, where the sum of terms of an arithmetic progression is introduced.

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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