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MathematicsClass 9Predicting What Comes Next: Exploring Sequences and Progressions

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

By ConceptScroll Team · Published on 17 July 2026 · 4 min read

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

8.3 RECURSIVE RULE FOR A SEQUENCE

A recursive rule defines each term of a sequence based on one or more preceding terms. Unlike explicit formulas, recursive formulas require knowledge of previous terms to find the next term. For example, the sequence 1, 4, 7, 10, 13, ... has the recursive rule t₁ = 1 and tₙ = tₙ₋₁ + 3 for n ≥ 2, meaning each term is obtained by adding 3 to the previous term. Another example is u₁ = 1, uₙ = 2uₙ₋₁ + 3 for n ≥ 2, where terms grow by doubling the previous term and adding 3. Calculating the first four terms gives 1, 5, 13, 29. Recursive rules can also involve more than one previous term, such as the Virahānka–Fibonacci sequence defined by V₁ = 1, V₂ = 2, and Vₙ = Vₙ₋₁ + Vₙ₋₂ for n ≥ 3, producing the sequence 1, 2, 3, 5, 8, 13, 21, 34, ... Each term is the sum of the two preceding terms. This sequence, first studied by Indian mathematicians Virahānka, Gopāla, and Hemachandra, and later by Fibonacci, appears in many areas of mathematics and science. Recursive rules are powerful for describing sequences where terms depend on previous values.

📊 Diagram: Fig. 8.1 (triangular numbers) and Fig. 8.2 (square numbers) relate to sequences; no new diagrams specific to recursive rules.

🧪 Activity: Exercises include finding terms of sequences defined by recursive rules and identifying terms.

🔗 Connection: Paves the way to study special sequences like arithmetic and geometric progressions in the following sections.

Frequently asked questions

Consider the sequence 1, 4, 7, 10, 13, ... Can you predict the next four terms? Can you derive the first 10 terms of the sequence obtained by adding all the terms up to a given term of this sequence? (Hint: The first term is 1. The second term is $1 + 4 = 5$, the third term is $1 + 4 + 7 = 12$, and so on.)

Step 1: Identify the pattern in the sequence 1, 4, 7, 10, 13, ... The sequence increases by 3 each time, so it is an arithmetic progression (AP) with first term a = 1 and common difference d = 3.

Step 2: Predict the next four terms: 6th term = 13 + 3 = 16 7th term = 16 + 3 = 19 8th term = 19 + 3 = 22 9th term = 22 + 3 = 25

So, the next four terms are 16, 19, 22, 25.

Step 3: Derive the first 10 terms of the sequence obtained by adding all terms up to a given term. This is the sequence of parti

Can you write $t_5, t_6, t_7$ and $t_8$ for the sequence of triangular numbers?

The sequence of triangular numbers is given by the formula $t_n = \frac{n(n+1)}{2}$. Using this, we calculate:

$t_5 = \frac{5 \times 6}{2} = \frac{30}{2} = 15$

$t_6 = \frac{6 \times 7}{2} = \frac{42}{2} = 21$

$t_7 = \frac{7 \times 8}{2} = \frac{56}{2} = 28$

$t_8 = \frac{8 \times 9}{2} = \frac{72}{2} = 36$

Thus, $t_5=15$, $t_6=21$, $t_7=28$, and $t_8=36$.

Using the explicit rule $u_n = 2n - 1$, find the $53^{\text{rd}}$ term, the $108^{\text{th}}$ term, and the $1170^{\text{th}}$ term of the odd number sequence.

Given the explicit rule $u_n = 2n - 1$, we find the terms as follows:

(i) For the 53rd term: $u_{53} = 2 \times 53 - 1 = 106 - 1 = 105$

(ii) For the 108th term: $u_{108} = 2 \times 108 - 1 = 216 - 1 = 215$

(iii) For the 1170th term: $u_{1170} = 2 \times 1170 - 1 = 2340 - 1 = 2339$

Thus, the 53rd term is 105, the 108th term is 215, and the 1170th term is 2339.

Can you find the rule describing the $n^{th}$ term of the sequence of square numbers?

The sequence of square numbers is: 1, 4, 9, 16, 25, ... The nth term of this sequence is given by $u_n = n^2$. This means the term at position n is the square of n.

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