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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.

8.2 EXPLICIT RULE FOR A SEQUENCE

An explicit rule for a sequence provides a direct formula to find the nth term using the term's position number n. This formula allows us to calculate any term in the sequence without knowing the previous terms. For example, the sequence of odd numbers can be described by the explicit formula uₙ = 2n - 1, where substituting n = 1, 2, 3, ... gives the terms 1, 3, 5, and so on. Having an explicit formula is useful because it enables us to find terms far along in the sequence quickly, check if a number belongs to the sequence, and determine its position. For instance, to find whether 137 is a term in the odd number sequence, we solve 2n - 1 = 137, resulting in n = 69, confirming that 137 is the 69th term. Another example is the sequence defined by sₙ = 5n - 2; the first six terms are 3, 8, 13, 18, 23, 28. To check if 308 is a term, solve 5n - 2 = 308, giving n = 62, so 308 is the 62nd term. However, 471 is not a term since solving 5n - 2 = 471 yields a non-integer n. This demonstrates that n must be a natural number in explicit formulas. Explicit formulas can be found for various sequences, including square numbers, though some sequences like prime numbers do not have simple explicit formulas.

📊 Diagram: No specific diagrams in this section, but references to sequences and their terms.

🧪 Activity: Exercises include finding specific terms using explicit formulas and determining if numbers belong to sequences.

🔗 Connection: Leads to understanding recursive rules, where terms are defined based on previous terms, discussed in the next section.

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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