NCERT Solutions Ganita Prakash (Part 1) Chapter 4 –97Algebraic Expressions to Describe Patterns — In-text Questions

Book page 96 Updated on2026-09-05

Q1.
Example 12: Somjit noticed a repeating pattern along the border of a saree (A B C D E F …). Somjit wonders if there is a way to describe all the positions where the (i) Design A occurs, (ii) Design B occurs, and (iii) Design C occurs.
Answer

The border repeats in blocks of three: A, B, C, A, B, C, …

DesignPositions where it occursFormula for the nth time
A1, 4, 7, 10, 13, …3n – 2
B2, 5, 8, 11, 14, …3n – 1
C3, 6, 9, 12, 15, …3n
Why it happens: D, E, F in the picture are just A, B, C repeating. Every third position carries C, so C's positions are the multiples of 3. B always sits one place before C, and A two places before.
Q2.
Where would design C appear for the nth time?
Answer

Design C first appears at position 3, then 6, then 9 — the multiples of 3.

nth occurrence of Design C is at position 3n
Check it yourself: The 20th C would be at position 3 × 20 = 60.
Q3.
Similarly, find the formula that gives the position where the other Designs appear for the nth time.
Answer

B occurs at 2, 5, 8, 11, 14, … — always one less than a multiple of 3.

nth occurrence of Design B: 3n – 1
nth occurrence of Design A: 3n – 2
Why it happens: In each block of three, C sits last, B just before it and A just before B. So their positions are 3n, 3n – 1 and 3n – 2.
Q4.
Given a position number can we find out the design that appears there? Which Design appears at Position 122?
Answer

Yes, and the design at position 122 is Design B.

122 = 3 × 41 – 1
So 122 is one less than 123, a multiple of 3
Positions of the form 3n – 1 carry Design B
Why it happens: Every position is a multiple of 3 (→ C), one less than a multiple of 3 (→ B), or two less (→ A). There is no fourth possibility.
Q5.
Can the remainder obtained by dividing the position number by 3 be used for this? Observe the table below. (99 → quotient 33, remainder 0; 122 → quotient 40, remainder 2; 148 → quotient 49, remainder 1)
Answer

Yes. The remainder alone decides the design.

Remainder on dividing by 3Form of the positionDesign
03nC
13n + 1, i.e. 3(n+1) – 2A
23n + 2, i.e. 3(n+1) – 1B
Check it yourself: Position 1 leaves remainder 1 and holds A ✔. Position 2 leaves remainder 2 and holds B ✔. Position 3 leaves remainder 0 and holds C ✔.
Q6.
Use this to find what design appears at positions 99, 122, and 148.
Answer
PositionDivision by 3RemainderDesign
9999 = 3 × 33 + 00C
122122 = 3 × 40 + 22B
148148 = 3 × 49 + 11A
99 = 3 × 33 → Design C
122 = 3 × 41 – 1 → Design B
148 = 3 × 50 – 2 → Design A
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