Can you predict the number of squares in Stages 5 and 6 of the pattern? In Stages 10, 11 and 12? In Stage 20? At any stage? How is this different from the growing pattern in Fig. 8.3? Stage 1 Stage 2 Stage 3 Stage 4 Fig. 8.6 — Stages 1 to 4 of a growing pattern of squares. Stage 1 Stage 2 Stage 3 St

Question

Accepted Answer

Each stage of Fig. 8.6 is a rectangle 3 squares wide whose height doubles: 1, 2, 4, 8 rows. So the counts double rather than grow by a fixed amount.

3, 6, 12, 24, 48, 96, …
t₁ = 3, t₂ = 3 × 2, t₃ = 3 × 2², t₄ = 3 × 2³
t_n = 3 × 2^n–1

Stage	5	6	10	11	12	20	n
Number of squares	48	96	1536	3072	6144	15,72,864	3 × 2^n–1

Stage 10: 3 × 2⁹ = 3 × 512 = 1536
Stage 20: 3 × 2¹⁹ = 3 × 524288 = 15,72,864

How it differs from Fig. 8.3:

	Fig. 8.3	Fig. 8.6
Step from one stage to the next	add 4	multiply by 2
Sequence	1, 5, 9, 13, 17, …	3, 6, 12, 24, 48, …
Constant that describes it	common difference d = 4	common ratio r = 2
nth term	4n – 3	3 × 2^n–1
Type	AP	GP
At Stage 20	77	15,72,864

Why it happens: repeated addition puts n in the formula as a plain multiplier, so the terms grow in proportion to n. Repeated multiplication puts n in the exponent, and an exponent compounds — each doubling acts on everything already accumulated. That is why two patterns which look similar at Stage 4 (13 against 24) are worlds apart at Stage 20 (77 against more than fifteen lakh).

Check it yourself: 3 × 2^n–1 must give 3 at n = 1. And 2⁰ = 1, so it does.

NCERT Solutions for Class 9th Maths Chapter 8 Think and Reflect — Geometric Progressions