Question 1
↑ Jump to NavigatorFind the 12th term of a GP with common ratio 2, whose 8th term is 192.
Solution
For a GP,
Given:
So,
Now,
Therefore, the 12th term is 3072.
Answer:t₁₂ = 3072
Chapter 8: Predicting What Comes Next: Exploring Sequences and ProgressionsPage 193–194All 7 Questions on This Page
Class 9 Maths Chapter 8 Exercise Set 8.3 Solutions
Complete step-by-step solutions for Class 9 Mathematics Ganita Manjari Chapter 8 Exercise Set 8.3. Covers Geometric Progressions (GP), evaluating nth terms, term position determination, recursive and explicit sequence formulas, bouncing ball distance applications, and Sierpiński square carpet fractal modeling.
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Key Concepts for Exercise Set 8.3
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Geometric Progression (GP)
A sequence where each term after the first is obtained by multiplying the preceding term by a fixed non-zero number called the common ratio (r).
a, ar, ar², ar³, …, arⁿ⁻¹
2
First Term (a) and Common Ratio (r)
The initial term is denoted by a, and the ratio between consecutive terms is constant: r = tₙ / tₙ₋₁.
r = t₂ / t₁ = t₃ / t₂ = …
3
General nth Term of a GP
The formula for the nth term of a geometric progression with first term a and common ratio r.
tₙ = arⁿ⁻¹
4
Finding Term Position (n)
Equate the general term tₙ = arⁿ⁻¹ to the given value and solve for n using equal bases (rⁿ⁻¹ = rᵏ ⟹ n − 1 = k).
arⁿ⁻¹ = Value ⟹ n − 1 = k ⟹ n = k + 1
5
Explicit vs Recursive GP Formulas
An explicit formula gives tₙ directly from n (tₙ = arⁿ⁻¹). A recursive formula links each term to its predecessor (t₁ = a, tₙ = r · tₙ₋₁ for n ≥ 2).
Explicit: tₙ = arⁿ⁻¹ | Recursive: t₁ = a, tₙ = r · tₙ₋₁ (n ≥ 2)
6
Sequences with Linear Recursive Rules
A recursive relation of the form tₙ₊₁ = p · tₙ + q can often be transformed into a GP by substituting uₙ = tₙ − c, where c is a fixed constant.
tₙ₊₁ − 1 = 3(tₙ − 1) ⟹ uₙ₊₁ = 3uₙ
7
Bouncing Ball Vertical Distance
When a ball drops from initial height H and rebounds each time to ratio r, the drop is travelled once, while every rebound height is travelled twice (upward and downward).
Total Distance = H + 2(h₁ + h₂ + … + h₅)
8
Sierpiński Carpet: Count vs Area
Fractal stages starting at Stage 0: the count of red squares multiplies by 8 each stage (tₙ = 8ⁿ), while the area is multiplied by 8/9 each stage (Aₙ = (8/9)ⁿ), approaching 0 as n increases.
Count: tₙ = 8ⁿ | Area: Aₙ = (8/9)ⁿ ⟶ 0 as n ⟶ ∞
Question 2
↑ Jump to NavigatorFind the 10th and nth terms of the GP: 5, 25, 125, … .
Solution
The GP is:
First term,
Common ratio,
For a GP,
For the 10th term:
Therefore:
For the nth term:
Therefore:
Answer:t₁₀ = 5¹⁰ = 9,765,625
tₙ = 5ⁿ
Question 3
↑ Jump to NavigatorA sequence is given by the recursive rule t₁ = 2, tₙ₊₁ = 3tₙ − 2, for n ≥ 1. Which term of the sequence is 730?
Solution
Given:
Subtract 1 from both sides:
Let:
Then:
Also,
Therefore:
and hence,
For 730:
Therefore:
Hence, 730 is the 7th term.
Answer:730 is the 7th term (t₇ = 730).
Question 4
↑ Jump to NavigatorWhich term of the GP: 2, 6, 18, ... is 4374? Write the explicit formula as well as the recursive formula for the nth term.
Solution
The GP is:
First term,
Common ratio,
Which term is 4374?
Therefore:
Hence, 4374 is the 8th term.
Explicit formula:
For a GP,
Therefore:
Recursive formula:
Each term is obtained by multiplying the previous term by 3:
Final result:
Explicit formula:
Recursive formula:
Answer:4374 is the 8th term (t₈ = 4374) | Explicit: tₙ = 2(3)ⁿ⁻¹ | Recursive: t₁ = 2, tₙ = 3tₙ₋₁ (n ≥ 2)
Question 5
↑ Jump to NavigatorA ball is dropped from a height of 80 metres. After hitting the ground, it bounces back to 60% of the height from which it fell. It continues bouncing in this way — each time rising to 60% of the previous height. (i) What height does the ball reach after the 5th bounce? (ii) What is the total vertical distance the ball has travelled by the time it hits the ground for the 6th time?
Solution
The heights reached after successive bounces form a GP.
First bounce:
So,
For a GP,
(i) Height after the 5th bounce:
Therefore, the ball reaches a height of:
6.2208 m
(ii) Total vertical distance:
The heights reached after the first five bounces are:
48, 28.8, 17.28, 10.368, 6.2208
Their sum is:
48 + 28.8 + 17.28 + 10.368 + 6.2208
The ball first falls through:
80 m
After every bounce, the ball travels upward and then downward through the same height.
Therefore:
Therefore, the total vertical distance travelled is:
301.3376 m
(Note: The initial 80 m fall is the first time the ball hits the ground. Therefore, the 6th ground contact occurs after five bounce heights have been travelled upward and downward.)
Answer:(i) 6.2208 m | (ii) 301.3376 m
Question 6
↑ Jump to NavigatorWhich term of the sequence 2, 2√2, 4, ... is 128?
Solution
The sequence is:
2, 2√2, 4, …
This is a GP.
First term,
Common ratio,
For a GP,
Therefore:
For 128:
Since,
we get,
Therefore:
Hence:
128 is the 13th term.
Answer:128 is the 13th term (t₁₃ = 128).
Question 7
↑ Jump to NavigatorFig. 8.12 shows Stages 0 to 3 of the Sierpiński square carpet. Stage 0 of this fractal is a square sheet of paper. To construct Stage 1, each side of the square is trisected and the points of trisection of opposite sides are joined to obtain nine smaller squares. The centre square is then removed and the 8 smaller squares are retained, leaving a square hole in the centre. The same process is repeated on the eight smaller shaded squares to obtain Stage 2 and so on. Look at Fig. 8.12 and try to answer the following questions.
Fig. 8.12: Sierpiński Square Carpet (Stages 0 to 3)
Stages 0 to 3 of the Sierpiński square carpet fractal: at each stage, each red square is trisected into 9 smaller squares and the centre square is removed.
- (i)How many red squares are there in Stages 0 to 3?
- (ii)Can you predict the number of red squares in Stages 4 and 5?
- (iii)Can you find a rule for the number of red squares at the nth stage? Write the explicit formula as well as the recursive formula for the number of red squares at any stage.
- (iv)Suppose the area of the square in Stage 0 is 1 square unit. What is the area of the red region in Stages 1, 2 and 3? What will be the area of the red region in Stages 4 and 5? Find the explicit as well as the recursive formula for the area of the red region at the nth stage. What happens to this area as n, the number of stages, goes on increasing?
Solution
(i) Number of red squares in Stages 0 to 3:
The number of red squares in Stages 0 to 3 is:
Therefore:
(ii) Number of red squares in Stages 4 and 5:
Each stage has 8 times as many red squares as the previous stage.
Stage 4:
Stage 5:
Therefore:
(iii) Rule for the number of red squares at the nth stage:
The number of red squares is:
Since Stage 0 has 1 red square:
Explicit formula:
Each stage has 8 times as many red squares as the previous stage.
Therefore:
For example:
which matches the pattern.
Recursive formula:
Starting with:
Each stage has 8 times as many red squares as the previous stage.
Therefore:
Final result:
Explicit formula:
Recursive formula:
(iv) Area of the red region at each stage:
Given:
At each stage, the square is divided into 9 equal smaller squares and the centre square is removed.
So, the remaining red area becomes 8/9 of the previous stage's area:
For Stage 1:
For Stage 2:
For Stage 3:
For Stage 4:
For Stage 5:
Therefore:
Explicit formula:
Recursive formula:
Since:
the red area keeps decreasing as the number of stages increases.
Therefore, the red area approaches 0 as n increases.
Answer:(i) 1, 8, 64, 512 | (ii) Stage 4 = 4096, Stage 5 = 32768 | (iii) Explicit: tₙ = 8ⁿ; Recursive: t₀ = 1, tₙ = 8tₙ₋₁ (n ≥ 1) | (iv) Explicit: Aₙ = (89)ⁿ; Recursive: A₀ = 1, Aₙ = (89)Aₙ₋₁ (n ≥ 1); Area approaches 0 as n increases.
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