Q1. Find the 12th term of a GP with common ratio 2, whose 8th term is 192.
Solution:
Identify the components of the given GP First, let’s look at the information given to us for this Geometric Progression (GP).
Common ratio, r = 2
8th term, t_8 = 192
The formula for the n^{th} term (t_n) of a GP is:
t_n = a \times r^{n - 1}
To find the first term (a), we plug n = 8 into our formula, along with our values for r and t_8.
r = 2
t_8 = 192
n = 8
t_n = a \times r^{n - 1}
192 = a \times 2^{8 - 1}
192 = a \times 2^7
192 = a \times 128
a = \frac{192}{128}
a = \frac{3}{2}
So, the first term of this GP is \frac{3}{2}.
Now, we repeat the exact same process to find the 12th term.
This time, we substitute n = 12.
a = \frac{3}{2}
r = 2
n = 12
t_n = a \times r^{n - 1}
t_{12} = \frac{3}{2} \times 2^{12 - 1}
t_{12} = \frac{3}{2} \times 2^{11}
t_{12} = \frac{3}{2} \times 2048
t_{12} = 3 \times 1024
t_{12} = 3072
So, the 12th term of this GP is 3072.
Final Answer: \boxed {t_{12} = 3072}
Q2. Find the 10^{th} and n^{th} terms of the GP: 5, 25, 125, …
Solution:
Identify the components of the given GP First, let’s look at the information given to us for this Geometric Progression (GP).
First term, a = 5
Common ratio, r = \frac{25}{5} = 5
The formula for the n^{th} term (t_n) of a GP is:
t_n = a \times r^{n - 1}
To find the 10th term, we plug n = 10 into our formula, along with our values for a and r.a = 5
r = 5
n = 10
t_n = a \times r^{n - 1}
t_{10} = 5 \times 5^{10 - 1}
t_{10} = 5 \times 5^9
t_{10} = 5^{10}
t_{10} = 9765625
So, the 10th term of this GP is 9765625.
Now, we repeat the exact same process to find the n^{th} term.
This time, we leave n as n.
a = 5
r = 5
t_n = a \times r^{n - 1}
t_n = 5 \times 5^{n - 1}
t_n = 5^{1 + (n - 1)}
t_n = 5^n
So, the n^{th} term of this GP is 5^n.
Final Answer: \boxed {t_{10} = 9765625, \ t_n = 5^n}
Q3. A sequence is given by the recursive rule t_1 = 2, t_{n+1} = 3t_n - 2 for n \ge 1. Which term of the sequence is 730?
Solution:
Identify the components of the given sequence First, let’s look at the information given to us for this recursive sequence.
First term, t_1 = 2 Recursive rule, t_{n+1} = 3t_n - 2
To find which term is 730, we will calculate each consecutive term using the rule until we reach 730.
For n = 1:
t_2 = 3(t_1) - 2
t_2 = 3(2) - 2
t_2 = 6 - 2
t_2 = 4
For n = 2:
t_3 = 3(t_2) - 2
t_3 = 3(4) - 2
t_3 = 12 - 2
t_3 = 10
For n = 3:
t_4 = 3(t_3) - 2
t_4 = 3(10) - 2
t_4 = 30 - 2
t_4 = 28
For n = 4:
t_5 = 3(t_4) - 2
t_5 = 3(28) - 2
t_5 = 84 - 2
t_5 = 82
For n = 5:
t_6 = 3(t_5) - 2
t_6 = 3(82) - 2
t_6 = 246 - 2
t_6 = 244
For n = 6:
t_7 = 3(t_6) - 2
t_7 = 3(244) - 2
t_7 = 732 - 2
t_7 = 730
So, 730 is the 7th term of this sequence.
Final Answer: \boxed {\text{7th term}}
Q4. Which term of the GP: 2, 6, 18, … is 4374? Write the explicit formula as well as the recursive formula for the n^{th} term.
Solution:
Identify the components of the given GP First, let’s look at the information given to us for this Geometric Progression (GP).
First term, a = 2
Common ratio, r = \frac{6}{2} = 3
The formula for the n^{th} term (t_n) of a GP is:
t_n = a \times r^{n - 1}
To find which term is 4374, we plug t_n = 4374 into our formula, along with our values for a and r.
a = 2
r = 3
t_n = 4374
t_n = a \times r^{n - 1}
4374 = 2 \times 3^{n - 1}
2187 = 3^{n - 1}
3^7 = 3^{n - 1}
7 = n - 1
n = 8
So, 4374 is the 8th term of this GP.
Now, we write the explicit and recursive formulas.
The explicit formula calculates the value directly from the position n.
Explicit formula: t_n = 2 \times 3^{n - 1}
The recursive formula calculates the value from the previous term.
Recursive formula: t_1 = 2, t_n = 3 \times t_{n-1} for n \ge 2
Final Answer: \boxed {\text{8th term; Explicit: } t_n = 2 \times 3^{n - 1} \text{; Recursive: } t_1 = 2, t_n = 3 \times t_{n-1}}
Q5. A 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 6^{th} time?
Solution:
Identify the components of the given GP First, let’s look at the information given to us for this bouncing ball problem.
Initial drop height, H = 80 \text{ m}
Common ratio (bounce back percentage), r = 60% = 0.6
The maximum heights of the bounces form a GP where the first term (height of the 1st bounce) is a = 80 \times 0.6 = 48 \text{ m}.
The formula for the height after the n^{th} bounce (h_n) is:
h_n = a \times r^{n - 1}
To find the height after the 5th bounce (i), we plug n = 5 into our formula.
a = 48
r = 0.6
n = 5
h_n = a \times r^{n - 1}
h_5 = 48 \times 0.6^{5 - 1}
h_5 = 48 \times 0.6^4
h_5 = 48 \times 0.1296
h_5 = 6.2208 \text{ m}
So, the height after the 5th bounce is 6.2208 metres.
Now, we find the total vertical distance travelled when it hits the ground for the 6th time (ii).
This distance includes the initial drop (80 m) plus going UP and DOWN for the first 5 bounces.
Total distance = 80 + 2(h_1 + h_2 + h_3 + h_4 + h_5)
We use the sum of a GP formula for the 5 bounces:
S_n = a \frac{1 - r^n}{1 - r}.
S_5 = 48 \frac{1 - 0.6^5}{1 - 0.6}
S_5 = 48 \frac{1 - 0.07776}{0.4}
S_5 = 120 \times 0.92224
S_5 = 110.6688 \text{ m}
Now, multiply by 2 (for the up and down movement of each bounce) and add the initial drop:
Total distance = 80 + 2(110.6688)
Total distance = 80 + 221.3376
Total distance = 301.3376 \text{ m}
So, the total distance is 301.3376 metres.
Final Answer: \boxed {\text{(i) } 6.2208 \text{ m, (ii) } 301.3376 \text{ m}}
Q6. Which term of the sequence 2\sqrt{2}, 4, … is 128? (Teacher’s Note: The PDF has a slight typo in the question text reading “2\sqrt{2}, is 128″, but the mathematical standard for this progression dictates the next term is 4.)
Solution:
Identify the components of the given GP First, let’s look at the information given to us for this Geometric Progression (GP). First term, a = 2\sqrt{2} Common ratio, r = \frac{4}{2\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}
The formula for the n^{th} term (t_n) of a GP is:
t_n = a \times r^{n - 1}
To find which term is 128, we plug t_n = 128 into our formula, along with our values for a and r.
a = 2\sqrt{2}
r = \sqrt{2}
t_n = 128
t_n = a \times r^{n - 1}
128 = 2\sqrt{2} \times (\sqrt{2})^{n - 1}
128 = (\sqrt{2})^2 \times (\sqrt{2})^1 \times (\sqrt{2})^{n - 1}
128 = (\sqrt{2})^3 \times (\sqrt{2})^{n - 1}
2^7 = (\sqrt{2})^{n - 1 + 3}
(\sqrt{2})^{14} = (\sqrt{2})^{n + 2}
14 = n + 2
n = 12
So, 128 is the 12th term of this sequence.
Final Answer: \boxed {\text{12th term}}
Q7. Fig shows Stages 0 to 3 of the Sierpiński square carpet.
(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 n^{th} 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…
Solution:
Identify the components of the given GP First, let’s look at the information given to us for the Sierpinski carpet fractal.
At each stage, 1 square is broken down, and the center is removed, leaving 8 smaller red squares.
First term (Stage 0), a = 1 Common ratio, r = 8
The formula for the number of squares at Stage n is:
t_n = a \times r^n (Since Stage 0 is the starting point, we use n instead of n-1).
(i) we list the total number of squares for Stages 0 to 3:
Stage 0: 1
Stage 1: 1 \times 8 = 8
Stage 2: 8 \times 8 = 64
Stage 3: 64 \times 8 = 512
(ii) we calculate the number of squares for Stages 4 and 5:
Stage 4: 512 \times 8 = 4096
Stage 5: 4096 \times 8 = 32768
(iii) we write the rules for the number of squares at the n^{th} stage:
Explicit formula: t_n = 8^n
Recursive formula: t_0 = 1,
t_n = 8 \times t_{n-1}
(iv) we track the area.
The total area gets multiplied by \frac{8}{9} at each stage because 1 out of 9 equal sections is completely removed.
Stage 1 Area: \frac{8}{9}
Stage 2 Area: (\frac{8}{9})^2 = \frac{64}{81}
Stage 3 Area: (\frac{8}{9})^3 = \frac{512}{729}
Stage 4 Area: (\frac{8}{9})^4 = \frac{4096}{6561}
Stage 5 Area: (\frac{8}{9})^5 = \frac{32768}{59049}
Explicit formula for area: Area_n = (\frac{8}{9})^n
Recursive formula for area:
Area_0 = 1,
Area_n = \frac{8}{9} \times Area_{n-1}
As n (the number of stages) goes on increasing, the area keeps multiplying by a fraction less than 1, meaning the remaining red area gradually approaches 0.
Final Answer: \boxed {\text{Explicit Area } = \left(\frac{8}{9} \right)^n, \text{ Area approaches 0 as stages increase}}