Q1. Exercise Set 8.3, Question 1 (Page 20): Find the 12th term of a GP with common ratio 2, whose 8th term is 192.
Answer:
Let a be the first term of the GP. The common ratio is r = 2.
We are given that the 8th term is t₈ = a × 2⁷ = 192.
a × 128 = 192 → a = 192/128 = 3/2.
The 12th term is: t₁₂ = a × 2¹¹ = (3/2) × 2048 = 3 × 1024 = 3072.
Thus, the 12th term is 3072.
Q2. Exercise Set 8.3, Question 2 (Page 20): Find the 10th and nth terms of the GP: 5, 25, 125, ...
Answer:
The GP is 5, 25, 125, ... with first term a = 5 and common ratio r = 25/5 = 5.
The nth term is: tn = 5 × 5n-1 = 5n.
The 10th term is: t₁₀ = 5¹⁰ = 9,765,625.
Q3. Exercise Set 8.3, Question 3 (Page 20): A sequence is given by the recursive rule t1 = 2, tn+1 = 3tn - 2 for n ≥ 1. Which term of the sequence is 730?
Answer:
Let's find the first few terms:
- t₁ = 2
- t₂ = 3(2) - 2 = 4
- t₃ = 3(4) - 2 = 10
- t₄ = 3(10) - 2 = 28
- t₅ = 3(28) - 2 = 82
- t₆ = 3(82) - 2 = 244
- t₇ = 3(244) - 2 = 730
So, 730 is the 7th term of the sequence.
Q4. Exercise Set 8.3, Question 4 (Page 20): Which term of the GP: 2, 6, 18, ... is 4374? Write the explicit formula as well as the recursive formula for the nth term.
Answer:
The GP is 2, 6, 18, ... with first term a = 2 and common ratio r = 3.
Explicit formula: tn = 2 × 3n-1
Recursive formula: t₁ = 2, tn = 3 × tn-1 for n ≥ 2.
Which term is 4374?
Set tn = 2 × 3n-1 = 4374.
3n-1 = 2187.
We know 3⁷ = 2187, so n-1 = 7 → n = 8.
Thus, 4374 is the 8th term of the GP.
Q5. Exercise Set 8.3, Question 5 (Page 20): 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 6th time?
Answer:
(i) Height after the 5th bounce:
Initial drop height = 80 m. Each bounce reaches 60% (or 0.6) of the previous height.
Height after 1st bounce = 80 × 0.6 = 48 m.
Height after 2nd bounce = 48 × 0.6 = 28.8 m.
Height after 3rd bounce = 28.8 × 0.6 = 17.28 m.
Height after 4th bounce = 17.28 × 0.6 = 10.368 m.
Height after 5th bounce = 10.368 × 0.6 = 6.2208 m.
Alternatively, height after nth bounce = 80 × (0.6)n. For n=5, height = 80 × 0.6⁵ = 80 × 0.07776 = 6.2208 m.
(ii) Total distance travelled by the time it hits the ground for the 6th time:
This is the sum of the distances for each fall and bounce up. The ball travels:
- 1st fall: 80 m (hits ground 1st time)
- 1st bounce up: 48 m; 2nd fall: 48 m (hits ground 2nd time)
- 2nd bounce up: 28.8 m; 3rd fall: 28.8 m (hits ground 3rd time)
- 3rd bounce up: 17.28 m; 4th fall: 17.28 m (hits ground 4th time)
- 4th bounce up: 10.368 m; 5th fall: 10.368 m (hits ground 5th time)
- 5th bounce up: 6.2208 m; 6th fall: 6.2208 m (hits ground 6th time)
Total distance = 80 + 2 × (48 + 28.8 + 17.28 + 10.368 + 6.2208).
Sum of bounce heights = 48 + 28.8 + 17.28 + 10.368 + 6.2208 = 110.6688 m.
Total distance = 80 + 2 × 110.6688 = 80 + 221.3376 = 301.3376 m.
Thus, the total distance travelled is 301.3376 metres.
Q6. Exercise Set 8.3, Question 6 (Page 21): Which term of the sequence 2, 2√2, 4, ... is 128?
Answer:
The sequence is 2, 2√2, 4, ...
Find the common ratio: r = (2√2)/2 = √2. Also, 4/(2√2) = 2/√2 = √2. So it's a GP with a = 2, r = √2.
The explicit formula is: tn = 2 × (√2)n-1.
We need tn = 128.
2 × (√2)n-1 = 128 → (√2)n-1 = 64.
Since 64 = 2⁶ and √2 = 21/2, we have (21/2)n-1 = 2⁶ → 2(n-1)/2 = 2⁶.
Equating exponents: (n-1)/2 = 6 → n-1 = 12 → n = 13.
Thus, 128 is the 13th term of the sequence.
Q7. Exercise Set 8.3, Question 7 (Page 21): Fig. 8.12 shows Stages 0 to 3 of the Sierpinski 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. (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?
Answer:
(i) Number of red squares in Stages 0 to 3:
- Stage 0: 1 square
- Stage 1: 8 squares (9 smaller squares, 1 removed)
- Stage 2: 8 × 8 = 64 squares
- Stage 3: 64 × 8 = 512 squares
(ii) Stages 4 and 5:
- Stage 4: 512 × 8 = 4096 squares
- Stage 5: 4096 × 8 = 32768 squares
(iii) Rule for the number of red squares at the nth stage:
The number of red squares at each stage is multiplied by 8. This is a GP with a = 1 and r = 8.
Explicit formula: tn = 1 × 8n = 8n (where Stage 0 is n=0). If using n starting from 1 for Stage 0, tn = 8n-1.
Recursive formula: t₀ = 1, tn = 8 × tn-1 for n ≥ 1.
(iv) Area of the red region:
Stage 0 area = 1. In each stage, the square is divided into 9 equal parts and 1 is removed, so the red area is multiplied by 8/9.
- Stage 1: 8/9
- Stage 2: (8/9)² = 64/81
- Stage 3: (8/9)³ = 512/729
- Stage 4: (8/9)⁴ = 4096/6561
- Stage 5: (8/9)⁵ = 32768/59049
Explicit formula for area: sn = (8/9)n
Recursive formula for area: s₀ = 1, sn = (8/9) × sn-1 for n ≥ 1.
What happens as n increases? As n increases, (8/9)n gets smaller and approaches 0. The total red area diminishes and gets infinitely close to 0, even though the number of squares increases exponentially.