Mathematics solution NCERT
Class 9 - Chapter 8: Predicting What Comes Next: Exploring Sequences and Progressions
Q1: Find the 12th Term of a GP
Given:
Common ratio, r = 2
8th term = 192
Formula: Tn = arn−1
T8 = ar7
192 = a(27)
192 = 128a
a = 192/128 = 3/2
T12 = a(211)
= (3/2)(2048)
= 3072
Answer: T12 = 3072
Q2: Find the 10th and nth Terms of the GP
Given GP:
5, 25, 125, ...
a = 5
r = 5
10th Term
T10 = 5(59)
= 510
= 9,765,625
Answer: T10 = 9,765,625
nth Term
Tn = arn−1
= 5 × 5n−1
= 5n
Answer:
Tn = 5n
Q3: Which Term of the Sequence is 730?
t1 = 2
tn+1 = 3tn − 2
First few terms:
| Term | Value |
|---|---|
| t1 | 2 |
| t2 | 4 |
| t3 | 10 |
| t4 | 28 |
| t5 | 82 |
| t6 | 244 |
| t7 | 730 |
Answer: 730 is the 7th term.
Q4: Which Term of the GP 2, 6, 18, ... is 4374?
a = 2
r = 3
Tn = 2(3n−1)
4374 = 2(3n−1)
2187 = 3n−1
2187 = 37
n − 1 = 7
n = 8
Answer: 4374 is the 8th term.
Explicit Formula
Tn = 2(3n−1)
Recursive Formula
T1 = 2
Tn+1 = 3Tn
Q5: Bouncing Ball
Initial height = 80 m
Common ratio = 60% = 0.6
(i) Height after the 5th Bounce
Height after n bounces:
Hn = 80(0.6)n
H5 = 80(0.6)5
= 80 × 0.07776
= 6.2208 m
Answer: 6.2208 m
(ii) Total Distance Before Hitting Ground for the 6th Time
Downward journeys: 6 times
Upward journeys: 5 times
| Bounce | Height (m) |
|---|---|
| 1 | 48 |
| 2 | 28.8 |
| 3 | 17.28 |
| 4 | 10.368 |
| 5 | 6.2208 |
Total Distance = 80 + 2(48 + 28.8 + 17.28 + 10.368 + 6.2208)
= 80 + 221.3376
= 301.3376 m
Answer: 301.3376 m
Q6: Which Term of the Sequence 2, 2√2, 4, ... is 128?
a = 2
r = √2
Tn = 2(√2)n−1
128 = 2(√2)n−1
64 = (√2)n−1
64 = 26
(21/2)n−1 = 26
(n−1)/2 = 6
n − 1 = 12
n = 13
Answer: 128 is the 13th term.
Q7: Sierpiński Square Carpet
(i) Number of Red Squares in Stages 0 to 3
| Stage | Number of Red Squares |
|---|---|
| 0 | 1 |
| 1 | 8 |
| 2 | 64 |
| 3 | 512 |
Answer: 1, 8, 64, 512
(ii) Predict Stages 4 and 5
Each stage multiplies by 8.
Stage 4 = 512 × 8 = 4096
Stage 5 = 4096 × 8 = 32768
Answer:
- Stage 4 = 4096
- Stage 5 = 32768
(iii) Formula for Number of Red Squares
This is a GP:
1, 8, 64, 512, ...
a = 1
r = 8
Explicit Formula
Rn = 8n
Recursive Formula
R0 = 1
Rn+1 = 8Rn
(iv) Area of the Red Region
Stage 0 Area = 1 square unit
At each stage, only 8/9 of the previous area remains.
| Stage | Area |
|---|---|
| 0 | 1 |
| 1 | 8/9 |
| 2 | (8/9)2 = 64/81 |
| 3 | (8/9)3 = 512/729 |
| 4 | (8/9)4 = 4096/6561 |
| 5 | (8/9)5 = 32768/59049 |
Explicit Formula
An = (8/9)n
Recursive Formula
A0 = 1
An+1 = (8/9)An
What Happens as n Increases?
Since 8/9 is less than 1,
(8/9)n → 0
Therefore the red area keeps decreasing and approaches 0 as the number of stages becomes very large.
Answer: The red area tends to 0.
Summary of Answers
| Question | Answer |
|---|---|
| Q1 | 3072 |
| Q2 | T10 = 9,765,625, Tn = 5n |
| Q3 | 730 is the 7th term |
| Q4 | 4374 is the 8th term |
| Q5(i) | 6.2208 m |
| Q5(ii) | 301.3376 m |
| Q6 | 13th term |
| Q7 | Red squares = 8n, Area = (8/9)n |