Mathematics solution NCERT
Class 9 - Chapter 8: Predicting What Comes Next: Exploring Sequences and Progressions
Q1: Find the 12th term of a GP with common ratio 2, whose 8th term is 192.
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: 5, 25, 125, … .
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? Write the explicit formula as well as the recursive formula for the nth term.
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: 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?
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: Fig. 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.
(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?
(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 |