Class 9 Maths Ganita Manjari Chapter 8 End of Chapter Exercises Predicting What Comes Next: Exploring Sequences and Progressions

Class 9 Maths Ganita Manjari Chapter 8 End of Chapter Exercises provide students with additional questions to revise and practise the concepts covered in “Predicting What Comes Next: Exploring Sequences and Progressions.” These exercises help students strengthen their understanding of sequences, identify patterns and apply mathematical reasoning to predict what comes next.

The Class 9 Maths Ganita Manjari Chapter 8 End of Chapter Exercises Solutions offer step-by-step explanations for the questions, making it easier for students to understand the correct approach and verify their answers. Students can use these solutions for homework, self-study, revision and exam preparation.

Class 9 Maths Ganita Manjari Chapter 8 End of Chapter Exercises Predicting What Comes Next: Exploring Sequences and Progressions

Predicting What Comes Next: Exploring Sequences and Progressions
Medium
Q.
If the sum of n terms of an A.P. is (pn + qn2), where p and q are constants, find the common difference.
Predicting What Comes Next: Exploring Sequences and Progressions
Medium
Q.
Which term of the AP : 21, 18, 15, … is – 81? Also, is 0 a term of this AP? Give reasons for your answer
Predicting What Comes Next: Exploring Sequences and Progressions
Difficult
Q.
Observe the Sierpiński triangle and try to answer the following questions
(a) How many black triangles are there in Stages 0 to 3 of Fig. 8.7?
(b) Can you predict the number of black triangles at Stages 4 and 5?
(c) Can you find a rule for the number of black triangles at the nth stage? 
(d) Suppose the area of the triangle (that is, the black region) in Stage 0 is 1 square unit. What is the area of the black region in Stages 1, 2 and 3? What will be the area of the black region in Stages 4 and 5? Find a rule
for the area of the black region at the nth stage. What happens to this area as n, the number of stages, goes on increasing?
Predicting What Comes Next: Exploring Sequences and Progressions
Difficult
Q.
Can you predict the number of squares in Stages 5 and 6 of the pattern? In Stages 10, 11 and 12? In Stage 20? At any stage?

 How is this different from the growing pattern in Fig. 8.3?

Predicting What Comes Next: Exploring Sequences and Progressions
Medium
Q.
A child arranges marbles in rows so that the first row has 1 marble, the second has 2 marbles, the third has 3, and so on up to 25 rows. How many marbles does the child use in all?
Predicting What Comes Next: Exploring Sequences and Progressions
Medium
Q.
Harish started work at an annual salary of ₹5,00,000 and received an increment of ₹20,000 each year. After how many years did his income reach ₹7,00,000?
Predicting What Comes Next: Exploring Sequences and Progressions
Difficult
Q.
How many 2-digit numbers are divisible by 3? What is the sum of all these 2-digit numbers?
Predicting What Comes Next: Exploring Sequences and Progressions
Difficult
Q.
An AP consists of 50 terms in which the 3rd term is 12 and the last term is 106. Find the 29th term.
Predicting What Comes Next: Exploring Sequences and Progressions
Medium
Q.
Find the nth term of the AP: 11, 8, 5, 2 … Write the recursive rule for this AP.
Predicting What Comes Next: Exploring Sequences and Progressions
Medium
Q.
Find the 10th and 26th terms of the AP: 3, 8, 13, 18, ….
Predicting What Comes Next: Exploring Sequences and Progressions
Medium
Q.
Find the 10th and nth terms of the GP: 5, 25, 125, … .
Predicting What Comes Next: Exploring Sequences and Progressions
Medium
Q.
Let us revisit the sequence tn of triangular numbers 1, 3, 6, 10, 15, … shown in Fig. 8.1. Note that the nth term of this sequence is the sum of the first n natural numbers. thus tn=n(n+1)2\rm\;t_n=\frac{n(n+1)}{2}​.
Can you use this to find the 10th, 17th and 80th triangular numbers?
Predicting What Comes Next: Exploring Sequences and Progressions
Difficult
Q.
Can you predict the number of squares in Stages 5 and 6 of the sequence? In Stages 10, 11 and 12? In Stage 20? At any stage?
Predicting What Comes Next: Exploring Sequences and Progressions
Medium
Q.
Let T1 = 1, T2 = 2, T3 = 4, and Tn = Tn–1 + Tn–2 + Tn–3 for n ≥ 4. Find T4, T5, T6, T7, and T8.
Predicting What Comes Next: Exploring Sequences and Progressions
Medium
Q.
A sequence is given by the recursive rule t1 = – 5, tn +1=  tn + 3 for n ≥ 1. Find the first five terms of the sequence. Is 52 a term of this sequence? If so, which term is it?
Predicting What Comes Next: Exploring Sequences and Progressions
Easy
Q.
Which term of the sequence tn = 5n – 3 for n ≥ 1 is 607?
Predicting What Comes Next: Exploring Sequences and Progressions
Medium
Q.
Determine whether 97 and 172 are terms of the sequence tn = 5n – 3 for n ≥ 1.
Predicting What Comes Next: Exploring Sequences and Progressions
Easy
Q.
Find the 10th and 15th terms of the sequence tn = 5n – 3 for n ≥ 1.
Predicting What Comes Next: Exploring Sequences and Progressions
Medium
Q.
Find the first five terms of the sequence in which the nth term is given by 
(i) tn = 3n – 4, 
(ii) tn = 2 – 5n, and 
(iii) tn = n2 – 2n + 3 for n ≥ 1.
Predicting What Comes Next: Exploring Sequences and Progressions
Medium
Q.
Find the 12th term of a GP with common ratio 2, whose 8th term is 192.
Predicting What Comes Next: Exploring Sequences and Progressions
Medium
Q.
A sequence is given by the recursive rule  t1= 2, tn+1 = 3tn – 2 for n ≥ 1. Which term of the sequence is 730?
Predicting What Comes Next: Exploring Sequences and Progressions
Difficult
Q.
The first term of a G.P. is 1. The sum of the third term and fifth term is 90. Find the common ratio of G.P.
Predicting What Comes Next: Exploring Sequences and Progressions
Medium
Q.
The number of bacteria in a certain culture doubles every hour. If there were 30 bacteria present in the culture originally, how many bacteria will be present at the end of the 2nd hour, 4th hour and nth hour?
Predicting What Comes Next: Exploring Sequences and Progressions
Difficult
Q.
Suppose W1 = 1, W2 = 2 and for n > 2,  Wn = W1 + W2 + … + Wn–2 + 2.  Find the values of W1, W2, …, W8.  Do you recognise this sequence?
Predicting What Comes Next: Exploring Sequences and Progressions
Difficult
Q.
Suppose P1 = 1,  P2 = 2 and for n > 2, Pn = P1 + P2 + … + Pn–1 + 1. Find the values of P1, P2, …, P8. Can you find a simpler recursive formula for Pn ? Can you give an explicit formula?
Predicting What Comes Next: Exploring Sequences and Progressions
Difficult
Q.
The sum of the first three terms of a geometric progression is 26, and the sum of their squares is 364. Find the terms of the GP.
Predicting What Comes Next: Exploring Sequences and Progressions
Medium
Q.
If the 4th, 10th and 16th terms of a GP are x, y and z respectively, prove that x, y, z are in GP.
Predicting What Comes Next: Exploring Sequences and Progressions
Difficult
Q.
The sum of the first three terms of a GP is 1312\cfrac{13}{12}​  and their product is –1. Find the common ratio and the terms.
Predicting What Comes Next: Exploring Sequences and Progressions
Difficult
Q.
Which term of the GP: 2, 8, 32, … is 131072? Write the explicit formula as well as the recursive formula for the nth term.
Predicting What Comes Next: Exploring Sequences and Progressions
Difficult
Q.
Find the smallest value of n such that the sum of the first n natural numbers is greater than 1,000.
Predicting What Comes Next: Exploring Sequences and Progressions
Difficult
Q.
The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the AP.
Predicting What Comes Next: Exploring Sequences and Progressions
Difficult
Q.
Find all possible ways of expressing 100 as the sum of consecutive natural numbers.
Predicting What Comes Next: Exploring Sequences and Progressions
Medium
Q.
Which term of the GP: 2, 6, 18, … is 4374? Write the explicit formula as well as the recursive formula for the nth term.
Predicting What Comes Next: Exploring Sequences and Progressions
Difficult
Q.
Find a GP for which the sum of the first two terms is – 4 and the fifth term is 4 times the third term.
Predicting What Comes Next: Exploring Sequences and Progressions
Medium
Q.
How many multiples of 4 lie between 10 and 250?
Predicting What Comes Next: Exploring Sequences and Progressions
Medium
Q.
How many three-digit numbers are divisible by 7?
Predicting What Comes Next: Exploring Sequences and Progressions
Medium
Q.
Determine the AP whose third term is 16 and whose 7th term exceeds the 5th term by 12.
Predicting What Comes Next: Exploring Sequences and Progressions
Medium
Q.
Find the 31st term of an AP whose 11th term is 38 and 16th term is 73.
Predicting What Comes Next: Exploring Sequences and Progressions
Medium
Q.
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 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?
Predicting What Comes Next: Exploring Sequences and Progressions
Difficult
Q.
Which term of the sequence 2, 22,\;2\sqrt2,​4, … is 128?
Predicting What Comes Next: Exploring Sequences and Progressions
Difficult
Q.
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?
Predicting What Comes Next: Exploring Sequences and Progressions
Medium
Q.
Can you find the rule describing the nth term of the sequence of square numbers?

Class 9 Maths Ganita Manjari Chapter 8 End of Chapter Exercises Predicting What Comes Next: Exploring Sequences and Progressions

NCERT Solutions for Class 9 Maths Chapter 8 End-of-Chapter Exercises

The end-of-chapter exercises include AP, GP, consecutive natural number sums, bacteria growth, recursive rules and sequence-based reasoning.

Question 1. Find the 31st term of an AP whose 11th term is 38 and 16th term is 73.

Answer:
Let first term be a and common difference be d.

11th term:

a + 10d = 38 …(1)

16th term:

a + 15d = 73 …(2)

Subtract (1) from (2):

5d = 35
d = 7

Substitute in (1):

a + 10(7) = 38
a + 70 = 38
a = -32

31st term:

t₃₁ = a + 30d
= -32 + 30(7)
= -32 + 210
= 178

Final answer:
31st term = 178

Question 2. Determine the AP whose third term is 16 and whose 7th term exceeds the 5th term by 12.

Answer:
Let first term be a and common difference be d.

Third term:

a + 2d = 16 …(1)

7th term exceeds 5th term by 12:

t₇ - t₅ = 12

(a + 6d) - (a + 4d) = 12
2d = 12
d = 6

Substitute in (1):

a + 2(6) = 16
a + 12 = 16
a = 4

Therefore, the AP is:

4, 10, 16, 22, 28, …

Final answer:
AP = 4, 10, 16, 22, 28, …

Question 3. How many three-digit numbers are divisible by 7?

Answer:
Smallest three-digit number divisible by 7:

105

Largest three-digit number divisible by 7:

994

The numbers form an AP:

105, 112, 119, …, 994

Here:

a = 105
d = 7
last term = 994

Find n:

994 = 105 + (n - 1)7
889 = 7(n - 1)
n - 1 = 127
n = 128

Final answer:
128 three-digit numbers are divisible by 7.

Question 4. How many multiples of 4 lie between 10 and 250?

Answer:
Smallest multiple of 4 greater than 10 is 12.

Largest multiple of 4 less than 250 is 248.

The multiples are:

12, 16, 20, …, 248

Here:

a = 12
d = 4
last term = 248

Find n:

248 = 12 + (n - 1)4
236 = 4(n - 1)
n - 1 = 59
n = 60

Final answer:
60 multiples of 4 lie between 10 and 250.

Question 5. Find a GP for which the sum of the first two terms is -4 and the fifth term is 4 times the third term.

Answer:
Let the GP be:

a, ar, ar², ar³, ar⁴, …

Given:

a + ar = -4

So:

a(1 + r) = -4 …(1)

Fifth term is 4 times third term:

ar⁴ = 4ar²

Assuming a ≠ 0:

r² = 4
r = 2 or r = -2

If r = 2:

a(1 + 2) = -4
3a = -4
a = -4/3

GP is:

-4/3, -8/3, -16/3, …

If r = -2:

a(1 - 2) = -4
-a = -4
a = 4

GP is:

4, -8, 16, -32, 64, …

Final answer:
Possible GPs are -4/3, -8/3, -16/3, … and 4, -8, 16, -32, …

Question 6. Find all possible ways of expressing 100 as the sum of consecutive natural numbers.

Answer:
Let 100 be written as the sum of k consecutive natural numbers starting from a.

100 = a + (a + 1) + … + (a + k - 1)

100 = k/2[2a + k - 1]

Check possible values of k.

One-term sum:

100

Two-term sum:

49 + 51 is not consecutive because 50 + 51 = 101 and 49 + 50 = 99. So no 2-term expression.

Four-term sum:

22 + 23 + 24 + 25 = 94
23 + 24 + 25 + 26 = 98
24 + 25 + 26 + 27 = 102
So no 4-term expression.

Five-term sum:

18 + 19 + 20 + 21 + 22 = 100

Eight-term sum:

9 + 10 + 11 + 12 + 13 + 14 + 15 + 16 = 100

Final answer:
100 = 100 = 18 + 19 + 20 + 21 + 22 = 9 + 10 + 11 + 12 + 13 + 14 + 15 + 16

Question 7. The number of bacteria in a culture doubles every hour. If there were 30 bacteria originally, how many bacteria will be present at the end of the 2nd hour, 4th hour and nth hour?

Answer:
Initial number = 30

The number doubles every hour.

After 1 hour:

30 × 2 = 60

After 2 hours:

30 × 2² = 120

After 4 hours:

30 × 2⁴ = 480

After n hours:

30 × 2ⁿ

Final answer:
After 2 hours = 120, after 4 hours = 480, after nth hour = 30 × 2ⁿ

Question 8. The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms.

Answer:
Let first term be a and common difference be d.

4th term:

a + 3d

8th term:

a + 7d

Given:

(a + 3d) + (a + 7d) = 24
2a + 10d = 24
a + 5d = 12 …(1)

6th term:

a + 5d

10th term:

a + 9d

Given:

(a + 5d) + (a + 9d) = 44
2a + 14d = 44
a + 7d = 22 …(2)

Subtract (1) from (2):

2d = 10
d = 5

Substitute in (1):

a + 5(5) = 12
a + 25 = 12
a = -13

First three terms:

a = -13
a + d = -8
a + 2d = -3

Final answer:
-13, -8, -3

Question 9. Find the smallest value of n such that the sum of the first n natural numbers is greater than 1000.

Answer:
Sum of first n natural numbers:

Sₙ = n(n + 1)/2

We need:

n(n + 1)/2 > 1000

n(n + 1) > 2000

Try n = 44:

44 × 45 = 1980
S₄₄ = 990

Try n = 45:

45 × 46 = 2070
S₄₅ = 1035

So, the smallest n is 45.

Final answer:
n = 45

Question 10. Which term of the GP 2, 8, 32, … is 131072? Write the explicit formula and recursive formula.

Answer:
First term:

a = 2

Common ratio:

r = 8/2 = 4

Explicit formula:

tₙ = 2 × 4ⁿ⁻¹

Find n:

2 × 4ⁿ⁻¹ = 131072

4ⁿ⁻¹ = 65536

Since:

65536 = 4⁸

n - 1 = 8
n = 9

Recursive formula:

t₁ = 2
tₙ = 4tₙ₋₁, for n ≥ 2

Final answer:
131072 is the 9th term. Explicit formula: tₙ = 2 × 4ⁿ⁻¹. Recursive formula: t₁ = 2, tₙ = 4tₙ₋₁.

Question 11. The sum of the first three terms of a GP is 13/12 and their product is -1. Find the common ratio and the terms.

Answer:
Let the three terms of the GP be:

a/r, a, ar

Their product is:

(a/r) × a × ar = a³

Given product = -1

So:

a³ = -1
a = -1

The terms are:

-1/r, -1, -r

Their sum is 13/12:

-1/r - 1 - r = 13/12

Multiply by 12r:

-12 - 12r - 12r² = 13r

12r² + 25r + 12 = 0

Factor:

12r² + 25r + 12 = (3r + 4)(4r + 3)

So:

r = -4/3 or r = -3/4

If r = -4/3, terms are:

3/4, -1, 4/3

If r = -3/4, terms are:

4/3, -1, 3/4

Final answer:
Common ratio = -4/3 or -3/4. Terms are 3/4, -1, 4/3 or 4/3, -1, 3/4.

Question 12. If the 4th, 10th and 16th terms of a GP are x, y and z respectively, prove that x, y, z are in GP.

Answer:
Let the GP have first term a and common ratio r.

4th term:

x = ar³

10th term:

y = ar⁹

16th term:

z = ar¹⁵

Now:

y/x = ar⁹/ar³ = r⁶

z/y = ar¹⁵/ar⁹ = r⁶

Since:

y/x = z/y

Therefore, x, y and z are in GP.

Final answer:
x, y, z are in GP because their consecutive ratios are equal.

Question 13. The sum of the first three terms of a geometric progression is 26, and the sum of their squares is 364. Find the terms of the GP.

Answer:
Let the three terms be:

a/r, a, ar

Their sum:

a/r + a + ar = 26 …(1)

Sum of squares:

a²/r² + a² + a²r² = 364 …(2)

Let:

x = a/r, y = a, z = ar

Then x, y, z are in GP and:

x + y + z = 26
x² + y² + z² = 364

Also:

(x + y + z)² = x² + y² + z² + 2(xy + yz + zx)

26² = 364 + 2(xy + yz + zx)

676 - 364 = 2(xy + yz + zx)

312 = 2(xy + yz + zx)

xy + yz + zx = 156

Checking simple GP triples with sum 26 and square sum 364 gives:

2, 6, 18

Sum = 2 + 6 + 18 = 26
Squares = 4 + 36 + 324 = 364

Final answer:
The terms are 2, 6 and 18.

Question 14. Suppose P₁ = 1, P₂ = 2 and for n > 2, Pₙ = P₁ + P₂ + … + Pₙ₋₁ + 1. Find P₁ to P₈. Can you find a simpler recursive formula and an explicit formula?

Answer:
Given:

P₁ = 1
P₂ = 2

For n > 2:

Pₙ = P₁ + P₂ + … + Pₙ₋₁ + 1

Now:

P₃ = P₁ + P₂ + 1
= 1 + 2 + 1
= 4

P₄ = P₁ + P₂ + P₃ + 1
= 1 + 2 + 4 + 1
= 8

P₅ = 1 + 2 + 4 + 8 + 1
= 16

So the sequence is:

P₁ = 1, P₂ = 2, P₃ = 4, P₄ = 8, P₅ = 16, P₆ = 32, P₇ = 64, P₈ = 128

Simpler recursive formula:

P₁ = 1
Pₙ = 2Pₙ₋₁, for n ≥ 2

Explicit formula:

Pₙ = 2ⁿ⁻¹

Final answer:
1, 2, 4, 8, 16, 32, 64, 128; Pₙ = 2Pₙ₋₁; Pₙ = 2ⁿ⁻¹

Question 15. Suppose W₁ = 1, W₂ = 2 and for n > 2, Wₙ = W₁ + W₂ + … + Wₙ₋₂ + 2. Find W₁ to W₈. Do you recognise this sequence?

Answer:
Given:

W₁ = 1
W₂ = 2

For n > 2:

Wₙ = W₁ + W₂ + … + Wₙ₋₂ + 2

Now:

W₃ = W₁ + 2 = 1 + 2 = 3

W₄ = W₁ + W₂ + 2 = 1 + 2 + 2 = 5

W₅ = W₁ + W₂ + W₃ + 2
= 1 + 2 + 3 + 2
= 8

W₆ = W₁ + W₂ + W₃ + W₄ + 2
= 1 + 2 + 3 + 5 + 2
= 13

W₇ = 1 + 2 + 3 + 5 + 8 + 2
= 21

W₈ = 1 + 2 + 3 + 5 + 8 + 13 + 2
= 34

Final answer:
W₁ to W₈ are 1, 2, 3, 5, 8, 13, 21, 34. This is the Virahānka-Fibonacci sequence.

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