NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 8 Predicting What Comes Next: Exploring Sequences and Progressions 2026-27

Think and Reflect (NCERT Textbook Page No. 174)

Can you describe the pattern in each of the above sequences? Can you predict the next few numbers in these sequences?

Solution:

Sequence: 1, 2, 3, 4, 5, 6, … — Natural Numbers

Pattern: Every term is obtained by adding 1 to the previous term.

Next few numbers: 7, 8, 9, 10, 11, …

Sequence: 1, 3, 5, 7, 9, 11, … — Odd Numbers

Pattern: Every term is obtained by adding 2 to the previous term.

Next few numbers: 13, 15, 17, 19, 21, …

Sequence: 1, 3, 6, 10, 15, 21, … — Triangular Numbers

Pattern: Each term is the sum of the natural numbers from 1 up to its position.

For example:

1st term = 1

2nd term = 1 + 2 = 3

3rd term = 1 + 2 + 3 = 6

4th term = 1 + 2 + 3 + 4 = 10

Continuing the same pattern, the next few numbers are:

28, 36, 45, 55, 66, …

Sequence: 1, 4, 9, 16, 25, 36, … — Square Numbers

Pattern: Each term is the square of a natural number.

1st term = 1² = 1

2nd term = 2² = 4

3rd term = 3² = 9

4th term = 4² = 16

The next few numbers are:

7² = 49, 8² = 64, 9² = 81, 10² = 100, 11² = 121

Exercise (Page 176)

1. Consider the sequence 1, 4, 7, 10, 13,… Can you predict the next four terms? Can you derive the first 10 terms of the sequence obtained by adding all the terms up to a given term of this sequence (Hint: The first term is 1. The second term is 1 + 4 = 5, the third term 1+4 + 7=12 and so on.)

Solution:

The given sequence is:

1, 4, 7, 10, 13, …

The difference between consecutive terms is 3.

4 - 1 = 3

7 - 4 = 3

10 - 7 = 3

13 - 10 = 3

Therefore, the next four terms are obtained by repeatedly adding 3:

13 + 3 = 16

16 + 3 = 19

19 + 3 = 22

22 + 3 = 25

Hence, the next four terms are:

16, 19, 22, 25

Now, add all terms up to each position:

T₁ = 1

T₂ = 1 + 4 = 5

T₃ = 1 + 4 + 7 = 12

T₄ = 1 + 4 + 7 + 10 = 22

T₅ = 1 + 4 + 7 + 10 + 13 = 35

T₆ = 35 + 16 = 51

T₇ = 51 + 19 = 70

T₈ = 70 + 22 = 92

T₉ = 92 + 25 = 117

T₁₀ = 117 + 28 = 145

Therefore, the first 10 terms of the new sequence are:

1, 5, 12, 22, 35, 51, 70, 92, 117, 145

2. Can you write t5, t6, t7 and t8, for the sequence of triangular numbers?

Solution:

A triangular number is obtained by adding the natural numbers from 1 up to the required position.

t₅ = 1 + 2 + 3 + 4 + 5 = 15

t₆ = 1 + 2 + 3 + 4 + 5 + 6 = 21

t₇ = 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28

t₈ = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = 36

Therefore:

t₅ = 15, t₆ = 21, t₇ = 28 and t₈ = 36

Think and Reflect (NCERT Textbook Page No. 176)

1. Can you think of any other kinds of sequences? List out five different types of sequences and discus their properties with your friends.

Solution:

Five examples of sequences are:

1. Even-number sequence: 2, 4, 6, 8, …

Each term increases by 2.

2. Multiples of 5: 5, 10, 15, 20, …

Each term increases by 5.

3. Cube-number sequence: 1, 8, 27, 64, …

Each term is the cube of a natural number.

4. Fibonacci sequence: 1, 1, 2, 3, 5, 8, …

Each term after the first two is the sum of the previous two terms.

5. Powers of 2: 1, 2, 4, 8, 16, …

Each term is twice the previous term.

Students may discuss how these sequences grow and whether they follow an additive or multiplicative pattern.

Think and Reflect (NCERT Textbook Page No. 177)

1. Why is it useful to have an explicit formula for the nth term of a sequence?

Solution:

An explicit formula gives the value of any term directly from its position.

It is useful because:

It helps us calculate a distant term without finding all the earlier terms.

It clearly describes the pattern followed by the sequence.

It saves time and makes calculations more efficient.

It also helps us check whether a particular number belongs to the sequence.

Exercise (Page 177)

Using the explicit rule uₙ = 2n – 1, find the 53rd term, 108th term, and the 1170th term of the odd number sequence.

Solution:

The given rule is:

uₙ = 2n - 1

For the 53rd term:

u₅₃ = 2(53) - 1

= 106 - 1

= 105

For the 108th term:

u₁₀₈ = 2(108) - 1

= 216 - 1

= 215

For the 1170th term:

u₁₁₇₀ = 2(1170) - 1

= 2340 - 1

= 2339

Therefore:

u₅₃ = 105, u₁₀₈ = 215 and u₁₁₇₀ = 2339

Think and Reflect (NCERT Textbook Page No. 177)

Can you find the rule describing the nth term of the sequence of square numbers?

Solution:

The sequence of square numbers is:

1, 4, 9, 16, 25, 36, 49, 64, 81, 100, …

These terms can be written as:

1², 2², 3², 4², 5², 6², 7², 8², 9², 10², …

Therefore, the nth term of the sequence is:

tₙ = n²

Exercise (Page 178)

Consider the expression tₙ = 3n – 7.

(i) Find its first, second, third, 12th, 18th, and 50th terms.

Solution:

The nth term is:

tₙ = 3n - 7

First term:

t₁ = 3(1) - 7 = -4

Second term:

t₂ = 3(2) - 7 = -1

Third term:

t₃ = 3(3) - 7 = 2

12th term:

t₁₂ = 3(12) - 7

= 36 - 7

= 29

18th term:

t₁₈ = 3(18) - 7

= 54 - 7

= 47

50th term:

t₅₀ = 3(50) - 7

= 150 - 7

= 143

Therefore, the required terms are:

-4, -1, 2, 29, 47 and 143

(ii) Which term of the sequence is 332?

Solution:

Set tₙ = 332.

3n - 7 = 332

3n = 339

n = 113

Therefore, 332 is the 113th term of the sequence.

(iii) Is 557 a term of this sequence? Why or why not?

Solution:

Set tₙ = 557.

3n - 7 = 557

3n = 564

n = 188

Since 188 is a positive whole number, 557 belongs to the sequence.

Therefore, 557 is the 188th term.

Think and Reflect (NCERT Textbook Page No. 180)

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?

Solution:

The number of squares forms the sequence:

1, 5, 9, 13, …

Four squares are added at every new stage.

The number of squares in Stage n is:

tₙ = 1 + 4(n - 1)

= 1 + 4n - 4

= 4n - 3

Now:

Stage 5:

t₅ = 4(5) - 3 = 17

Stage 6:

t₆ = 4(6) - 3 = 21

Stage 10:

t₁₀ = 4(10) - 3 = 37

Stage 11:

t₁₁ = 4(11) - 3 = 41

Stage 12:

t₁₂ = 4(12) - 3 = 45

Stage 20:

t₂₀ = 4(20) - 3 = 77

Therefore, the number of squares at any stage n is:

tₙ = 4n - 3

Think and Reflect (NCERT Textbook Page No. 181)

Consider all the sequences we have discussed so far in this chapter. Which ones are arithmetic progressions and which ones are not? Can you justify your claim?

Solution:

A sequence is an arithmetic progression when the difference between every pair of consecutive terms is constant.

Arithmetic progressions include:

1, 2, 3, 4, … with common difference 1

1, 3, 5, 7, … with common difference 2

1, 4, 7, 10, … with common difference 3

1, 5, 9, 13, … with common difference 4

The following are not arithmetic progressions:

1, 3, 6, 10, 15, … because the differences are 2, 3, 4, 5, …

1, 4, 9, 16, 25, … because the differences are 3, 5, 7, 9, …

A sequence in which each term is multiplied by a fixed number is also not generally an arithmetic progression.

Exercise (Page 182)

Verify that the following sequences are arithmetic progressions and write their terms. What do you observe when you plot the ordered pairs emerging from them?

(i) 2, 5, 8, 11,…

Solution:

Find the differences between consecutive terms:

5 - 2 = 3

8 - 5 = 3

11 - 8 = 3

Since the difference is always 3, the sequence is an arithmetic progression.

Here:

First term, a = 2

Common difference, d = 3

Using:

tₙ = a + (n - 1)d

tₙ = 2 + (n - 1)3

= 2 + 3n - 3

= 3n - 1

The ordered pairs are:

(1, 2), (2, 5), (3, 8), (4, 11), (5, 14), …

When these points are plotted, they lie on a straight line. This is a characteristic property of an arithmetic progression.

(ii) -5, -1, 3, 7,…

Solution:

Find the differences between consecutive terms:

-1 - (-5) = 4

3 - (-1) = 4

7 - 3 = 4

The common difference is 4. Therefore, the sequence is an arithmetic progression.

Here:

a = -5

d = 4

Using:

tₙ = a + (n - 1)d

tₙ = -5 + 4(n - 1)

= -5 + 4n - 4

= 4n - 9

The ordered pairs are:

(1, -5), (2, -1), (3, 3), (4, 7), …

When plotted, these points lie on a straight line.

Exercise (Page 182)

Using the formula tₙ = a + (n – 1) × d, find the nth term of the following arithmetic progressions.

(i) 1/2, 5/2, 9/2, 13/2,…

Solution:

The first term is:

a = 1/2

The common difference is:

d = 5/2 - 1/2 = 2

Using:

tₙ = a + (n - 1)d

tₙ = 1/2 + 2(n - 1)

= 1/2 + 2n - 2

= 2n - 3/2

Therefore:

tₙ = (4n - 3)/2

(ii) 1.5, 3.5, 5.5, 7.5,…

Solution:

The first term is:

a = 1.5

The common difference is:

d = 3.5 - 1.5 = 2

Using:

tₙ = a + (n - 1)d

tₙ = 1.5 + 2(n - 1)

= 1.5 + 2n - 2

= 2n - 0.5

Therefore:

tₙ = 2n - 0.5

Exercise (Page 183)

Find recursive rules for the APs in the previous exercises.

Solution:

A recursive rule for an arithmetic progression is:

tₙ = tₙ₋₁ + d, for n ≥ 2

(i) 2, 5, 8, 11, …

First term:

t₁ = 2

Common difference:

d = 3

Recursive rule:

t₁ = 2

tₙ = tₙ₋₁ + 3, for n ≥ 2

(ii) -5, -1, 3, 7, …

First term:

t₁ = -5

Common difference:

d = 4

Recursive rule:

t₁ = -5

tₙ = tₙ₋₁ + 4, for n ≥ 2

(iii) 1/2, 5/2, 9/2, 13/2, …

First term:

t₁ = 1/2

Common difference:

d = 2

Recursive rule:

t₁ = 1/2

tₙ = tₙ₋₁ + 2, for n ≥ 2

(iv) 1.5, 3.5, 5.5, 7.5, …

First term:

t₁ = 1.5

Common difference:

d = 2

Recursive rule:

t₁ = 1.5

tₙ = tₙ₋₁ + 2, for n ≥ 2

Think and Reflect (NCERT Textbook Page No. 184)

Can the same approach be used to find the sum of 1 + 2 + 3 + … + 100?

Solution:

Let:

S = 1 + 2 + 3 + … + 100

Write the same sum in reverse order:

S = 100 + 99 + 98 + … + 1

Add the two equations term by term:

2S = (1 + 100) + (2 + 99) + (3 + 98) + … + (100 + 1)

Each pair has a sum of 101, and there are 100 such pairs.

Therefore:

2S = 100 × 101

2S = 10100

S = 5050

Hence, the sum of the first 100 natural numbers is 5050.

Think and Reflect (NCERT Textbook Page No. 185)

Can you use this formula to find S20, S50 or S1000?

Solution:

The sum of the first n natural numbers is:

Sₙ = n(n + 1)/2

For n = 20:

S₂₀ = 20(21)/2

= 210

For n = 50:

S₅₀ = 50(51)/2

= 1275

For n = 1000:

S₁₀₀₀ = 1000(1001)/2

= 500500

Therefore:

S₂₀ = 210

S₅₀ = 1275

S₁₀₀₀ = 500500

Let us revisit the sequence tₙ of triangular numbers 1, 3, 6, 10, 15,… shown in Figure. Note that the nth term of this sequence is the sum of the first n natural numbers. Thus tₙ = n(n+1)/2

Can you use this to find the 10th, 17th and 80th triangular numbers?

Solution:

The nth triangular number is:

tₙ = n(n + 1)/2

For the 10th triangular number:

t₁₀ = 10(11)/2

= 55

For the 17th triangular number:

t₁₇ = 17(18)/2

= 153

For the 80th triangular number:

t₈₀ = 80(81)/2

= 3240

Therefore, the 10th, 17th and 80th triangular numbers are 55, 153 and 3240.

Think and Reflect (NCERT Textbook Page No. 186)

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?

Solution:

The number of squares doubles at every stage.

Stage 1 = 3 squares

Stage 2 = 3 × 2 = 6 squares

Stage 3 = 3 × 2² = 12 squares

Stage 4 = 3 × 2³ = 24 squares

Therefore, the number of squares in Stage n is:

tₙ = 3 × 2ⁿ⁻¹

Now:

Stage 5:

t₅ = 3 × 2⁴ = 48

Stage 6:

t₆ = 3 × 2⁵ = 96

Stage 10:

t₁₀ = 3 × 2⁹ = 1536

Stage 11:

t₁₁ = 3 × 2¹⁰ = 3072

Stage 12:

t₁₂ = 3 × 2¹¹ = 6144

Stage 20:

t₂₀ = 3 × 2¹⁹ = 1572864

In this pattern, the number of squares is multiplied by 2 at every stage, so it shows geometric or exponential growth.

In Fig. 8.3, a fixed number of squares is added at each stage. Therefore, that pattern shows arithmetic or linear growth.

Exercise (Page 188)

Check whether the following sequences are geometric progressions and find their nth terms.

(i) 2, 10, 50, 250, ………

Solution:

The given sequence is:

2, 10, 50, 250, …

Find the ratio between consecutive terms:

10/2 = 5

50/10 = 5

250/50 = 5

Since the ratio is constant, the sequence is a geometric progression.

First term:

a = 2

Common ratio:

r = 5

The nth term of a geometric progression is:

tₙ = arⁿ⁻¹

Therefore:

tₙ = 2 × 5ⁿ⁻¹

(ii) 4, 8/3, 16/9, 32/27,…

Solution:

The given sequence is:

4, 8/3, 16/9, 32/27, …

Find the ratio between consecutive terms:

(8/3) ÷ 4 = 2/3

(16/9) ÷ (8/3) = 2/3

(32/27) ÷ (16/9) = 2/3

Since the ratio is constant, the sequence is a geometric progression.

First term:

a = 4

Common ratio:

r = 2/3

Therefore:

tₙ = 4(2/3)ⁿ⁻¹

(iii) 3, −3/2, 3/4, −3/8,…

Solution:

The given sequence is:

3, -3/2, 3/4, -3/8, …

Find the ratio between consecutive terms:

(-3/2) ÷ 3 = -1/2

(3/4) ÷ (-3/2) = -1/2

(-3/8) ÷ (3/4) = -1/2

Since the ratio is constant, the sequence is a geometric progression.

First term:

a = 3

Common ratio:

r = -1/2

Therefore:

tₙ = 3(-1/2)ⁿ⁻¹

Exercise (Page 188)

Can you find a recursive rule for the formula tₙ = 3 × 10ⁿ⁻¹ that generates the geometric progression 3, 30, 300, 3000, ?

Solution:

The sequence is:

3, 30, 300, 3000, …

The first term is:

t₁ = 3

Each term is obtained by multiplying the previous term by 10.

Therefore, the recursive rule is:

t₁ = 3

tₙ = 10tₙ₋₁, for n ≥ 2

Think and Reflect (NCERT Textbook Page No. 189)

Observe the Sierpinski triangle and try to answer the following questions

(a) How many black triangles are there in Stages 0 to 3 of Fig. 8.7?

Solution:

The number of black triangles is multiplied by 3 at every stage.

Stage 0:

3⁰ = 1 black triangle

Stage 1:

3¹ = 3 black triangles

Stage 2:

3² = 9 black triangles

Stage 3:

3³ = 27 black triangles

Therefore, the numbers of black triangles in Stages 0, 1, 2 and 3 are:

1, 3, 9 and 27

(b) Can you predict the number of black triangles at Stages 4 and 5?

Solution:

Stage 4:

3⁴ = 81 black triangles

Stage 5:

3⁵ = 243 black triangles

Therefore, Stage 4 has 81 black triangles and Stage 5 has 243 black triangles.

(c) Can you find a rule for the number of black triangles at the nth stage?

Solution:

The number of black triangles forms a geometric progression with first term 1 and common ratio 3.

Therefore, the number of black triangles at Stage n is:

tₙ = 3ⁿ

Here, the stage numbering begins from Stage 0.

(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?

Solution:

At each stage, one-fourth of every black triangle is removed. Therefore, three-fourths of the previous black area remains.

Stage 0:

A₀ = 1 square unit

Stage 1:

A₁ = 3/4 square unit

Stage 2:

A₂ = (3/4)² = 9/16 square unit

Stage 3:

A₃ = (3/4)³ = 27/64 square unit

Stage 4:

A₄ = (3/4)⁴ = 81/256 square unit

Stage 5:

A₅ = (3/4)⁵ = 243/1024 square unit

Therefore, the area of the black region at Stage n is:

Aₙ = (3/4)ⁿ square units

As n increases, the area becomes smaller and approaches zero. Thus, the black area decreases geometrically with every new stage.

Exercise Set 8.1 

Question 1. 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.

Solution:

(i) tₙ = 3n – 4

Substitute n = 1, 2, 3, 4 and 5.

t₁ = 3(1) – 4 = –1

t₂ = 3(2) – 4 = 2

t₃ = 3(3) – 4 = 5

t₄ = 3(4) – 4 = 8

t₅ = 3(5) – 4 = 11

Therefore, the first five terms are:

–1, 2, 5, 8, 11

(ii) tₙ = 2 – 5n

Substitute n = 1, 2, 3, 4 and 5.

t₁ = 2 – 5(1) = –3

t₂ = 2 – 5(2) = –8

t₃ = 2 – 5(3) = –13

t₄ = 2 – 5(4) = –18

t₅ = 2 – 5(5) = –23

Therefore, the first five terms are:

–3, –8, –13, –18, –23

(iii) tₙ = n² – 2n + 3

Substitute n = 1, 2, 3, 4 and 5.

t₁ = 1² – 2(1) + 3 = 2

t₂ = 2² – 2(2) + 3 = 3

t₃ = 3² – 2(3) + 3 = 6

t₄ = 4² – 2(4) + 3 = 11

t₅ = 5² – 2(5) + 3 = 18

Therefore, the first five terms are:

2, 3, 6, 11, 18

Question 2. Find the 10th and 15th terms of the sequence tn = 5n – 3 for n ≥ 1.

Solution:

The given rule is:

tₙ = 5n – 3

For the 10th term:

t₁₀ = 5(10) – 3

= 50 – 3

= 47

For the 15th term:

t₁₅ = 5(15) – 3

= 75 – 3

= 72

Therefore, the 10th term is 47 and the 15th term is 72.

Question 3. Determine whether 97 and 172 are terms of the sequence tn = 5n – 3 for n ≥ 1.

Solution:

The given rule is:

tₙ = 5n – 3

To check whether 97 is a term, set:

5n – 3 = 97

5n = 100

n = 20

Since n is a positive whole number, 97 belongs to the sequence. It is the 20th term.

Now, check whether 172 is a term:

5n – 3 = 172

5n = 175

n = 35

Since n is a positive whole number, 172 also belongs to the sequence. It is the 35th term.

Question 4. Which term of the sequence tn = 5n – 3 for n ≥ 1 is 607?

Solution:

The given rule is:

tₙ = 5n – 3

Set tₙ = 607.

5n – 3 = 607

5n = 610

n = 122

Therefore, 607 is the 122nd term of the sequence.

Question 5. 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?

Solution:

Given:

t₁ = –5

tₙ₊₁ = tₙ + 3

The first five terms are:

t₁ = –5

t₂ = t₁ + 3 = –5 + 3 = –2

t₃ = t₂ + 3 = –2 + 3 = 1

t₄ = t₃ + 3 = 1 + 3 = 4

t₅ = t₄ + 3 = 4 + 3 = 7

Therefore, the first five terms are:

–5, –2, 1, 4, 7

The explicit rule for this arithmetic sequence is:

tₙ = –5 + (n – 1)3

To check whether 52 is a term:

52 = –5 + 3(n – 1)

57 = 3(n – 1)

19 = n – 1

n = 20

Therefore, 52 is a term of the sequence, and it is the 20th term.

Question 6. 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.

Solution:

Given:

T₁ = 1

T₂ = 2

T₃ = 4

Each new term is obtained by adding the previous three terms.

T₄ = T₃ + T₂ + T₁

= 4 + 2 + 1

= 7

T₅ = T₄ + T₃ + T₂

= 7 + 4 + 2

= 13

T₆ = T₅ + T₄ + T₃

= 13 + 7 + 4

= 24

T₇ = T₆ + T₅ + T₄

= 24 + 13 + 7

= 44

T₈ = T₇ + T₆ + T₅

= 44 + 24 + 13

= 81

Therefore:

T₄ = 7, T₅ = 13, T₆ = 24, T₇ = 44 and T₈ = 81

Exercise Set 8.2 

Question 1. Find the 10th and 26th terms of the AP: 3, 8, 13, 18, …

Solution:

The given arithmetic progression is:

3, 8, 13, 18, …

First term:

a = 3

Common difference:

d = 8 – 3 = 5

The nth term of an AP is:

tₙ = a + (n – 1)d

For the 10th term:

t₁₀ = 3 + (10 – 1)5

= 3 + 45

= 48

For the 26th term:

t₂₆ = 3 + (26 – 1)5

= 3 + 125

= 128

Therefore, the 10th term is 48 and the 26th term is 128.

Question 2. Which term of the AP : 21, 18, 15,… is – 81? Also, is 0 a term of this AP? Give reasons for your answer.

Solution:

The given AP is:

21, 18, 15, …

Here:

a = 21

d = 18 – 21 = –3

Using:

tₙ = a + (n – 1)d

To find the position of –81:

–81 = 21 + (n – 1)(–3)

–102 = –3(n – 1)

34 = n – 1

n = 35

Therefore, –81 is the 35th term of the AP.

Now, check whether 0 is a term:

0 = 21 + (n – 1)(–3)

–21 = –3(n – 1)

7 = n – 1

n = 8

Since n is a positive whole number, 0 is a term of the AP. It is the 8th term.

Question 3. Find the 77th term of the AP: 11, 8, 5, 2 … Write the recursive rule for this AP.

Solution:

The given AP is:

11, 8, 5, 2, …

Here:

a = 11

d = 8 – 11 = –3

Using the nth-term formula:

tₙ = a + (n – 1)d

tₙ = 11 + (n – 1)(–3)

= 11 – 3n + 3

= 14 – 3n

For the 77th term:

t₇₇ = 14 – 3(77)

= 14 – 231

= –217

Therefore, the 77th term is –217.

The recursive rule is:

t₁ = 11

tₙ = tₙ₋₁ – 3, for n ≥ 2

Question 4. An AP consists of 50 terms in which the 3rd term is 12 and the last term is 106. Find the 29th term.
(Hint: If ‘a’ is the first term and ‘d’ the common difference, then we arrive at the equations a + 2d = 12 and a + 49d = 106. Solve this pair of linear equations for ‘a’ and ‘d’.)

Solution:

Let the first term be a and the common difference be d.

Given:

t₃ = 12

t₅₀ = 106

Using:

tₙ = a + (n – 1)d

For the third term:

a + 2d = 12 …(i)

For the 50th term:

a + 49d = 106 …(ii)

Subtract equation (i) from equation (ii):

47d = 94

d = 2

Substitute d = 2 in equation (i):

a + 2(2) = 12

a + 4 = 12

a = 8

Now, find the 29th term:

t₂₉ = a + (29 – 1)d

= 8 + 28(2)

= 8 + 56

= 64

Therefore, the 29th term is 64.

Question 5. How many 2-digit numbers are divisible by 3? What is the sum of all these 2-digit numbers?

Solution:

The two-digit numbers divisible by 3 are:

12, 15, 18, …, 99

This is an arithmetic progression with:

a = 12

d = 3

Last term = 99

Using:

tₙ = a + (n – 1)d

99 = 12 + (n – 1)3

87 = 3(n – 1)

29 = n – 1

n = 30

Therefore, there are 30 two-digit numbers divisible by 3.

The sum of an AP is:

Sₙ = n/2 × (first term + last term)

S₃₀ = 30/2 × (12 + 99)

= 15 × 111

= 1665

Therefore, there are 30 two-digit numbers divisible by 3, and their total sum is 1665.

Question 6. 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?

Solution:

Harish’s annual salaries form an arithmetic progression.

First salary:

a = ₹5,00,000

Annual increment:

d = ₹20,000

Required salary:

tₙ = ₹7,00,000

Using:

tₙ = a + (n – 1)d

7,00,000 = 5,00,000 + (n – 1)(20,000)

2,00,000 = (n – 1)(20,000)

n – 1 = 10

n = 11

Therefore, Harish’s salary reached ₹7,00,000 in the 11th year, that is, after completing 10 years of work.

Question 7. 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?

Solution:

The numbers of marbles in the rows are:

1, 2, 3, …, 25

The total number of marbles is the sum of the first 25 natural numbers.

Using:

Sₙ = n(n + 1)/2

S₂₅ = 25(25 + 1)/2

= 25 × 26/2

= 25 × 13

= 325

Therefore, the child uses 325 marbles in all.

Exercise Set 8.3 

Question 1. Find the 12th term of a GP with common ratio 2, whose 8th term is 192.

Solution:

The nth term of a geometric progression is:

tₙ = arⁿ⁻¹

Given:

t₈ = 192

r = 2

Therefore:

192 = a × 2⁸⁻¹

192 = a × 2⁷

192 = 128a

a = 192/128

a = 3/2

Now, the 12th term is:

t₁₂ = a × r¹²⁻¹

= 3/2 × 2¹¹

= 3 × 2¹⁰

= 3 × 1024

= 3072

Therefore, the 12th term of the GP is 3072.

Question 2. Find the 10th and nth terms of the GP: 5, 25, 125,….

Solution:

The given geometric progression is:

5, 25, 125, …

Here:

First term, a = 5

Common ratio:

r = 25/5 = 5

The nth term of a GP is:

tₙ = arⁿ⁻¹

For the 10th term:

t₁₀ = 5 × 5¹⁰⁻¹

= 5 × 5⁹

= 5¹⁰

= 9765625

The nth term is:

tₙ = 5 × 5ⁿ⁻¹

= 5ⁿ

Therefore:

10th term = 9765625

nth term = 5ⁿ

Question 3. A sequence is given by the recursive rule t1 = 2, tn+1 = 3tn-2 for n ≥ 1. Which term of the sequence is 730?

Solution:

The recursive rule is:

t₁ = 2

tₙ₊₁ = 3tₙ - 2

Calculate the terms one by one:

t₁ = 2

t₂ = 3t₁ - 2

= 3(2) - 2

= 4

t₃ = 3t₂ - 2

= 3(4) - 2

= 10

t₄ = 3t₃ - 2

= 3(10) - 2

= 28

t₅ = 3t₄ - 2

= 3(28) - 2

= 82

t₆ = 3t₅ - 2

= 3(82) - 2

= 244

t₇ = 3t₆ - 2

= 3(244) - 2

= 730

Therefore, 730 is the 7th term of the sequence.

Question 4. Which term of the GP: 2, 6, 18, … is 4374? Write the explicit formula as well as the recursive formula for the 17th term.

Solution:

The given geometric progression is:

2, 6, 18, …

Here:

First term, a = 2

Common ratio:

r = 6/2 = 3

The nth term of a GP is:

tₙ = arⁿ⁻¹

Therefore:

tₙ = 2 × 3ⁿ⁻¹

To find the position of 4374:

4374 = 2 × 3ⁿ⁻¹

4374/2 = 3ⁿ⁻¹

2187 = 3ⁿ⁻¹

3⁷ = 3ⁿ⁻¹

n - 1 = 7

n = 8

Therefore, 4374 is the 8th term of the GP.

Explicit formula:

tₙ = 2 × 3ⁿ⁻¹

Recursive formula:

t₁ = 2

tₙ₊₁ = 3tₙ, for n ≥ 1

Using the explicit formula, the 17th term is:

t₁₇ = 2 × 3¹⁶

= 86093442

Therefore, the 17th term is 86093442.

Question 5. 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?

Solution:

The height reached after each bounce forms a geometric progression.

Initial height = 80 metres

Common ratio = 60%

= 0.6

The height after the first bounce is:

80 × 0.6 = 48 metres

Therefore, the height after the fifth bounce is:

80 × (0.6)⁵

= 80 × 0.07776

= 6.2208 metres

Therefore, the ball reaches a height of 6.2208 metres, or approximately 6.22 metres, after the fifth bounce.

(ii) What is the total vertical distance the ball has travelled by the time it hits the ground for the 6th time?

Solution:

The ball first falls 80 metres before hitting the ground for the first time.

The successive bounce heights are:

First bounce:

80 × 0.6 = 48 metres

Second bounce:

48 × 0.6 = 28.8 metres

Third bounce:

28.8 × 0.6 = 17.28 metres

Fourth bounce:

17.28 × 0.6 = 10.368 metres

Fifth bounce:

10.368 × 0.6 = 6.2208 metres

Before hitting the ground for the sixth time, the ball travels upward and downward through each of these five bounce heights.

Therefore, the total distance is:

80 + 2(48 + 28.8 + 17.28 + 10.368 + 6.2208)

= 80 + 2(110.6688)

= 80 + 221.3376

= 301.3376 metres

Therefore, the total vertical distance travelled is approximately 301.34 metres.

Question 6. Which term of the sequence 2, 2√2, 4, … is 128?

Solution:

The given sequence is:

2, 2√2, 4, …

The first term is:

a = 2

The common ratio is:

r = 2√2/2

= √2

The nth term of a GP is:

tₙ = arⁿ⁻¹

Therefore:

tₙ = 2(√2)ⁿ⁻¹

To find the position of 128:

128 = 2(√2)ⁿ⁻¹

64 = (√2)ⁿ⁻¹

Write both sides as powers of √2:

64 = 2⁶

= (√2)¹²

Therefore:

(√2)¹² = (√2)ⁿ⁻¹

n - 1 = 12

n = 13

Therefore, 128 is the 13th term of the sequence.

Question 7. Figure 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.

Figure: Stages 0, 1, 2 and 3 of the Sierpiński square carpet

Look at Figure and try to answer the following questions.

(i) How many red squares are there in Stages 0 to 3?

Solution:

At each new stage, every retained red square produces 8 smaller red squares.

Stage 0:

1 red square

Stage 1:

1 × 8 = 8 red squares

Stage 2:

8 × 8 = 64 red squares

Stage 3:

64 × 8 = 512 red squares

Therefore, the numbers of red squares in Stages 0, 1, 2 and 3 are:

1, 8, 64 and 512

The total number across these four stages is:

1 + 8 + 64 + 512 = 585

(ii) Can you predict the number of red squares in Stages 4 and 5?

Solution:

The number of red squares is multiplied by 8 at every stage.

Stage 4:

512 × 8 = 4096

Stage 5:

4096 × 8 = 32768

Therefore:

Stage 4 has 4096 red squares.

Stage 5 has 32768 red squares.

(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.

Solution:

The sequence of red squares is:

1, 8, 64, 512, …

This is a geometric progression with:

First term = 1 at Stage 0

Common ratio = 8

Therefore, the explicit formula for the number of red squares at Stage n is:

tₙ = 8ⁿ

The recursive formula is:

t₀ = 1

tₙ₊₁ = 8tₙ, for n ≥ 0

(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?

Solution:

At every stage, each square is divided into 9 equal smaller squares, and the central square is removed.

Therefore, 8/9 of the area from the previous stage remains.

Stage 0:

A₀ = 1 square unit

Stage 1:

A₁ = 8/9 square unit

Stage 2:

A₂ = 8/9 × 8/9

= (8/9)²

= 64/81 square unit

Stage 3:

A₃ = (8/9)³

= 512/729 square unit

Stage 4:

A₄ = (8/9)⁴

= 4096/6561 square unit

Stage 5:

A₅ = (8/9)⁵

= 32768/59049 square unit

The explicit formula for the area at Stage n is:

Aₙ = (8/9)ⁿ

The recursive formula is:

A₀ = 1

Aₙ₊₁ = 8/9 Aₙ, for n ≥ 0

Since 8/9 is less than 1, the area of the red region becomes smaller with every stage.

As n increases without limit:

(8/9)ⁿ approaches 0

Therefore, the area of the red region approaches zero as the number of stages increases.

End of Chapter Exercises Solutions

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

Solution:

Let the first term be a and the common difference be d.

The nth term of an AP is:

tₙ = a + (n – 1)d

Given:

t₁₁ = 38

a + 10d = 38 …(i)

Also:

t₁₆ = 73

a + 15d = 73 …(ii)

Subtract equation (i) from equation (ii):

5d = 35

d = 7

Substitute d = 7 in equation (i):

a + 10(7) = 38

a + 70 = 38

a = –32

Now:

t₃₁ = a + 30d

= –32 + 30(7)

= –32 + 210

= 178

Therefore, the 31st term is 178.

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

Solution:

Let the first term be a and the common difference be d.

Given:

t₃ = 16

a + 2d = 16 …(i)

Also:

t₇ – t₅ = 12

Now:

t₇ = a + 6d

and

t₅ = a + 4d

Therefore:

(a + 6d) – (a + 4d) = 12

2d = 12

d = 6

Substitute d = 6 in equation (i):

a + 2(6) = 16

a + 12 = 16

a = 4

Therefore, the required AP is:

4, 10, 16, 22, 28, 34, 40, …

Question 3. How many three-digit numbers are divisible by 7? (Hint: All three-digit numbers divisible by 7 form an AP. Find the smallest and largest such three-digit numbers.)

Solution:

The smallest three-digit number divisible by 7 is 105, and the largest is 994.

The numbers form the AP:

105, 112, 119, …, 994

Here:

a = 105

d = 7

tₙ = 994

Using:

tₙ = a + (n – 1)d

994 = 105 + (n – 1)7

889 = 7(n – 1)

127 = n – 1

n = 128

Therefore, there are 128 three-digit numbers divisible by 7.

Question 4. How many multiples of 4 lie between 10 and 250? (Hint: All multiples of 4 form an AP. Find the smallest and largest multiples of 4 between 10 and 250.)

Solution:

The smallest multiple of 4 greater than 10 is 12, and the largest multiple of 4 less than 250 is 248.

Thus, the AP is:

12, 16, 20, …, 248

Here:

a = 12

d = 4

tₙ = 248

Using:

248 = 12 + (n – 1)4

236 = 4(n – 1)

59 = n – 1

n = 60

Therefore, 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.

Solution:

Let the first term be a and the common ratio be r.

The first, second, third and fifth terms are:

a, ar, ar² and ar⁴

Given:

a + ar = –4 …(i)

Also:

ar⁴ = 4ar²

Assuming a and r are non-zero:

r² = 4

Therefore:

r = 2 or r = –2

Case 1: r = 2

Using equation (i):

a + 2a = –4

3a = –4

a = –4/3

Therefore, one possible GP is:

–4/3, –8/3, –16/3, –32/3, …

Case 2: r = –2

Using equation (i):

a – 2a = –4

–a = –4

a = 4

Therefore, the other possible GP is:

4, –8, 16, –32, …

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

Solution:

Let there be k consecutive natural numbers beginning with a:

a, a + 1, a + 2, …, a + k – 1

Their sum is:

a + (a + 1) + … + (a + k – 1) = 100

Using the sum formula:

ka + k(k – 1)/2 = 100

Multiplying by 2:

2ka + k(k – 1) = 200

k(2a + k – 1) = 200

Checking suitable factor pairs of 200 gives the following possibilities.

For k = 8:

2a + 7 = 25

2a = 18

a = 9

Therefore:

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

For k = 5:

2a + 4 = 40

2a = 36

a = 18

Therefore:

100 = 18 + 19 + 20 + 21 + 22

For k = 1:

a = 100

Therefore:

100 = 100

Hence, all possible representations are:

100

18 + 19 + 20 + 21 + 22

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

Question 7. 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?

Solution:

The number of bacteria doubles every hour.

Initial number of bacteria = 30

Common ratio = 2

After n hours, the number of bacteria is:

30 × 2ⁿ

At the end of the second hour:

30 × 2²

= 30 × 4

= 120

At the end of the fourth hour:

30 × 2⁴

= 30 × 16

= 480

Therefore:

After the second hour = 120 bacteria

After the fourth hour = 480 bacteria

After the nth hour = 30 × 2ⁿ bacteria

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 of the AP.

Solution:

Let the first term be a and the common difference be d.

Given:

t₄ + t₈ = 24

Now:

t₄ = a + 3d

t₈ = a + 7d

Therefore:

a + 3d + a + 7d = 24

2a + 10d = 24

a + 5d = 12 …(i)

Also:

t₆ + t₁₀ = 44

t₆ = a + 5d

t₁₀ = a + 9d

Therefore:

a + 5d + a + 9d = 44

2a + 14d = 44

a + 7d = 22 …(ii)

Subtract equation (i) from equation (ii):

2d = 10

d = 5

Substitute d = 5 in equation (i):

a + 25 = 12

a = –13

The first three terms are:

a = –13

a + d = –8

a + 2d = –3

Therefore, the first three terms are –13, –8 and –3.

Question 9. Find the smallest value of n such that the sum of the first n natural numbers is greater than 1,000.

Solution:

The sum of the first n natural numbers is:

Sₙ = n(n + 1)/2

We need:

n(n + 1)/2 > 1000

Therefore:

n(n + 1) > 2000

Check the values near √2000.

For n = 44:

S₄₄ = 44 × 45/2

= 990

This is less than 1000.

For n = 45:

S₄₅ = 45 × 46/2

= 1035

This is greater than 1000.

Therefore, the smallest possible value of n is 45.

Question 10. Which term of the GP: 2, 8, 32, … is 131072? Write the explicit formula as well as the recursive formula for the nth term.

Solution:

The given GP is:

2, 8, 32, …

Here:

a = 2

r = 8/2 = 4

The explicit formula is:

Tₙ = arⁿ⁻¹

Therefore:

Tₙ = 2 × 4ⁿ⁻¹

Now, set Tₙ = 131072:

131072 = 2 × 4ⁿ⁻¹

65536 = 4ⁿ⁻¹

Since:

65536 = 4⁸

we get:

n – 1 = 8

n = 9

Therefore, 131072 is the 9th term.

Explicit formula:

Tₙ = 2 × 4ⁿ⁻¹

Recursive formula:

T₁ = 2

Tₙ = 4Tₙ₋₁, for n ≥ 2

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.

Solution:

Let the first three terms of the GP be:

a, ar and ar²

Given:

a + ar + ar² = 13/12 …(i)

Also:

a × ar × ar² = –1

a³r³ = –1

(ar)³ = –1

Therefore:

ar = –1

So:

a = –1/r

Substitute this in equation (i):

–1/r – 1 – r = 13/12

Multiplying by 12r:

–12 – 12r – 12r² = 13r

12r² + 25r + 12 = 0

Factorising:

(3r + 4)(4r + 3) = 0

Therefore:

r = –4/3 or r = –3/4

Case 1: r = –4/3

Since ar = –1:

a = 3/4

The terms are:

3/4, –1, 4/3

Case 2: r = –3/4

Since ar = –1:

a = 4/3

The terms are:

4/3, –1, 3/4

Therefore, the common ratio is –4/3 or –3/4, and the 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.

Solution:

Let the first term of the original GP be a and the common ratio be r.

The nth term is:

Tₙ = arⁿ⁻¹

Therefore:

x = T₄ = ar³

y = T₁₀ = ar⁹

z = T₁₆ = ar¹⁵

Now:

y/x = ar⁹/ar³

= r⁶

Also:

z/y = ar¹⁵/ar⁹

= r⁶

Thus:

y/x = z/y

Therefore, x, y and z have a common ratio r⁶.

Hence, x, y and z are in GP.

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.

Solution:

Let the first three terms be:

a, ar and ar²

Given:

a + ar + ar² = 26 …(i)

Also:

a² + a²r² + a²r⁴ = 364 …(ii)

Square equation (i):

(a + ar + ar²)² = 26²

a² + a²r² + a²r⁴ + 2(a²r + a²r² + a²r³) = 676

Using equation (ii):

364 + 2ar[a(1 + r + r²)] = 676

From equation (i):

a(1 + r + r²) = 26

Therefore:

364 + 2ar(26) = 676

52ar = 312

ar = 6

Thus, the middle term is 6.

Since ar = 6:

a = 6/r

Using equation (i):

6/r + 6 + 6r = 26

Divide by 2:

3/r + 3 + 3r = 13

Multiplying by r:

3 + 3r + 3r² = 13r

3r² – 10r + 3 = 0

Factorising:

(3r – 1)(r – 3) = 0

Therefore:

r = 1/3 or r = 3

If r = 3, the terms are:

2, 6, 18

If r = 1/3, the terms are:

18, 6, 2

Therefore, the terms of the GP are 2, 6 and 18, or the same terms in reverse order.

Question 14. 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?

Solution:

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

P₆ = 1 + 2 + 4 + 8 + 16 + 1

= 32

P₇ = 1 + 2 + 4 + 8 + 16 + 32 + 1

= 64

P₈ = 1 + 2 + 4 + 8 + 16 + 32 + 64 + 1

= 128

Therefore:

P₁, P₂, …, P₈ = 1, 2, 4, 8, 16, 32, 64, 128

For n > 2:

Pₙ = P₁ + P₂ + … + Pₙ₋₂ + Pₙ₋₁ + 1

Also:

Pₙ₋₁ = P₁ + P₂ + … + Pₙ₋₂ + 1

Therefore:

Pₙ = (Pₙ₋₁ – 1) + Pₙ₋₁ + 1

Pₙ = 2Pₙ₋₁

A simpler recursive formula is:

P₁ = 1

Pₙ = 2Pₙ₋₁, for n ≥ 2

The sequence consists of powers of 2.

Therefore, the explicit formula is:

Pₙ = 2ⁿ⁻¹

Question 15. 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?

Solution:

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 = W₁ + W₂ + W₃ + W₄ + W₅ + 2

= 1 + 2 + 3 + 5 + 8 + 2

= 21

W₈ = W₁ + W₂ + W₃ + W₄ + W₅ + W₆ + 2

= 1 + 2 + 3 + 5 + 8 + 13 + 2

= 34

Therefore:

W₁, W₂, …, W₈ = 1, 2, 3, 5, 8, 13, 21, 34

To obtain a simpler recursive rule, observe that:

Wₙ₋₁ = W₁ + W₂ + … + Wₙ₋₃ + 2

Therefore:

Wₙ = Wₙ₋₁ + Wₙ₋₂

So, the simpler recursive rule is:

W₁ = 1

W₂ = 2

Wₙ = Wₙ₋₁ + Wₙ₋₂, for n ≥ 3

This is the Fibonacci-type sequence beginning with 1 and 2.

Key Formulas Used in Class 9 Maths Chapter 8 Sequences and Progressions

The whole chapter rests on a few formulas worth memorising: the nth term of an AP, tₙ = a + (n − 1)d; the nth term of a GP, tₙ = arⁿ⁻¹; the sum of the first n natural numbers, n(n + 1)/2; and the AP sum, n/2 × (first term + last term). The quickest way to lock these in is to practise "which term is this number" questions, like the ones in Exercise Sets 8.2 and 8.3, then test yourself with Vedantu's Chapter 8 Important Questions.

Access Exercise Wise NCERT Solutions for Chapter 8 Maths Class 9

CBSE Class 9 Maths Chapter 8 Predicting What Comes Next: Exploring Sequences and Progressions Other Study Materials

Chapter-Specific NCERT Solutions for Class 9 Maths

Given below are the chapter-wise NCERT Solutions for Class 9 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

Additional Study Materials for Class 9 Maths