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Updated 6 August 2026

NCERT Solutions

Predicting What Comes Next : Exploring Sequences and Progressions

CBSE · Class 9 · Mathematics

NCERT Solutions for Predicting What Comes Next : Exploring Sequences and Progressions — CBSE Class 9 Mathematics.

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Think and Reflect

1Can you describe the pattern in each of the above sequences? Can you predict the next few numbers in these sequences?Show solution

- Natural numbers: each term is 1 more than the previous term, so the next few are 7, 8, 9, ...
- Odd numbers: each term increases by 2, so the next few are 13, 15, 17, ...
- Triangular numbers: the differences increase by 1 each time; the next few are 28, 36, 45, ...
- Square numbers: each term is the square of a natural number; the next few are 49, 64, 81, ...

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2Can you think of other finite sequences that you see in your daily life?Show solution

Yes. Examples of finite sequences in daily life include:
- a number of people in each row of seats in a classroom
- pages in a chapter
- floors in a building
- scores in a game over a fixed number of rounds
- steps in a staircase with a fixed number of steps

Any such list with a fixed number of terms is a finite sequence.

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3Why is it useful to have an explicit formula for the

n^{th}

term of a sequence?Show solution

An explicit formula is useful because it lets us find any term directly by substituting the position number

n

.

So we can find very far terms like the 20th, 53rd, or 300th term without calculating all the earlier terms first. We can also check whether a number belongs to the sequence and find its position.

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4Can you find the rule describing the

n^{th}

term of the sequence of square numbers?Show solution

The sequence of square numbers is

1,4,9,16,25,\dots

These are the squares of natural numbers:

1=1^2,\;4=2^2,\;9=3^2,\;16=4^2,\;25=5^2,\dots

So the rule for the

n^{\text{th}}

term is

\boxed{t_n=n^2}.

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5Can you think of a rule that can predict the next few prime numbers?Show solution

The chapter says prime numbers do not show a clear regular pattern like APs or GPs. So there is no simple rule given here that can predict the next few prime numbers exactly.

The prime sequence begins:

2,3,5,7,11,13,17,19,23,29,\dots

But unlike APs or GPs, there is no easy formula in this chapter for generating the next prime.

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6Can 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?Show solution

The sequence for the number of squares is a GP with first term

3

and common ratio

2

.

So:
- Stage 5:

3\times 2^{5-1}=3\times 16=48

- Stage 6:

3\times 2^{6-1}=3\times 32=96

- Stage 10:

3\times 2^9=3\times 512=1536

- Stage 11:

3\times 2^{10}=3072

- Stage 12:

3\times 2^{11}=6144

- Stage 20:

3\times 2^{19}=1,572,864

In general, at stage

n

, the number of squares is

\boxed{3\times 2^{n-1}}.

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7Consider 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?Show solution

The arithmetic progressions (APs) discussed in the chapter are the sequences in which the difference between consecutive terms is constant.

Examples of APs from the chapter:
-

1,4,7,10,\dots

1,5,9,13,\dots

11,7,3,-1,-5,\dots

2,5,8,11,\dots

-5,-1,3,7,\dots

Not APs:
- natural numbers are an AP, yes, with difference 1
- odd numbers are an AP, yes, with difference 2
- triangular numbers are not an AP
- square numbers are not an AP
- prime numbers are not an AP
- geometric progressions are not APs because their ratio, not difference, is constant

So, only those sequences with a constant difference are APs.

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8Can the same approach be used to find the sum of

1 + 2 + 3 + \dots + 100

?Show solution

Yes. The same method can be used.

Let

S=1+2+3+\cdots+100.

Reversing it,

S=100+99+98+\cdots+1.

Adding gives

2S=(1+100)+(2+99)+\cdots=(101)\text{ added 100 times}.

2S=100\times 101

and therefore

S=\frac{100\times 101}{2}=5050.

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9Can you use this formula to find

S_{20}

S_{50}

S_{1000}

?Show solution

Using

S_n=\frac{n(n+1)}{2},

we get:

-

S_{20}=\frac{20\times 21}{2}=10\times 21=210

S_{50}=\frac{50\times 51}{2}=25\times 51=1275

S_{1000}=\frac{1000\times 1001}{2}=500\times 1001=500500

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10Can you use this to find the

10^{th}

17^{th}

and

80^{th}

triangular numbers?Show solution

The

n^{\text{th}}

triangular number is

t_n=\frac{n(n+1)}{2}.

So:
-

t_{10}=\frac{10\times 11}{2}=55

t_{17}=\frac{17\times 18}{2}=153

t_{80}=\frac{80\times 81}{2}=40\times 81=3240

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11Can 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?Show solution

This is a geometric progression. The number of squares doubles at each stage, so the sequence is:

3,6,12,24,48,96,\dots

Thus:
- Stage 5: 48
- Stage 6: 96
- Stage 10: 1536
- Stage 11: 3072
- Stage 12: 6144
- Stage 20: 1,572,864

In contrast to Fig. 8.3, where each stage increases by a fixed difference of 4 and forms an AP, this pattern has a fixed ratio of 2 and forms a GP.

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12a rule that can predict the next few prime numbers?Show solution

No simple rule is given in the chapter for predicting prime numbers. The chapter only lists the first ten primes and notes that they do not show a clear regular pattern like APs or GPs.

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13Can you find a recursive rule for the formula

t_n = 3 \times 10^{n-1}

that generates the geometric progression 3, 30, 300, 3000, ... ?Show solution

The GP is

3,30,300,3000,\dots

Each term is obtained by multiplying the previous term by

10

.
So the recursive rule is:

\boxed{t_1=3,\; t_n=10t_{n-1}\text{ for }n\ge 2.}

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

The number of black triangles at Stages 0 to 3 is:
- Stage 0: 1
- Stage 1: 3
- Stage 2: 9
- Stage 3: 27

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

The number of black triangles triples at each stage, so:
- Stage 4:

27\times 3=81

- Stage 5:

81\times 3=243

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3(c)Can you find a rule for the number of black triangles at the

n^{th}

stage?Show solution

The number of black triangles follows the GP

1,3,9,27,81,243,\dots

which is powers of 3.
So the rule for the

n^{\text{th}}

stage is

\boxed{t_n=3^n}.

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3(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

n^{th}

stage. What happens to this area as

n

, the number of stages, goes on increasing?Show solution

If the black region in Stage 0 has area 1 square unit, then:
- Stage 1:

\frac{3}{4}

- Stage 2:

\left(\frac{3}{4}\right)^2

- Stage 3:

\left(\frac{3}{4}\right)^3

- Stage 4:

\left(\frac{3}{4}\right)^4

- Stage 5:

\left(\frac{3}{4}\right)^5

In general, the area of the black region at Stage

n

\left(\frac{3}{4}\right)^n.

n

increases, this area gets smaller and smaller and tends to 0.

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18Can you think of other kinds of sequences? List out five different types of sequences and discuss their properties with your friends.Show solution

Five different kinds of sequences could be:
1. Natural number sequence:

1,2,3,4,\dots

2. Odd number sequence:

1,3,5,7,\dots

3. Even number sequence:

2,4,6,8,\dots

4. Square number sequence:

1,4,9,16,\dots

5. Geometric sequence:

3,6,12,24,\dots

Their properties differ: some increase by a fixed difference, some by a fixed ratio, and some follow special patterns like squares or triangular numbers.

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Exercise

1Can 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?Show solution

The sequence is

1,4,7,10,13,\dots

Each term increases by 3, so the next four terms are

16,19,22,25.

The sequence of sums up to each term is:
-

1

1+4=5

1+4+7=12

1+4+7+10=22

1+4+7+10+13=35

1+4+7+10+13+16=51

\dots

Thus the first ten terms of the sequence of running totals are:

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

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8.2Can you write

t_5, t_6, t_7

and

t_8

for the sequence of triangular numbers?Show solution

The triangular numbers are

1,3,6,10,15,21,28,36,\dots

So:
-

t_5=15

t_6=21

t_7=28

t_8=36

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

The sequence is given by

t_n=3n-7.

Now substitute the required values of

n

:
-

t_1=3(1)-7=-4

t_2=6-7=-1

t_3=9-7=2

t_{12}=36-7=29

t_{18}=54-7=47

t_{50}=150-7=143

So the required terms are

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

These are not among the listed options because the question asks for a computed answer from the given rule.

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1(ii)Which term of the sequence is 332?Show solution

We need to find

n

such that

3n-7=332.

So,

3n=339

n=113.

Thus 332 is the 113th term.

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

Check whether 557 is a term by solving

3n-7=557.

Then

3n=564

n=188.

Since

188

is a natural number, 557 is a term of the sequence. It is the 188th term.

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6Verify that the following sequences are arithmetic progressions and write their

n^{th}

terms. What do you observe when you plot the ordered pairs emerging from them?Show solution

Yes, both sequences are arithmetic progressions because the difference between consecutive terms is constant.

- For

2,5,8,11,\dots

, the common difference is

3

, so

t_n=2+(n-1)\cdot 3=3n-1.

- For

-5,-1,3,7,\dots

, the common difference is

4

, so

t_n=-5+(n-1)\cdot 4=4n-9.

When the ordered pairs are plotted, they lie on a straight line.

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8.11(i)Using the formula

t_n = a + (n - 1) \times d

, find the

n^{th}

term of the following arithmetic progressions.Show solution

For an arithmetic progression (AP) with first term **

a

and common difference

d

**, the

n^{\text{th}}

term is

t_n = a + (n - 1)d.

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8.12Find recursive rules for the APs in the previous exercises.Show solution

For an AP, each term is obtained by adding the common difference

d

to the previous term. So the recursive rule is

t_1 = a, \qquad t_n = t_{n-1} + d \text{ for } n \geq 2.

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EXERCISE SET 8.1

1Find the first five terms of the sequence in which the

n^{\text{th}}

term is given by (i)

t_n = 3n - 4

, (ii)

t_n = 2 - 5n

, and (iii)

t_n = n^2 - 2n + 3

for

n \geq 1

.Show solution

Substitute

n=1,2,3,4,5

.

(i)

t_n=3n-4

t_1=3(1)-4=-1

t_2=3(2)-4=2

t_3=3(3)-4=5

t_4=3(4)-4=8

t_5=3(5)-4=11

So the first five terms are **

-1, 2, 5, 8, 11

**.

(ii)

t_n=2-5n

t_1=2-5=-3

t_2=2-10=-8

t_3=2-15=-13

t_4=2-20=-18

t_5=2-25=-23

So the first five terms are **

-3, -8, -13, -18, -23

**.

(iii)

t_n=n^2-2n+3

t_1=1-2+3=2

t_2=4-4+3=3

t_3=9-6+3=6

t_4=16-8+3=11

t_5=25-10+3=18

So the first five terms are **

2, 3, 6, 11, 18

**.

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2Find the

10^{\text{th}}

and

15^{\text{th}}

terms of the sequence

t_n = 5n - 3

for

n \geq 1

.Show solution

Use

t_n=5n-3

.

-

t_{10}=5(10)-3=50-3=47

t_{15}=5(15)-3=75-3=72

So the required terms are 47 and 72.

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3Determine whether 97 and 172 are terms of the sequence

t_n = 5n - 3

for

n \geq 1

.Show solution

For a number to be a term of the sequence, it must satisfy

t_n=5n-3

for a natural number

n

.

For 97:

5n-3=97 \Rightarrow 5n=100 \Rightarrow n=20

Since

n=20

is a natural number, 97 is a term.

For 172:

5n-3=172 \Rightarrow 5n=175 \Rightarrow n=35

Since

n=35

is a natural number, actually 172 is also a term.

So both numbers are terms of the sequence.

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4Which term of the sequence

t_n = 5n - 3

for

n \geq 1

is 607?Show solution

Set

t_n=607

t_n=5n-3

5n-3=607

5n=610

n=122

So 607 is the 122nd term.

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5A sequence is given by the recursive rule

t_1 = -5

t_{n+1} = t_n + 3

for

n \geq 1

. Find the first five terms of the sequence. Is 52 a term of this sequence? If so, which term is it?Show solution

The recursive rule is

t_1=-5,\qquad t_{n+1}=t_n+3.

So the terms are found by adding 3 each time:
-

t_1=-5

t_2=-5+3=-2

t_3=-2+3=1

t_4=1+3=4

t_5=4+3=7

Hence the first five terms are **

-5,-2,1,4,7

**.

To check whether 52 is a term, write the AP formula:

t_n=-5+(n-1)\cdot 3=3n-8.

Set

t_n=52

3n-8=52

3n=60

n=20

So 52 is a term, and it is the 20th term.

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6Let

T_1 = 1

T_2 = 2

T_3 = 4

, and

T_n = T_{n-1} + T_{n-2} + T_{n-3}

for

n \geq 4

. Find

T_4

T_5

T_6

T_7

, and

T_8

.Show solution

Given

T_1=1,\ T_2=2,\ T_3=4,

and

T_n=T_{n-1}+T_{n-2}+T_{n-3} \quad (n\ge 4).

Now compute successively:

T_4=T_3+T_2+T_1=4+2+1=7

T_5=T_4+T_3+T_2=7+4+2=13

T_6=T_5+T_4+T_3=13+7+4=24

T_7=T_6+T_5+T_4=24+13+7=44

T_8=T_7+T_6+T_5=44+24+13=81

So the values are 7, 13, 24, 44, 81.

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EXERCISE SET 8.2

1Find the

10^{th}

and

26^{th}

terms of the AP: 3, 8, 13, 18, ...Show solution

For the AP

3,8,13,18,\dots

, the first term is

a=3

and common difference is

d=5

.

Use

t_n=a+(n-1)d

.

-

t_{10}=3+(10-1)\cdot 5=3+45=48

t_{26}=3+(26-1)\cdot 5=3+125=128

So the required terms are 48 and 128.

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

For the AP

21,18,15,\dots

, we have first term

a=21

and common difference

d=-3

.

The

n^{\text{th}}

term is

t_n=a+(n-1)d=21+(n-1)(-3)=24-3n.

To find which term is

-81

24-3n=-81

-3n=-105

n=35

So **

-81

is the 35th term**.

Now check whether 0 is a term:

24-3n=0

3n=24

n=8

Since

n=8

is a natural number, 0 is also a term of the AP.

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3Find the

n^{th}

term of the AP: 11, 8, 5, 2 ... Write the recursive rule for this AP.Show solution

For the AP

11,8,5,2,\dots

, the first term is

a=11

and common difference is

d=-3

.

So

t_n=a+(n-1)d=11+(n-1)(-3)=11-3n+3=14-3n.

Thus the

n^{\text{th}}

term is **

t_n=14-3n

**.

For the recursive rule, start with the first term and add the common difference each time:

t_1=11, \qquad t_n=t_{n-1}-3 \text{ for } n\ge2.

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4An AP consists of 50 terms in which the

3^{rd}

term is 12 and the last term is 106. Find the

29^{th}

term.Show solution

For an AP, if the 3rd term is 12 and the 50th term is 106, let first term be

a

and common difference be

d

.

Then

a+2d=12

a+49d=106

Subtract:

47d=94 \Rightarrow d=2

Then

a+2(2)=12 \Rightarrow a=8

Now the 29th term is

t_{29}=a+28d=8+28\cdot 2=8+56=64

So the 29th term is 64.

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

The two-digit multiples of 3 form an AP:

12,15,18,\dots,99

Here

a=12

d=3

, and

l=99

.

Number of terms:

n=\frac{l-a}{d}+1=\frac{99-12}{3}+1=29+1=30.

So there are 30 such numbers.

Sum:

S_n=\frac{n}{2}(a+l)=\frac{30}{2}(12+99)=15\cdot 111=1665.

So the sum is 1665.

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6Harish 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?Show solution

Harish's salary forms an AP:

- first salary

a=5,00,000

- yearly increment

d=20,000

After

n

years, salary is

a+(n-1)d

Set it equal to ₹7,00,000:

5,00,000+(n-1)20,000=7,00,000

20,000(n-1)=2,00,000

n-1=10

n=11

So ₹7,00,000 is reached in the 11th year of salary, which means after 10 years of increments. The textbook wording expects the answer 10 years.

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7A 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?Show solution

The marbles are arranged in 25 rows with 1, 2, 3, ..., 25 marbles.

So total marbles =

1+2+3+\cdots+25

Use

S_n=\frac{n(n+1)}{2}

with

n=25

S_{25}=\frac{25\cdot 26}{2}=25\cdot 13=325.

So the child uses 325 marbles.

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Example 10

8.13(i)Can you write the sequence of numbers obtained from the heights attained by the ball in five successive bounces?Show solution

The height after each bounce is multiplied by

\frac{3}{4}=0.75

.

Starting from 24 feet:
- 1st bounce:

24\times 0.75=18

- 2nd bounce:

18\times 0.75=13.5

- 3rd bounce:

13.5\times 0.75=10.125

- 4th bounce:

10.125\times 0.75=7.59375

- 5th bounce:

7.59375\times 0.75=5.6953125\approx 5.695

So the sequence of heights is 18, 13.5, 10.125, 7.59375, 5.695....

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8.13(ii)How many bounces are required for the ball to remain below a height of

\frac{1}{6}

of the original height from which it was dropped?Show solution

From the chapter, the ball remains below

\frac16

of the original height after the 7th bounce.

So the required number of bounces is 7.

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EXERCISE SET 8.3

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

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

3A sequence is given by the recursive rule

t_1 = 2

t_{n+1} = 3t_n - 2

for

n \geq 1

. Which term of the sequence is 730?

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

5(i)What height does the ball reach after the 5th bounce?

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

6Which term of the sequence

2, 2\sqrt{2}, 4, \dots

is 128?

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

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

7(iii)Can you find a rule for the number of red squares at the

n^{th}

stage? Write the explicit formula as well as the recursive formula for the number of red squares at any stage.

7(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

n^{th}

stage. What happens to this area as

n

, the number of stages, goes on increasing?

5A 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.

7Look at Fig. 8.12 and try to answer the following questions.

END-OF-CHAPTER EXERCISES

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

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

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

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

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

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

7The 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 2

^{nd}

hour, 4

^{th}

hour and

n^{th}

hour?

8The sum of the 4

^{th}

and 8

^{th}

terms of an AP is 24 and the sum of the 6

^{th}

and 10

^{th}

terms is 44. Find the first three terms of the AP.

9Find the smallest value of

n

such that the sum of the first

n

natural numbers is greater than 1,000.

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

n^{th}

term.

11The sum of the first three terms of a GP is

\frac{13}{12}

and their product is -1. Find the common ratio and the terms.

12If the 4

^{th}

, 10

^{th}

and 16

^{th}

terms of a GP are

x, y

and

z

respectively, prove that

x, y, z

are in GP.

13The 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.

14Suppose

P_1 = 1, P_2 = 2

and for

n > 2, P_n = P_1 + P_2 + \dots + P_{n-1} + 1

. Find the values of

P_1, P_2, \dots, P_8

. Can you find a simpler recursive formula for

P_n

? Can you give an explicit formula?

15Suppose

W_1 = 1, W_2 = 2

and for

n > 2, W_n = W_1 + W_2 + \dots + W_{n-2} + 2

. Find the values of

W_1, W_2, \dots, W_8

. Do you recognise this sequence?

8.1 INTRODUCTION TO SEQUENCES

1Can you draw the patterns for the next two terms of the sequence?

2Can you explain the relationship? You may recall some of these ideas from Grade 6, Chapter 1!

8.1Consider 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?

4Can you think of any other kinds of sequences? List out five different types of sequences and discuss their properties with your friends.

8.2 EXPLICIT RULE FOR A SEQUENCE

1Using the explicit rule

u_n = 2n - 1

, find the

53^{rd}

term, the

108^{th}

term, and the

1170^{th}

term of the odd number sequence.

8.3 RECURSIVE RULE FOR A SEQUENCE

1Find the first four terms of the sequence given by the recursive rule

2Find the first four terms of the sequence given by the recursive rule

s_1 = 3

s_n = s_{n-1}(s_{n-1} - 1)

for

n \geq 2

3Can you write the next two terms of this sequence?

8.4 ARITHMETIC PROGRESSIONS

1Can you find the sum of the first ten natural numbers without actually adding all of them?

2Let us revisit the sequence

t_n

of triangular numbers 1, 3, 6, 10, 15, ... shown in Fig. 8.1. Note that the

n^{th}

term of this sequence is the sum of the first

n

natural numbers. Thus $t_n = \frac{n(n+1)}{2}.

8.6 GEOMETRIC PROGRESSIONS

1Can you check whether the following sequences are geometric progressions and find their

n^{th}

terms.

Exercises

8.10(i)Verify that the following sequences are arithmetic progressions and write their

n^{th}

terms.

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Think and Reflect

Exercise

EXERCISE SET 8.1

EXERCISE SET 8.2

Example 10

EXERCISE SET 8.3

END-OF-CHAPTER EXERCISES

8.1 INTRODUCTION TO SEQUENCES

8.2 EXPLICIT RULE FOR A SEQUENCE

8.3 RECURSIVE RULE FOR A SEQUENCE

8.4 ARITHMETIC PROGRESSIONS

8.6 GEOMETRIC PROGRESSIONS

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