Important Questions and Answers for Class 9 Maths Chapter 8 – Predicting What Comes Next: Exploring Sequences and Progressions(Ganita Manjari Part I) | 2026–27

The following 25 original maths Class 9 Chapter 8 important questions include five MCQs, five very short answers, eight short answers and seven long answers. They focus on understanding mathematical patterns, choosing the right formula, solving application-based problems and interpreting diagrams.

Class 9 Maths Chapter 8 Important Questions MCQ

Q1. A sequence follows the rule $$t_n=4n-7$$. What is its 8th term?

(a) 21

(b) 25

(c) 28

(d) 32

Answer: (b) 25

Solution:

Given:

$$t_n=4n-7$$

Substitute n=8:

$$t_8=4(8)-7=32-7=\boxed{25}$$

Explanation: An explicit rule allows us to calculate any term directly from its position without finding the preceding terms.

Q2. Which of the following sequences is an arithmetic progression?

(a) 2,4,8,16,...

(b) 1,4,9,16,...

(c) 4,9,14,19,...

(d) 1,3,6,10,...

Answer: (c) 4,9,14,19,...

Explanation:

Calculate the differences:

9-4=5, 14-9=5, 19-14=5

The common difference is constant.

d=5

Therefore, the sequence is an AP.

Q3. What is the common ratio of the geometric progression 2,6,18,54,...?

(a) 2

(b) 3

(c) 4

(d) 6

Answer: (b) 3

Solution:

$$r=\dfrac{6}{2}=\dfrac{18}{6}=\dfrac{54}{18}=\boxed{3}$$

Explanation: A geometric progression is formed by multiplying each term by the same non-zero number.

Q4. What is the sixth triangular number?

(a) 15

(b) 18

(c) 21

(d) 25

Answer: (c) 21

Solution:

The formula for the nth triangular number is:

$$T_n=\dfrac{n(n+1)}{2}$$

For n=6:

$$T_6=\dfrac{6(7)}{2}=\boxed{21}$$

Q5. A Sierpiński triangle starts with an area of 1 square unit. At each stage, one-fourth of the remaining area is removed. What area remains after two stages?

$$ \begin{aligned} &\text{(a) } \dfrac{1}{4}\\[8pt] &\text{(b) } \dfrac{3}{8}\\[8pt] &\text{(c) } \dfrac{9}{16}\\[8pt] &\text{(d) } \dfrac{15}{16}\\[14pt] &\textbf{Answer: } \text{(c) } \dfrac{9}{16}\\[14pt] &\textbf{Solution:}\\[8pt] &\text{The remaining fraction at every stage is:}\\[8pt] &1-\dfrac{1}{4}=\dfrac{3}{4} \end{aligned} $$

After two stages:

$$A_2=1\times\dfrac34\times\dfrac34$$

$$\boxed{A_2=\dfrac9{16}}$$

Explanation: The remaining areas form a geometric progression with common ratio $$\dfrac{3}{4}$$.

Class 9 Maths Chapter 8 Very Short Answer Questions

Q6. What is the difference between a finite sequence and an infinite sequence? Give an example of each.

Answer:

A finite sequence contains a fixed number of terms.

Example:

3,6,9,12,15

An infinite sequence continues without a last term.

Example:

3,6,9,12,15,...

The first has five terms, while the second continues indefinitely.

Q7. How is an explicit rule different from a recursive rule?

Answer:

An explicit rule calculates a term directly using its position.

Example:

$$t_n=3n+1$$

A recursive rule calculates a term using one or more earlier terms.

Example:

$$t_1=4,\qquad t_n=t_{n-1}+3,\quad n\ge2$$

Both describe the same sequence 4,7,10,13,....

Q8. Find the next four terms of the Virahānka–Fibonacci sequence 1,2,3,5,8,13,....

Answer:

Each new term is the sum of the previous two.

8+13=21

13+21=34

21+34=55

34+55=89

Next four terms:

21,34,55,89

Q9. Write the nth term of an AP whose first term is -2 and common difference is 5.

Answer:

$$t_n=a+(n-1)d$$

Substitute a=-2 and d=5:

$$t_n=-2+5(n-1)$$

$$\boxed{t_n=5n-7}$$

Q10. Identify the common ratio of the GP 81,27,9,3,....

Answer:

$$r=\dfrac{27}{81}=\dfrac{9}{27}=\dfrac{3}{9}$$

$$\boxed{r=\dfrac13}$$

Since the ratio is less than 1, the positive terms decrease.

Class 9 Maths Chapter 8 Short Answer Questions

Q11. Find the nth term and 20th term of the AP 5,12,19,26,....

Solution:

Given:

a=5, d=12-5=7

The nth term of an AP is:

$$t_n=a+(n-1)d$$

Substituting:

$$t_n=5+(n-1)7$$

$$t_n=5+7n-7$$

$$\boxed{t_n=7n-2}$$

For the 20th term:

$$t_{20}=7(20)-2$$

$$t_{20}=140-2$$

$$\boxed{t_{20}=138}$$

Q12. The recursive rule of a sequence is $$t_1=3$$ and $$t_n=2t_{n-1}-1$$, for n≥2. Find its first five terms.

Solution:

First term:

$$t_1=3$$

Second term:

$$t_2=2(3)-1=5$$

Third term:

$$t_3=2(5)-1=9$$

Fourth term:

$$t_4=2(9)-1=17$$

Fifth term:

$$t_5=2(17)-1=33$$

Answer:

3,5,9,17,33

Explanation: This is not an arithmetic progression because the differences 2,4,8,16 are not constant.

Q13. The nth term of a sequence is $$t_n=6n+2$$. Check whether 206 belongs to the sequence. If yes, find its position.

Solution:

Set:

6n+2=206

Subtract 2:

6n=204

Divide by 6:

$$n=\dfrac{204}{6}=34$$

Since 34 is a positive integer, 206 belongs to the sequence.

\boxed{206 is the 34th term}

Explanation: A number belongs to a sequence given by an explicit rule if its position is a valid positive integer.

Q14. Find the sum of all natural numbers from 31 to 70 using a suitable formula.

Solution:

The sequence is:

31,32,33,...,70

Number of terms:

n=70-31+1=40

The terms form an AP.

Sum:

$$S=\dfrac n2(a+l)$$

where a=31 and l=70.

$$S=\dfrac{40}{2}(31+70)$$

S=20×101

S=2020

Alternative: Subtract the sum of the first 30 natural numbers from the sum of the first 70.

Q15. A triangular dot pattern has 1 dot in the first row, 2 in the second row, and so on. Find the total number of dots in the first 18 rows.

Solution:

Each row adds one more dot.

Total dots in n rows:

$$T_n=1+2+3+\cdots+n$$

Using:

$$T_n=\dfrac{n(n+1)}2$$

For n=18:

$$T_{18}=\dfrac{18\times19}{2}$$

$$T_{18}=9\times19$$

$\boxed{171 dots}$

Explanation: Triangular numbers are formed by adding consecutive natural numbers.

Q16. Two patterns contain the following numbers of squares at successive stages:

Pattern A: 7,11,15,19,...

Pattern B: 3,6,12,24,...

Find the number of squares at Stage 8 in each pattern.

Solution:

Pattern A

The differences are:

11-7=4, 15-11=4

This is an AP with a=7 and d=4.

$$t_8=7+(8-1)4$$

$$t_8=7+28=\boxed{35}$$

Pattern B

Each term is twice the previous term.

This is a GP with a=3 and r=2.

$$t_8=ar^{8-1}$$

$$t_8=3(2)^7$$

$$t_8=3\times128=\boxed{384}$$

Answer: Pattern A has 35 squares; Pattern B has 384 squares.

The comparison shows how multiplication can produce faster growth than adding a fixed number.

Q17. The first four terms of a sequence are 2,5,8,11. Write its explicit rule and explain why the graph of its terms forms a straight-line pattern.

Solution:

First term:

a=2

Common difference:

d=5-2=3

Explicit rule:

$$t_n=2+(n-1)3$$

$$\boxed{t_n=3n-1}$$

Plot the points:

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

Each time the position increases by 1, the term increases by 3.

The relationship is linear:

y=3x-1

Therefore, the plotted points lie on a straight line.

Q18. A ball is dropped from a height of 64 metres. After each bounce, it reaches half its previous height. Find the height reached after its fourth bounce.

Solution:

Initial height:

H=64 m

Common ratio:

$$r=\dfrac12$$

Height after n bounces:

$$H_n=64\left(\dfrac12\right)^n$$

After the first bounce:

$$H_1=32\text{ m}$$

After the second bounce:

$$H_2=16\text{ m}$$

After the third bounce:

$$H_3=8\text{ m}$$

After the fourth bounce:

$$H_4=64\left(\dfrac12\right)^4$$

$$H_4=\dfrac{64}{16}$$

$$\boxed{H_4=4\text{ metres}}$$

The heights form a decreasing geometric progression.

Class 9 Maths Chapter 8 Long Answer Questions

Q19. Derive the formula for the sum of the first n natural numbers. Hence, find the sum of the first 75 natural numbers.

Solution:

We want to calculate:

$$S_n=1+2+3+\cdots+n$$

Step 1: Write the sum in reverse order.

$$S_n=n+(n-1)+(n-2)+\cdots+1$$

Step 2: Add both expressions vertically.

$$[ \begin{aligned} S_n &=1+2+3+\cdots+n\ S_n &=n+(n-1)+(n-2)+\cdots+1 \end{aligned} ]$$

Each corresponding pair adds up to n+1.

There are n such pairs.

Therefore:

$$2S_n=n(n+1)$$

Step 3: Divide by 2.

$$\boxed{S_n=\dfrac{n(n+1)}2}$$

Step 4: Find the sum of the first 75 natural numbers.

$$S_{75}=\dfrac{75(76)}2$$

$$S_{75}=75\times38$$

$$\boxed{S_{75}=2850}$$

Answer: The sum of the first 75 natural numbers is 2850.

This method is useful because it avoids adding all 75 terms separately.

Q20. In an arithmetic progression, the fourth term is 17, and the twelfth term is 49. Find the first term, common difference and 25th term.

Solution:

The formula is:

$$t_n=a+(n-1)d$$

Step 1: Use the fourth term.

$$t_4=a+3d=17$$

So:

a+3d=17 (1)

Step 2: Use the twelfth term.

$$t_{12}=a+11d=49$$

So:

a+11d=49 (2)

Step 3: Subtract Equation (1) from Equation (2).

8d=32

d=4

Step 4: Find the first term.

Substitute d=4 into Equation (1):

a+3(4)=17

a+12=17

a=5

Step 5: Calculate the 25th term.

$$t_{25}=5+(25-1)4$$

$$t_{25}=5+96$$

$$\boxed{t_{25}=101}$$

Final Answer:

First term = 5

Common difference = 4

25th term = 101

Q21. The second term of a geometric progression is 12, and its fifth term is 324. If the common ratio is positive, find the first term, common ratio and seventh term.

Solution:

The nth term of a GP is:

$$t_n=ar^{n-1}$$

Step 1: Use the second term.

ar=12 (1)

Step 2: Use the fifth term.

$$ar^4=324\qquad(2)$$

Step 3: Divide Equation (2) by Equation (1). 

$$\dfrac{ar^4}{ar}=\dfrac{324}{12}$$

$$r^3=27$$

Since the common ratio is positive:

r=3

Step 4: Find the first term.

a(3)=12

a=4

Step 5: Find the seventh term.

$$t_7=ar^6$$

$$t_7=4(3)^6$$

$$t_7=4(729)$$

$$\boxed{t_7=2916}$$

Final Answer:

First term = 4

Common ratio = 3

Seventh term = 2916

The GP is:

4,12,36,108,324,972,2916,...

Q22. A Sierpiński triangle begins with one black equilateral triangle. At every stage, each remaining triangle is replaced with three smaller black triangles. Find the number of triangles and shaded areas at Stages 4 and 5, assuming the starting area is 1 square unit.

Solution:

Two patterns develop simultaneously:

• The number of black triangles increases.

• The total black area decreases.

Step 1: Find the number of black triangles.

Each stage triples the number.

Therefore:

$$N_n=3^n$$

For Stage 4:

$$N_4=3^4=\boxed{81}$$

For Stage 5:

$$N_5=3^5=\boxed{243}$$

Step 2: Find the shaded area.

At each stage, we remove the middle one-fourth of every remaining triangle.

Therefore, the remaining area is multiplied by:

$$\dfrac{3}{4}$$

The area at Stage n is:

$$A_n=\left(\dfrac34\right)^n$$

Step 3: Find the area at Stage 4.

$$A_4=\left(\dfrac34\right)^4$$

$$\boxed{A_4=\dfrac{81}{256}\text{ square units}}$$

Step 4: Find the area at Stage 5.

$$A_5=\left(\dfrac34\right)^5$$

$$\boxed{A_5=\dfrac{243}{1024}\text{ square units}}$$

Final Answer:

The number of triangles increases, while their combined shaded area decreases.

Q23. A square carpet follows a repeating pattern. At Stage 0, the complete square has area 1 square unit. At every stage, each remaining square is divided into nine equal smaller squares, and the central square is removed. Find the number of remaining squares and shaded areas at Stage 3.

Solution:

This pattern is called the Sierpiński square carpet.

Step 1: Find the number of remaining squares.

Every square produces eight smaller shaded squares.

The number of shaded squares follows:

1,8,64,512,...

The explicit formula is:

$$N_n=8^n$$

At Stage 3:

$$N_3=8^3$$

$$\boxed{N_3=512}$$

Step 2: Find the shaded area.

Each stage retains eight-ninths of the previous shaded area.

The common ratio is:

$$r=\dfrac89$$

The area at Stage n is:

$$A_n=\left(\dfrac89\right)^n$$

At Stage 3:

$$A_3=\left(\dfrac89\right)^3$$

$$\boxed{A_3=\dfrac{512}{729}\text{ square units}}$$

Step 3: Write the recursive rules.

For the number of squares:

$$N_0=1,\qquad N_n=8N_{n-1}$$

For the remaining area:

$$A_0=1,\qquad A_n=\dfrac89A_{n-1}$$

Final Answer:

Number of shaded squares at Stage 3 = 512

Total shaded area at Stage 3:

$$\boxed{\dfrac{512}{729}\text{ square units}}$$

Q24. A taxi charges a fixed booking fee of ₹150 and ₹35 per kilometre. Another situation involves a bacterial culture that initially contains 5 bacteria and doubles every hour. Compare the two situations using AP and GP, and calculate the taxi fare for 20 km and the bacterial count after 6 hours.

Solution:

The first situation involves repeated addition, while the second involves repeated multiplication.

Part A: Taxi Fare

Fixed fee:

₹150

Charge per kilometre:

₹35

Total fare after travelling n kilometres:

$$F_n=150+35n$$

For the first few whole-kilometre distances, the fares are:

185,220,255,290,...

These form an AP with:

a=185, d=35

For 20 km:

$$F_{20}=150+35(20)$$

$$F_{20}=150+700$$

The total fare is ₹850.

Part B: Bacterial Growth

Initial bacteria:

$$B_0=5$$

The population doubles every hour.

So:

$$B_n=5(2)^n$$

The counts at the end of successive hours are:

10,20,40,80,160,320,...

This is a GP with common ratio:

r=2

After 6 hours:

$$B_6=5(2)^6$$

$$B_6=5(64)$$

$$\boxed{B_6=320}$$

Part C: Compare the Two Patterns

Explanation: APs model repeated changes by a fixed amount, while GPs model repeated changes by a fixed multiplier.

Q25. A student notices two number patterns:

• Pattern A: 1,4,9,16,25,...

• Pattern B: 1,3,6,10,15,...

Explain how both patterns are formed. Derive their nth-term formulas, find their 15th terms, and calculate the difference between those terms.

Solution:

Step 1: Identify Pattern A.

Pattern A represents square numbers:

$$1^2,\,2^2,\,3^2,\,4^2,\,5^2,\,\ldots$$

The nth term is:

$$\boxed{A_n=n^2}$$

Each square number is also the sum of the first n odd numbers:

$$n^2=1+3+5+\cdots+(2n-1)$$

For example:

16=1+3+5+7

Step 2: Identify Pattern B.

Pattern B represents triangular numbers.

Each triangular number is the sum of the first n natural numbers:

$$B_n=1+2+3+\cdots+n$$

Therefore:

$$\boxed{B_n=\dfrac{n(n+1)}2}$$

Step 3: Find the 15th square number.

$$A_{15}=15^2$$

$$\boxed{A_{15}=225}$$

Step 4: Find the 15th triangular number.

$$B_{15}=\dfrac{15(16)}2$$

$$B_{15}=15(8)$$

$$\boxed{B_{15}=120}$$

Step 5: Calculate the difference.

$$A_{15}-B_{15}=225-120$$

105

Final Answer:

15th square number = 225

15th triangular number = 120

Difference = 105

The two patterns are not APs because their consecutive differences are not constant, but each follows an explicit formula that allows us to calculate any term directly.

Important Topics and Preparation Tips for Class 9 Maths Chapter 8

For students looking for Chapter 8 Class 9 Maths Important Questions, this revision guide follows Ganita Manjari Part I. It explains how number patterns work, how formulas predict future terms and how APs and GPs appear in diagrams and real-life situations.

How to Prepare Class 9 Maths Chapter 8 Important Questions

• Identify the pattern: Check whether terms increase by a fixed difference or multiplier.

• Learn the formulas: Practise explicit rules, recursive rules, APs, GPs and sums.

• Solve visual questions: Use dot patterns, coordinate graphs and fractals to understand sequences.

• Show calculation steps: Write the given values, formula, substitution and answer clearly.

• Practice applications: Revise growth, fares, savings and bouncing-ball problems.

Download Class 9 Maths Chapter 8 Important Questions PDF

The Class 9 Maths Chapter 8 Important Questions PDF, once available, will help students revise sequences and progressions through solved MCQs, short answers, long answers and diagram-based practice.

Students using Ganita Manjari Part I can practise these Class 9 Chapter 8 maths important questions to improve their understanding of mathematical patterns and prepare independently for examinations during the 2026–27 academic session.

Important Study Materials for Class 9 Maths Chapter 8 – Predicting What Comes Next: Exploring Sequences and Progressions

CBSE Class 9 Maths Chapter-wise Important Questions

CBSE Class 9 Maths Chapter-wise Important Questions and Answers cover key concepts from the Ganita Manjari Part I and Part II textbooks for the 2026–27 academic session. These chapter-wise resources help students strengthen their mathematical understanding, practise different question types and organise their revision effectively.

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