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NCERT Solutions for Class 9 Maths Chapter 8 Predicting What Comes next: Exploring Sequences and Progressions
Build a strong academic foundation with Chapter-wise NCERT Solutions for Class 9 Maths, created as per the latest NCERT syllabus and fully CBSE aligned. With expert-created, step-by-step answers, students can understand concepts clearly, solve textbook questions accurately, and prepare effectively for examinations. Download FREE PDF solutions, revise with important questions, and explore trusted learning resources for Maths, Physics, Chemistry, and Biology. Every resource is thoughtfully curated by ALLEN Experts to support consistent learning and academic success.
1.0Download NCERT Class 9 Maths Chapter 8 Sequences and Progressions
Class 9 Maths Chapter 8: Sequences and Progressions helps students recognise patterns, understand sequences, and explore arithmetic progressions.Download NCERT Solutions for Class 9 Maths Chapter 8 -Sequences and Progressions PDF , prepared according to the latest NCERT syllabus, with detailed step-by-step answers for every exercise.
2.0Learning Outcomes – NCERT Class 9 Maths Chapter 8 Sequences and Progressions
- Identify patterns in a sequence and predict its next few terms.
- Distinguish between explicit rules and recursive rules for a sequence.
- Test whether a given sequence is an Arithmetic Progression (AP) by checking for a constant common difference.
- Find the nth term and the sum of terms of an AP using the standard formulas.
- Test whether a given sequence is a Geometric Progression (GP) by checking for a constant common ratio.
- Find the nth term of a GP and apply it to solve real-life problems, such as compound growth or bouncing-ball scenarios.
- Interpret the graphical difference between an AP (linear) and a GP (exponential) when plotted.
- Apply sequences and progressions to real-world contexts such as salary increments, population growth, and patterns like the Fibonacci sequence and Tower of Hanoi.
3.0Detailed NCERT Class 9 Maths Chapter 8 Sequences and Progressions Solutions
Exercise : 8.1
1. Find the first five terms of the sequence in which the term is given by (i) (ii) (iii) for
Sol. (i) Since
(ii)
2.Determine whether 97 and 172 are terms of the sequence for .
Sol. If 97 and 172 are terms of the sequence then ' ' should be natural number or positive integer for these given values.
Now, 5n = 100 n = 20 Yes 97 is term of given sequence. again 5n n = 35 Yes 172 is also a term of given sequence.
3. Which term of the sequence for is 607 ?
Sol.
So, If 607 is term of given sequence thus, given term will occur at place.
4. A sequence is given by the recursive rule for . Find the first five terms of the sequence. Is 52 a term of this sequence? If so, which term is it? (iii) for
Sol. Since for if then If then if then if then So first five terms are This is an arithmetic sequence with first term common difference
Since Formula Now checking whether 52 is a term Let n = 20 It is a natural number. Therefore, 52 is the term.
5. Let and (for ). Find , and . Sol. for Now (Put in above sequence format) (use ) (use )
Exercise : 8.2
1.Find the and terms of the AP: 3, 8, 13, 18, ...... Sol. Given A.P. I term a Common difference
2.Which term of the AP : , is -81? Also, is 0 a term of this AP? Give reasons for your answer. Sol. Given A.P. Now so, 0 is term
3.Find the term of the AP: Write the recursive rule for this AP. Sol. Given A.P. Recursive Rule for
4.An AP consists of 50 terms in which the term is 12 and the last term is 106. Find the term. Sol. Let first term & common difference
AP has 50 terms and last term = 106 term
Equation (2) - (1) 47d = 94 from equation (1) now find term
5. How many 2-digit numbers are divisible by 3 ? What is the sum of all these 2-digit numbers?
Sol. AP will be 99 , last term = 99 Now ? So, So, such total numbers are 30 and their sum is 1665.
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?
Sol. Initial salary ₹ 5,00,000 Annual Increment A.P.
Using formula n - 1 = 10 n = 11 So, Harish’s income will be ₹ 7,00000 Per annum in year. So, It took 10 year or 10 yearly increments. 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?
Sol. Number of marbles total marbles used by child
Exercise : 8.3
1.Find the term of a GP with common ratio 2 , whose term is 192.
Sol. common ratio term So Now 2. Find the and terms of the GP: . Sol. G.P. Hence nth term 10 term
3. A sequence is given by the recursive rule for . Which term of the sequence is 730 ?
Sol. If Same as
4. Which term of the GP: is 4374? Write the explicit formula as well as the recursive formula for the term.
Sol. Given G.P. Use formula: Recursive formula : for
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 of the previous height. (i) What height does the ball reach after the bounce? (ii) What is the total vertical distance the ball has travelled by the time it hits the ground for the time?
Sol. (i) Initial height Bounce Ratio Height office bounce Height after bounce . Height after bounce Height after bounce Height after bounce (ii) The ball first balls 80 m to hit the ground for the time. Then it rises and falls after each bounce.
Height after bounces: bounce bounce bounce bounce bounce To hit the ground for the time, it travels the total distance 6.2208) = 301.3376 Therefore, the total vertical distance travelled in 301.3376 m .
6. Which term of the sequence is 128 ?
Sol. G.P. So, Use n = 13 So, 128 is term.
7. Figure 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. (i) How many red squares are there in Stages 0 to 3? (ii) Can you predict the number of red squares in Stages 4 and 5? (iii) Can you find a rule for the number of red squares at the 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 stage. What happens to this area as n , the number of stages, goes on increasing?
Sol. (i) Stage red square Stage red squares Stage red squares Stage red squares Therefore stages 0 to 3 have and 512 red squares respectively. (ii) The number of red squares is multiplied by 8 at each stage. Stage Stage Therefore Stage 4 has 4096 red squares. Stage 5 has 32768 red squares. (iii) At each stage, every red square is replaced by 8 smaller red squares. So the number of red squares forms the sequence. 1, 8, 64, 512...... Explicit the formula . Here stage 0 corresponds to Recursive formula , for (iv) At each stage the square is divided into 9 equal parts and the centre part is removed.
So, the remaining red area becomes of the previous area
Stage 0 area square unit Stage 1 area Stage 2 area Stage 3 area Stage 4 area Stage 5 area Explicit formula Here stage 0 corresponding to Recursive formula As ' ' increases, the area keeps decreasing and gets closer to 0 , but it never becomes exactly 0 .
End of Chapter Exercise
1.Find the term of an AP whose term is 38 and term is 73 .
Sol. First term , common difference
Subtracting Equation. (1) from (2)
Now
2.Determine the AP whose third term is 16 and whose term exceeds the term by 12 .
Sol. First term , common difference
According to question term term Now a So A.P. 4, 10, 16, 22, 28
3. How many three-digit numbers are divisible by 7 ?
Sol. Smallest three-digit number divisible by
Largest three-digit number divisible by
Required A.P. 105, 112, 119,........ 994 Let n - n = 128 Therefore 128, three-digit numbers are divisible by 7 .
4. How many multiples of 4 lie between 10 and 250 ?
Sol. The smallest and largest required multiples 12, 248 required A.P.
12, 16, 20, ...... 248 last term n = 60 Therefore 60 multiples will lie between 10 and 250.
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.
Sol. required G.P. Given sum of first two terms
Also term is 4 times the term. term ar term ar So Case 1: when From Eq (1) Hence G.P. . Case 2: when From Eq. (1) Hence G.P. Therefore, possible GP are and
6. Find all possible ways of expressing 100 as the sum of consecutive natural numbers.
Sol. We need consecutive natural numbers whose sum is 100 .
Possible ways 1 term: 100 So, 100 = 100 5 terms Let the numbers be Sum So, 8 terms Let the numbers be Sum Thus, Therefore, all possible ways 100 = 100 7. The number of bacteria in a certain culture doubles every hour. If there were 30 bacteria present in the culture originally, now many bacteria will be present at the end of the hour, hour and hour?
Sol. Initial numbers of bacteria The bacteria double every hour at the end of hour at the end of hour at the end of hour at the end of hour There At the end of hour bacteria At the end of hour bacteria At the end of hour bacteria 8. The sum of the and terms of an AP is 24 and the sum of the and terms is 44 . Find the first three terms of the AP.
Sol. smd_data$\PNCF\2026-27\Print Module\SET-1\NCERT\Mathematics\9th So equation (2) - (1) we get and First three terms are
9. Find the smallest value of n such that the sum of the first n natural numbers is greater than 1,000 .
Sol. Sum of first ' n ' natural numbers Now checking So, So, Since 1035 > 1000 Therefore, smallest value of n is 45 .
10. Which term of the GP: is 131072? Write the explicit formula as well as the recursive formula for the term.
Sol. G.P. Here ; since New Recursive formula for
11. The sum of the first three of a GP is and their product is -1 . Find the common ratio and the terms.
Sol. Let three terms are
So, terms Sum So or If Terms are If Terms are Therefore the common ratio is or and the terms are or 12. If the and terms of a GP are x , y and z respectively, prove that are in G.P.
Sol. Let first term of GP is a and common ratio be r Then term term term Now also Therefore So, in G.P. 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.
Sol. Let the three terms be a , ar , According to first condition also
Now we have
Using {Dividing by ( )} If Hence the three terms a, ar, ar are 18, 6, 2. If from eq. (1) we have So, three terms are . 14. Suppose and for , . Find the values of . Can you find a simpler recursive formula for ? Can you give an explicit formula?
Sol. Given For Now So are , 128
Simpler recursive formula for Explicit formula 15. Suppose and for , . Find the values of . Do you recognise this sequence?
Sol. Given For Therefore are , 8, 13, 21, 34 Yes This is the virahanka - Fibonacci sequence.
Important Key Concepts of NCERT Class 9 Maths Chapter 8 Sequences and Progressions
4.0Exercise-wise NCERT Solutions for Class 9 Maths Sequences and Progressions Chapter 8
Chapter 8 has three exercises and an End-of-Chapter set, moving from general number patterns to Arithmetic and Geometric Progressions.
5.0Quick Revision on Class 9 Maths Chapter 8 - Sequences and Progressions
6.0Related Study Materials Class 9 Maths
Explore ALLEN’s Class 9 Maths study material to strengthen the conceptual base of each chapter from the latest NCERT syllabus. Along with NCERT Solutions, you will get an updated syllabus, NCERT Textbooks, chapter-wise notes, revision notes, sample papers and previous years’ question papers to strengthen concepts, revise effectively and prepare confidently for school and CBSE examinations.
7.0Advantages of Chapter 8 Maths Class 9 NCERT Solutions
- Develops Data Collection Skills: Explains how to collect and organise data systematically.
- Strengthens Data Representation: Helps students present data using tables and appropriate graphs.
- Improves Data Interpretation: Teaches how to analyse graphs and tables to draw meaningful conclusions.
- Builds Statistical Understanding: Introduces basic statistical ideas through practical examples and data sets.
- Enhances Analytical Thinking: Develops the ability to compare, interpret, and evaluate information from data.
- Prepares for Advanced Statistics: Builds a strong foundation for higher-level data handling and statistical concepts.