MathwiseNCERT Maths Help Get the app

NCERT · Ganita Manjari Part I · Chapter 8

Predicting What Comes Next: Exploring Sequences and Progressions Class 9 Solutions

One textbook question at a time, with its complete diagram or table and a detailed sequence of working steps.

102 solved items18 textbook pagesReviewed for 2026-27
Questions from Class 9 Maths Chapter 8, Predicting What Comes Next: Exploring Sequences and Progressions
Question preview

Question-by-question working

Read the prompt, attempt it, then check each step

Every exercise subpart and table row is a separate item. Shared figures are repeated in the crop whenever they are needed to understand that question.

Detailed worked answers

Textbook page 174

Textbook page 174 · solved item 1

NCERT Class 9 Maths Chapter 8, solved question 1
Question from the current NCERT textbook

Think and Reflect: Describe the four familiar sequences and predict their next three terms.

Show detailed solution
  1. Step 1: Natural numbers increase by 1, so 1,2,3,4,5,6,\ldots continues with (7,8,9).
  2. Step 2: Odd numbers increase by 2, so the next terms after 11 are (13,15,17).
  3. Step 3: Triangular numbers add consecutive natural numbers. After 21, add (7,8,9) to get (28,36,45).
  4. Step 4: Square numbers are n^2. After 6^2=36 come 7^2=49, 8^2=64, and 9^2=81.
  5. Answer: The next triples are 7, 8, 9; 13, 15, 17; 28, 36, 45; and 49, 64, 81.

Textbook page 174 · solved item 2

NCERT Class 9 Maths Chapter 8, solved question 2
Question from the current NCERT textbook

Give examples of finite sequences found in daily life.

Show detailed solution
  1. Step 1: A finite sequence has a first term, a last term, and only a fixed number of terms.
  2. Step 2: The seven days of one week form an ordered finite sequence from Monday through Sunday.
  3. Step 3: The twelve months of one calendar year form another ordered finite sequence from January through December.
  4. Answer: Days in a week, months in a year, a school timetable, and a countdown are valid examples because each list ends.

Detailed worked answers

Textbook page 175

Textbook page 175 · solved item 3

NCERT Class 9 Maths Chapter 8, solved question 3
Question from the current NCERT textbook

Draw the dot patterns for the sixth and seventh triangular numbers.

Show detailed solution
  1. Step 1: The (n)th triangular pattern has rows containing 1,2,\ldots,n dots.
  2. Step 2: The sixth pattern adds a row of 6 dots to 15, giving T_6=15+6=21 dots.
  3. Step 3: The seventh adds a row of 7 dots, giving T_7=21+7=28 dots.
  4. Answer: Draw triangular arrays with 6 rows and 21 dots, then 7 rows and 28 dots.

Textbook page 175 · solved item 4

NCERT Class 9 Maths Chapter 8, solved question 4
Question from the current NCERT textbook

Explain the relationship between odd numbers and square numbers in Fig. 8.2.

Show detailed solution
  1. Step 1: Begin with 1=1^2. Adding the next odd number gives 1+3=4=2^2.
  2. Step 2: Adding 5 gives 9=3^2, and adding 7 gives 16=4^2. Each new odd number forms an L-shaped border around the previous square.
  3. Step 3: In general, 1+3+5+\cdots+(2n-1)=n^2.
  4. Answer: The sum of the first n odd numbers is the nth square number.

Detailed worked answers

Textbook page 176

Textbook page 176 · solved item 5

NCERT Class 9 Maths Chapter 8, solved question 5
Question from the current NCERT textbook

For the sequence 1, 4, 7, 10, 13, ... predict the next four terms.

Show detailed solution
  1. Step 1: Subtract consecutive terms: each difference is 3.
  2. Step 2: Continue adding 3 to 13: 16, 19, 22, 25.
  3. Step 3: The explicit check is t_n=1+3(n-1)=3n-2; t_6,t_7,t_8,t_9 give the same values.
  4. Answer: 16, 19, 22, 25.

Textbook page 176 · solved item 6

NCERT Class 9 Maths Chapter 8, solved question 6
Question from the current NCERT textbook

Find the first ten cumulative sums of 1, 4, 7, 10, 13, ... .

Show detailed solution
  1. Step 1: The original term is t_n=3n-2. Let S_k=t_1+t_2+\cdots+t_k.
  2. Step 2: Add each new term successively: 1, 1+4, 5+7, 12+10, and so on.
  3. Step 3: This gives (1,5,12,22,35,51,70,92,117,145). A check is S_k=k(3k-1)/2.
  4. Answer: 1, 5, 12, 22, 35, 51, 70, 92, 117, 145.

Textbook page 176 · solved item 7

NCERT Class 9 Maths Chapter 8, solved question 7
Question from the current NCERT textbook

Write t5, t6, t7 and t8 for the triangular-number sequence.

Show detailed solution
  1. Step 1: The (n)th triangular number is t_n=n(n+1)/2.
  2. Step 2: t_5=5\times6/2=15 and t_6=6\times7/2=21.
  3. Step 3: t_7=7\times8/2=28 and t_8=8\times9/2=36.
  4. Answer: t_5=15, t_6=21, t_7=28, t_8=36.

Textbook page 176 · solved item 8

NCERT Class 9 Maths Chapter 8, solved question 8
Question from the current NCERT textbook

Think and Reflect: List five different types of sequences and discuss their properties.

Show detailed solution
  1. Step 1: Even numbers 2,4,6,\ldots form an increasing AP with rule 2n.
  2. Step 2: Powers of 2, 1,2,4,8,\ldots, form a GP with rule 2^{n-1}; reciprocals 1,1/2,1/3,\ldots decrease toward 0.
  3. Step 3: The alternating sequence 1,-1,1,-1,\ldots has rule (-1)^{n-1}, while 5,5,5,\ldots is constant.
  4. Answer: APs, GPs, reciprocal, alternating, and constant sequences are five valid types; their rules describe different kinds of change.

Detailed worked answers

Textbook page 177

Textbook page 177 · solved item 9

NCERT Class 9 Maths Chapter 8, solved question 9
Question from the current NCERT textbook

Think and Reflect: Why is an explicit formula for the nth term useful?

Show detailed solution
  1. Step 1: An explicit formula connects a term directly to its position n.
  2. Step 2: It finds a distant term without calculating every earlier term.
  3. Step 3: Setting the formula equal to a given number can also test membership and find the number's position.
  4. Answer: It gives direct access to any term and helps decide whether a value belongs to the sequence.

Textbook page 177 · solved item 10

NCERT Class 9 Maths Chapter 8, solved question 10
Question from the current NCERT textbook

Using u_n=2n-1, find the 53rd, 108th and 1170th odd numbers.

Show detailed solution
  1. Step 1: u_{53}=2(53)-1=106-1=105.
  2. Step 2: u_{108}=2(108)-1=216-1=215.
  3. Step 3: u_{1170}=2(1170)-1=2340-1=2339.
  4. Answer: 105, 215 and 2339.

Textbook page 177 · solved item 11

NCERT Class 9 Maths Chapter 8, solved question 11
Question from the current NCERT textbook

Why must n be a natural number when it represents a term's position?

Show detailed solution
  1. Step 1: Positions in a sequence are counted as first, second, third, and so on.
  2. Step 2: A fractional or negative position, such as the 2.5th or -4th term, has no meaning in an ordinary sequence.
  3. Step 3: Therefore solving an equation for n proves membership only when n is a positive integer.
  4. Answer: n must be a positive whole-number index; otherwise the value is not a term at a valid position.

Textbook page 177 · solved item 12

NCERT Class 9 Maths Chapter 8, solved question 12
Question from the current NCERT textbook

Think and Reflect: Find the nth-term rule for the square-number sequence.

Show detailed solution
  1. Step 1: The terms are 1,4,9,16,\ldots.
  2. Step 2: They equal 1^2,2^2,3^2,4^2,\ldots.
  3. Step 3: The term at position n is therefore the square of n.
  4. Answer: t_n=n^2 for n\ge1.

Textbook page 177 · solved item 13

NCERT Class 9 Maths Chapter 8, solved question 13
Question from the current NCERT textbook

Do the prime numbers show a simple pattern, and what are the next few primes after 29?

Show detailed solution
  1. Step 1: Consecutive prime gaps are not constant: the early gaps are 1, 2, 2, 4, 2, 4, 2, 4, 6.
  2. Step 2: Test integers after 29 for divisibility by smaller primes. 30, 32, 33, 34, 35, 36, 38, 39 and 40 are composite.
  3. Step 3: The next numbers with no divisors except 1 and themselves are 31, 37 and 41.
  4. Answer: There is no simple constant-step rule like an AP; the next three primes are 31, 37 and 41.

Detailed worked answers

Textbook page 178

Textbook page 178 · solved item 14

NCERT Class 9 Maths Chapter 8, solved question 14
Question from the current NCERT textbook

For t_n=3n-7, find the first, second, third, 12th, 18th and 50th terms.

Show detailed solution
  1. Step 1: Substitute n=1,2,3: t_1=-4, t_2=-1, t_3=2.
  2. Step 2: t_{12}=36-7=29 and t_{18}=54-7=47.
  3. Step 3: t_{50}=150-7=143.
  4. Answer: -4, -1, 2, 29, 47 and 143.

Textbook page 178 · solved item 15

NCERT Class 9 Maths Chapter 8, solved question 15
Question from the current NCERT textbook

For t_n=3n-7, which term is 332?

Show detailed solution
  1. Step 1: Set 3n-7=332.
  2. Step 2: Add 7: 3n=339. Divide by 3: n=113.
  3. Step 3: Check: t_{113}=3(113)-7=332.
  4. Answer: 332 is the 113th term.

Textbook page 178 · solved item 16

NCERT Class 9 Maths Chapter 8, solved question 16
Question from the current NCERT textbook

For t_n=3n-7, is 557 a term?

Show detailed solution
  1. Step 1: Set 3n-7=557.
  2. Step 2: Then 3n=564, so n=188, a positive integer.
  3. Step 3: Check: 3(188)-7=564-7=557.
  4. Answer: Yes. It is the 188th term.

Detailed worked answers

Textbook page 179

Textbook page 179 · solved item 17

NCERT Class 9 Maths Chapter 8, solved question 17
Question from the current NCERT textbook

Write the next two terms of the Virahanka-Fibonacci sequence 1, 2, 3, 5, 8, 13, 21, 34, ... .

Show detailed solution
  1. Step 1: Each term is the sum of the preceding two terms.
  2. Step 2: The next term is 21+34=55.
  3. Step 3: The following term is 34+55=89.
  4. Answer: 55 and 89.

Textbook page 179 · solved item 18

NCERT Class 9 Maths Chapter 8, solved question 18
Question from the current NCERT textbook

Exercise 8.1, 1(i): Find the first five terms when t_n=3n-4.

Show detailed solution
  1. Step 1: Substitute n=1,2,3,4,5.
  2. Step 2: The values are 3-4, 6-4, 9-4, 12-4, 15-4.
  3. Step 3: Simplify each value.
  4. Answer: -1, 2, 5, 8, 11.

Textbook page 179 · solved item 19

NCERT Class 9 Maths Chapter 8, solved question 19
Question from the current NCERT textbook

Exercise 8.1, 1(ii): Find the first five terms when t_n=2-5n.

Show detailed solution
  1. Step 1: Substitute n=1,2,3,4,5.
  2. Step 2: Compute 2-5, 2-10, 2-15, 2-20, 2-25.
  3. Step 3: The terms decrease by 5 each time.
  4. Answer: -3, -8, -13, -18, -23.

Textbook page 179 · solved item 20

NCERT Class 9 Maths Chapter 8, solved question 20
Question from the current NCERT textbook

Exercise 8.1, 1(iii): Find the first five terms when t_n=n^2-2n+3.

Show detailed solution
  1. Step 1: t_1=1-2+3=2 and t_2=4-4+3=3.
  2. Step 2: t_3=9-6+3=6 and t_4=16-8+3=11.
  3. Step 3: t_5=25-10+3=18.
  4. Answer: 2, 3, 6, 11, 18.

Textbook page 179 · solved item 21

NCERT Class 9 Maths Chapter 8, solved question 21
Question from the current NCERT textbook

Exercise 8.1, 2: Find the 10th and 15th terms of t_n=5n-3.

Show detailed solution
  1. Step 1: t_{10}=5(10)-3=47.
  2. Step 2: t_{15}=5(15)-3=72.
  3. Step 3: Both follow directly from the explicit rule.
  4. Answer: t_{10}=47 and t_{15}=72.

Textbook page 179 · solved item 22

NCERT Class 9 Maths Chapter 8, solved question 22
Question from the current NCERT textbook

Exercise 8.1, 3: Determine whether 97 and 172 are terms of t_n=5n-3.

Show detailed solution
  1. Step 1: For 97, solve 5n-3=97, giving 5n=100 and n=20.
  2. Step 2: For 172, solve 5n-3=172, giving 5n=175 and n=35.
  3. Step 3: Both positions are positive integers, so both values belong to the sequence.
  4. Answer: Yes. 97 is the 20th term and 172 is the 35th term.

Textbook page 179 · solved item 23

NCERT Class 9 Maths Chapter 8, solved question 23
Question from the current NCERT textbook

Exercise 8.1, 4: Which term of t_n=5n-3 is 607?

Show detailed solution
  1. Step 1: Set 5n-3=607.
  2. Step 2: Then 5n=610, so n=122.
  3. Step 3: Check: 5(122)-3=610-3=607.
  4. Answer: 607 is the 122nd term.

Textbook page 179 · solved item 24

NCERT Class 9 Maths Chapter 8, solved question 24
Question from the current NCERT textbook

Exercise 8.1, 5(a): Find the first five terms when t_1=-5 and t_{n+1}=t_n+3.

Show detailed solution
  1. Step 1: Start with -5 and add 3 to get -2.
  2. Step 2: Continue adding 3: 1, then 4, then 7.
  3. Step 3: Equivalently, t_n=-5+3(n-1)=3n-8.
  4. Answer: -5, -2, 1, 4, 7.

Textbook page 179 · solved item 25

NCERT Class 9 Maths Chapter 8, solved question 25
Question from the current NCERT textbook

Exercise 8.1, 5(b): Is 52 a term of that recursive sequence, and which term is it?

Show detailed solution
  1. Step 1: Use the explicit form t_n=3n-8.
  2. Step 2: Solve 3n-8=52, so 3n=60 and n=20.
  3. Step 3: Check: t_{20}=60-8=52.
  4. Answer: Yes. 52 is the 20th term.

Detailed worked answers

Textbook page 180

Textbook page 180 · solved item 26

NCERT Class 9 Maths Chapter 8, solved question 26
Question from the current NCERT textbook

Exercise 8.1, 6: Find T4 through T8 when each term is the sum of the previous three.

Show detailed solution
  1. Step 1: T_4=T_3+T_2+T_1=4+2+1=7.
  2. Step 2: T_5=7+4+2=13 and T_6=13+7+4=24.
  3. Step 3: T_7=24+13+7=44 and T_8=44+24+13=81.
  4. Answer: T_4=7, T_5=13, T_6=24, T_7=44, T_8=81.

Textbook page 180 · solved item 27

NCERT Class 9 Maths Chapter 8, solved question 27
Question from the current NCERT textbook

Fig. 8.3: Predict the number of squares in Stages 5 and 6.

Show detailed solution
  1. Step 1: The counts (1,5,9,13) increase by 4.
  2. Step 2: Stage 5 has 13+4=17 squares.
  3. Step 3: Stage 6 has 17+4=21 squares.
  4. Answer: 17 and 21 squares.

Textbook page 180 · solved item 28

NCERT Class 9 Maths Chapter 8, solved question 28
Question from the current NCERT textbook

Fig. 8.3: Predict the number of squares in Stages 10, 11 and 12.

Show detailed solution
  1. Step 1: The nth-stage rule is t_n=1+4(n-1)=4n-3.
  2. Step 2: t_{10}=40-3=37 and t_{11}=44-3=41.
  3. Step 3: t_{12}=48-3=45.
  4. Answer: 37, 41 and 45 squares.

Textbook page 180 · solved item 29

NCERT Class 9 Maths Chapter 8, solved question 29
Question from the current NCERT textbook

Fig. 8.3: Predict the number of squares in Stage 20.

Show detailed solution
  1. Step 1: Use t_n=4n-3.
  2. Step 2: Substitute n=20: t_{20}=4(20)-3.
  3. Step 3: 80-3=77.
  4. Answer: Stage 20 has 77 squares.

Textbook page 180 · solved item 30

NCERT Class 9 Maths Chapter 8, solved question 30
Question from the current NCERT textbook

Fig. 8.3: Find the number of squares at any stage n.

Show detailed solution
  1. Step 1: Stage 1 has one square.
  2. Step 2: Each move to the next stage adds 4, so Stage n has n-1 additions.
  3. Step 3: t_n=1+4(n-1)=4n-3.
  4. Answer: Stage n has 4n-3 squares.

Detailed worked answers

Textbook page 181

Textbook page 181 · solved item 31

NCERT Class 9 Maths Chapter 8, solved question 31
Question from the current NCERT textbook

Think and Reflect: Which sequences discussed so far are arithmetic progressions?

Show detailed solution
  1. Step 1: An AP must have the same difference between every pair of consecutive terms.
  2. Step 2: Natural numbers, odd numbers, 1, 4, 7, 10, ..., -7, -3, 1, 5, ..., and the Fig. 8.3 counts are APs, with differences 1, 2, 3, 4, and 4 respectively.
  3. Step 3: Triangular numbers, square numbers, unit fractions, primes, cumulative sums, powers, and the Virahanka-Fibonacci sequence do not have a constant consecutive difference.
  4. Answer: Exactly the constant-difference sequences are APs; checking differences justifies each classification.

Detailed worked answers

Textbook page 182

Textbook page 182 · solved item 32

NCERT Class 9 Maths Chapter 8, solved question 32
Question from the current NCERT textbook

Verify that 2, 5, 8, 11, ... is an AP, find its nth term, and describe its graph.

Show detailed solution
  1. Step 1: Consecutive differences are 5-2=8-5=11-8=3, so it is an AP.
  2. Step 2: With a=2,d=3, t_n=2+3(n-1)=3n-1.
  3. Step 3: The ordered pairs (n,t_n) satisfy y=3x-1, a linear equation.
  4. Answer: It is an AP with nth term 3n - 1, and its plotted points lie on a straight line.

Textbook page 182 · solved item 33

NCERT Class 9 Maths Chapter 8, solved question 33
Question from the current NCERT textbook

Verify that -5, -1, 3, 7, ... is an AP, find its nth term, and describe its graph.

Show detailed solution
  1. Step 1: Each consecutive difference is 4, so it is an AP.
  2. Step 2: With a=-5,d=4, t_n=-5+4(n-1)=4n-9.
  3. Step 3: Its ordered pairs satisfy y=4x-9, so they are collinear.
  4. Answer: It is an AP with nth term 4n - 9; its graph consists of points on a straight line.

Textbook page 182 · solved item 34

NCERT Class 9 Maths Chapter 8, solved question 34
Question from the current NCERT textbook

Find the nth term of the AP 1/2, 5/2, 9/2, 13/2, ... .

Show detailed solution
  1. Step 1: The first term is a=1/2.
  2. Step 2: The common difference is 5/2-1/2=2.
  3. Step 3: t_n=1/2+2(n-1)=2n-3/2=(4n-3)/2.
  4. Answer: t_n=(4n-3)/2.

Textbook page 182 · solved item 35

NCERT Class 9 Maths Chapter 8, solved question 35
Question from the current NCERT textbook

Find the nth term of the AP 1.5, 3.5, 5.5, 7.5, ... .

Show detailed solution
  1. Step 1: Here a=1.5 and d=2.
  2. Step 2: t_n=1.5+2(n-1).
  3. Step 3: Simplifying gives t_n=2n-0.5=(4n-1)/2.
  4. Answer: t_n=2n-0.5.

Detailed worked answers

Textbook page 183

Textbook page 183 · solved item 36

NCERT Class 9 Maths Chapter 8, solved question 36
Question from the current NCERT textbook

Find a recursive rule for the AP 2, 5, 8, 11, ... .

Show detailed solution
  1. Step 1: The first term is 2.
  2. Step 2: Every later term is 3 more than the preceding term.
  3. Step 3: State both the starting value and recurrence: t_1=2, t_n=t_{n-1}+3 for n\ge2.
  4. Answer: t_1=2,\ t_n=t_{n-1}+3.

Textbook page 183 · solved item 37

NCERT Class 9 Maths Chapter 8, solved question 37
Question from the current NCERT textbook

Find a recursive rule for the AP -5, -1, 3, 7, ... .

Show detailed solution
  1. Step 1: The sequence starts at -5.
  2. Step 2: Its common difference is 4.
  3. Step 3: Add 4 to the previous term: t_1=-5, t_n=t_{n-1}+4 for n\ge2.
  4. Answer: t_1=-5,\ t_n=t_{n-1}+4.

Textbook page 183 · solved item 38

NCERT Class 9 Maths Chapter 8, solved question 38
Question from the current NCERT textbook

Find a recursive rule for the AP 1/2, 5/2, 9/2, 13/2, ... .

Show detailed solution
  1. Step 1: The first term is 1/2.
  2. Step 2: Each term is 2 more than the previous term.
  3. Step 3: Therefore t_1=1/2, t_n=t_{n-1}+2 for n\ge2.
  4. Answer: t_1=1/2,\ t_n=t_{n-1}+2.

Textbook page 183 · solved item 39

NCERT Class 9 Maths Chapter 8, solved question 39
Question from the current NCERT textbook

Find a recursive rule for the AP 1.5, 3.5, 5.5, 7.5, ... .

Show detailed solution
  1. Step 1: The first term is 1.5.
  2. Step 2: The common difference is 2.
  3. Step 3: Thus t_1=1.5, t_n=t_{n-1}+2 for n\ge2.
  4. Answer: t_1=1.5,\ t_n=t_{n-1}+2.

Textbook page 183 · solved item 40

NCERT Class 9 Maths Chapter 8, solved question 40
Question from the current NCERT textbook

Find the sum of the first ten natural numbers without adding them one by one.

Show detailed solution
  1. Step 1: Write S=1+2+\cdots+10 and reverse it as S=10+9+\cdots+1.
  2. Step 2: Adding corresponding terms produces ten pairs, each equal to 11.
  3. Step 3: Hence 2S=10\times11=110, so S=55.
  4. Answer: The sum is 55.

Detailed worked answers

Textbook page 184

Textbook page 184 · solved item 41

NCERT Class 9 Maths Chapter 8, solved question 41
Question from the current NCERT textbook

Think and Reflect: Use the pairing method to find 1 + 2 + 3 + ... + 100.

Show detailed solution
  1. Step 1: Pair the first and last terms: 1 + 100 = 101, 2 + 99 = 101, and so on.
  2. Step 2: There are 50 such pairs.
  3. Step 3: The sum is 50\times101=5050, equivalently 100(101)/2.
  4. Answer: 5050.

Detailed worked answers

Textbook page 185

Textbook page 185 · solved item 42

NCERT Class 9 Maths Chapter 8, solved question 42
Question from the current NCERT textbook

Think and Reflect: Use S_n=n(n+1)/2 to find S20, S50 and S1000.

Show detailed solution
  1. Step 1: S_{20}=20(21)/2=210.
  2. Step 2: S_{50}=50(51)/2=25\times51=1275.
  3. Step 3: S_{1000}=1000(1001)/2=500\times1001=500500.
  4. Answer: S_{20}=210,\ S_{50}=1275,\ S_{1000}=500500.

Textbook page 185 · solved item 43

NCERT Class 9 Maths Chapter 8, solved question 43
Question from the current NCERT textbook

Think and Reflect: Find the 10th, 17th and 80th triangular numbers.

Show detailed solution
  1. Step 1: Use T_n=n(n+1)/2.
  2. Step 2: T_{10}=10(11)/2=55 and T_{17}=17(18)/2=153.
  3. Step 3: T_{80}=80(81)/2=40\times81=3240.
  4. Answer: T_{10}=55,\ T_{17}=153,\ T_{80}=3240.

Textbook page 185 · solved item 44

NCERT Class 9 Maths Chapter 8, solved question 44
Question from the current NCERT textbook

Exercise 8.2, 1: Find the 10th and 26th terms of 3, 8, 13, 18, ... .

Show detailed solution
  1. Step 1: Here a=3 and d=5.
  2. Step 2: t_{10}=3+9(5)=48.
  3. Step 3: t_{26}=3+25(5)=128.
  4. Answer: t_{10}=48 and t_{26}=128.

Textbook page 185 · solved item 45

NCERT Class 9 Maths Chapter 8, solved question 45
Question from the current NCERT textbook

Exercise 8.2, 2(a): Which term of 21, 18, 15, ... is -81?

Show detailed solution
  1. Step 1: Here a=21,d=-3, so t_n=21-3(n-1).
  2. Step 2: Solve 21-3(n-1)=-81. Then -3(n-1)=-102, so n-1=34.
  3. Step 3: Therefore n = 35; checking gives 21 - 102 = -81.
  4. Answer: -81 is the 35th term.

Textbook page 185 · solved item 46

NCERT Class 9 Maths Chapter 8, solved question 46
Question from the current NCERT textbook

Exercise 8.2, 2(b): Is 0 a term of 21, 18, 15, ...?

Show detailed solution
  1. Step 1: Set 21-3(n-1)=0.
  2. Step 2: Then 3(n-1)=21, so n-1=7 and n=8.
  3. Step 3: Since 8 is a positive integer, this is a valid term position.
  4. Answer: Yes. 0 is the 8th term.

Textbook page 185 · solved item 47

NCERT Class 9 Maths Chapter 8, solved question 47
Question from the current NCERT textbook

Exercise 8.2, 3: Find the nth term and recursive rule for 11, 8, 5, 2, ... .

Show detailed solution
  1. Step 1: The first term is a=11 and common difference is d=-3.
  2. Step 2: t_n=11-3(n-1)=14-3n.
  3. Step 3: Recursively, begin with 11 and subtract 3: t_1=11,\ t_n=t_{n-1}-3 for n\ge2.
  4. Answer: t_n=14-3n; recursively t_1=11,\ t_n=t_{n-1}-3.

Textbook page 185 · solved item 48

NCERT Class 9 Maths Chapter 8, solved question 48
Question from the current NCERT textbook

Exercise 8.2, 4: An AP has 50 terms, t3 = 12 and t50 = 106. Find t29.

Show detailed solution
  1. Step 1: a+2d=12 and a+49d=106.
  2. Step 2: Subtract the equations: 47d=94, so d=2. Then a=12-4=8.
  3. Step 3: t_{29}=a+28d=8+56=64.
  4. Answer: The 29th term is 64.

Detailed worked answers

Textbook page 186

Textbook page 186 · solved item 49

NCERT Class 9 Maths Chapter 8, solved question 49
Question from the current NCERT textbook

Exercise 8.2, 5(a): How many two-digit numbers are divisible by 3?

Show detailed solution
  1. Step 1: The first two-digit multiple of 3 is 12 and the last is 99.
  2. Step 2: These form an AP with a=12,d=3,l=99.
  3. Step 3: n=(99-12)/3+1=29+1=30.
  4. Answer: 30 two-digit numbers are divisible by 3.

Textbook page 186 · solved item 50

NCERT Class 9 Maths Chapter 8, solved question 50
Question from the current NCERT textbook

Exercise 8.2, 5(b): Find the sum of all two-digit numbers divisible by 3.

Show detailed solution
  1. Step 1: From the first part, there are 30 terms from 12 to 99.
  2. Step 2: Use S_n=n(a+l)/2.
  3. Step 3: S=30(12+99)/2=15(111)=1665.
  4. Answer: The sum is 1665.

Textbook page 186 · solved item 51

NCERT Class 9 Maths Chapter 8, solved question 51
Question from the current NCERT textbook

Exercise 8.2, 6: When does a salary starting at Rs 5,00,000 with yearly increments of Rs 20,000 reach Rs 7,00,000?

Show detailed solution
  1. Step 1: In year n, the salary is 500000+(n-1)20000.
  2. Step 2: Set this equal to 700000: (n-1)20000=200000.
  3. Step 3: Thus n-1=10, so n=11.
  4. Answer: It reaches Rs 7,00,000 in the 11th year, after 10 annual increments.

Textbook page 186 · solved item 52

NCERT Class 9 Maths Chapter 8, solved question 52
Question from the current NCERT textbook

Exercise 8.2, 7: How many marbles are used in rows of 1, 2, 3, ... up to 25?

Show detailed solution
  1. Step 1: The total is 1+2+\cdots+25.
  2. Step 2: Use S_n=n(n+1)/2.
  3. Step 3: S_{25}=25(26)/2=25(13)=325.
  4. Answer: 325 marbles.

Textbook page 186 · solved item 53

NCERT Class 9 Maths Chapter 8, solved question 53
Question from the current NCERT textbook

Fig. 8.6: Predict the number of green squares in Stages 5 and 6.

Show detailed solution
  1. Step 1: The counts 3, 6, 12, 24 double at each stage.
  2. Step 2: Stage 5 has 24\times2=48 squares.
  3. Step 3: Stage 6 has 48\times2=96 squares.
  4. Answer: 48 and 96 squares.

Textbook page 186 · solved item 54

NCERT Class 9 Maths Chapter 8, solved question 54
Question from the current NCERT textbook

Fig. 8.6: Predict the number of green squares in Stages 10, 11 and 12.

Show detailed solution
  1. Step 1: The rule is t_n=3\times2^{n-1}.
  2. Step 2: t_{10}=3\times2^9=1536 and t_{11}=3\times2^{10}=3072.
  3. Step 3: t_{12}=3\times2^{11}=6144.
  4. Answer: 1536, 3072 and 6144 squares.

Textbook page 186 · solved item 55

NCERT Class 9 Maths Chapter 8, solved question 55
Question from the current NCERT textbook

Fig. 8.6: Predict the number of green squares in Stage 20.

Show detailed solution
  1. Step 1: Use t_n=3\times2^{n-1}.
  2. Step 2: t_{20}=3\times2^{19}.
  3. Step 3: Since 2^{19}=524288, the product is 1572864.
  4. Answer: Stage 20 has 1,572,864 green squares.

Textbook page 186 · solved item 56

NCERT Class 9 Maths Chapter 8, solved question 56
Question from the current NCERT textbook

Fig. 8.6: Find the number of green squares at any stage n.

Show detailed solution
  1. Step 1: Stage 1 has 3 squares.
  2. Step 2: Each stage doubles the previous count, so the common ratio is 2.
  3. Step 3: A GP with first term 3 and ratio 2 has t_n=3\times2^{n-1}.
  4. Answer: Stage n has 3\times2^{n-1} green squares.

Textbook page 186 · solved item 57

NCERT Class 9 Maths Chapter 8, solved question 57
Question from the current NCERT textbook

How is the growing pattern in Fig. 8.6 different from Fig. 8.3?

Show detailed solution
  1. Step 1: Fig. 8.3 adds 4 squares at every stage, producing an AP and linear growth.
  2. Step 2: Fig. 8.6 multiplies the number of squares by 2, producing a GP and exponential growth.
  3. Step 3: Their rules are 4n-3 and 3\times2^{n-1}, respectively.
  4. Answer: Fig. 8.3 grows by a constant difference; Fig. 8.6 grows by a constant ratio.

Detailed worked answers

Textbook page 188

Textbook page 188 · solved item 58

NCERT Class 9 Maths Chapter 8, solved question 58
Question from the current NCERT textbook

Check whether 2, 10, 50, 250, ... is a GP and find its nth term.

Show detailed solution
  1. Step 1: Consecutive ratios are 10/2=50/10=250/50=5.
  2. Step 2: The constant ratio proves it is a GP with a = 2 and r = 5.
  3. Step 3: Use t_n=ar^{n-1}.
  4. Answer: It is a GP and t_n=2\times5^{n-1}.

Textbook page 188 · solved item 59

NCERT Class 9 Maths Chapter 8, solved question 59
Question from the current NCERT textbook

Check whether 4, 8/3, 16/9, 32/27, ... is a GP and find its nth term.

Show detailed solution
  1. Step 1: (8/3)/4=2/3, and the later consecutive ratios are also 2/3.
  2. Step 2: Thus a=4 and r=2/3.
  3. Step 3: Substitute in t_n=ar^{n-1}.
  4. Answer: It is a GP with t_n=4(2/3)^{n-1}.

Textbook page 188 · solved item 60

NCERT Class 9 Maths Chapter 8, solved question 60
Question from the current NCERT textbook

Check whether 3, -3/2, 3/4, -3/8, ... is a GP and find its nth term.

Show detailed solution
  1. Step 1: Each term divided by the preceding term equals -1/2.
  2. Step 2: Therefore a=3 and r=-1/2; the negative ratio makes signs alternate.
  3. Step 3: Use the standard GP formula t_n=ar^{n-1}.
  4. Answer: It is a GP with t_n=3(-1/2)^{n-1}.

Textbook page 188 · solved item 61

NCERT Class 9 Maths Chapter 8, solved question 61
Question from the current NCERT textbook

Find a recursive rule for t_n=3\times10^{n-1}, which generates 3, 30, 300, 3000, ... .

Show detailed solution
  1. Step 1: The starting term is 3.
  2. Step 2: Every term is 10 times the preceding term.
  3. Step 3: Record both parts: t_1=3, t_n=10t_{n-1} for n\ge2.
  4. Answer: t_1=3,\ t_n=10t_{n-1}.

Detailed worked answers

Textbook page 189

Textbook page 189 · solved item 62

NCERT Class 9 Maths Chapter 8, solved question 62
Question from the current NCERT textbook

Sierpinski triangle (a): How many black triangles are in Stages 0 to 3?

Show detailed solution
  1. Step 1: Stage 0 begins with one black triangle.
  2. Step 2: Each black triangle is replaced by 3 smaller black triangles at the next stage.
  3. Step 3: The counts are 1,1\times3,1\times3^2,1\times3^3.
  4. Answer: 1, 3, 9 and 27 black triangles.

Textbook page 189 · solved item 63

NCERT Class 9 Maths Chapter 8, solved question 63
Question from the current NCERT textbook

Sierpinski triangle (b): Predict the number of black triangles in Stages 4 and 5.

Show detailed solution
  1. Step 1: Continue the GP 1,3,9,27,\ldots with common ratio 3.
  2. Step 2: Stage 4 has 27\times3=81.
  3. Step 3: Stage 5 has 81\times3=243.
  4. Answer: 81 and 243 black triangles.

Textbook page 189 · solved item 64

NCERT Class 9 Maths Chapter 8, solved question 64
Question from the current NCERT textbook

Sierpinski triangle (c): Find a rule for the number of black triangles at Stage n.

Show detailed solution
  1. Step 1: Stage 0 has one triangle, which is 3^0.
  2. Step 2: Moving forward one stage multiplies the count by 3.
  3. Step 3: Therefore B_n=3^n; recursively, B_0=1 and B_n=3B_{n-1}.
  4. Answer: B_n=3^n for Stage n.

Textbook page 189 · solved item 65

NCERT Class 9 Maths Chapter 8, solved question 65
Question from the current NCERT textbook

Sierpinski triangle, part (d), question 1: If Stage 0 has area 1, find the black area in Stages 1, 2 and 3.

Show detailed solution
  1. Step 1: At each stage, three of four equal subtriangles remain, so the area is multiplied by 3/4.
  2. Step 2: Stage 1 area is 3/4, and Stage 2 area is (3/4)^2=9/16.
  3. Step 3: Stage 3 area is (3/4)^3=27/64.
  4. Answer: 3/4, 9/16, 27/64 square units.

Textbook page 189 · solved item 66

NCERT Class 9 Maths Chapter 8, solved question 66
Question from the current NCERT textbook

Sierpinski triangle, part (d), question 2: Find the black area in Stages 4 and 5.

Show detailed solution
  1. Step 1: The area at Stage n is (3/4)^n.
  2. Step 2: Stage 4 area is (3/4)^4=81/256.
  3. Step 3: Stage 5 area is (3/4)^5=243/1024.
  4. Answer: 81/256 and 243/1024 square units.

Textbook page 189 · solved item 67

NCERT Class 9 Maths Chapter 8, solved question 67
Question from the current NCERT textbook

Sierpinski triangle, part (d), question 3: Find a rule for the black area at Stage n.

Show detailed solution
  1. Step 1: The initial area is A_0=1.
  2. Step 2: Each construction retains 3/4 of the preceding black area.
  3. Step 3: Thus A_n=(3/4)^n; recursively, A_0=1, A_n=(3/4)A_{n-1}.
  4. Answer: A_n=(3/4)^n square units.

Textbook page 189 · solved item 68

NCERT Class 9 Maths Chapter 8, solved question 68
Question from the current NCERT textbook

Sierpinski triangle, part (d), question 4: What happens to the black area as the number of stages increases?

Show detailed solution
  1. Step 1: The area is A_n=(3/4)^n.
  2. Step 2: Because 3/4 lies between 0 and 1, repeated multiplication makes the value smaller.
  3. Step 3: Mathematically, \lim_{n\to\infty}(3/4)^n=0.
  4. Answer: The black area approaches 0, although it remains positive at every finite stage.

Textbook page 189 · solved item 69

NCERT Class 9 Maths Chapter 8, solved question 69
Question from the current NCERT textbook

Why does the number of black triangles increase so quickly with the stage number?

Show detailed solution
  1. Step 1: Every black triangle produces three black triangles in the next stage.
  2. Step 2: Therefore the count is multiplied, not merely increased by a fixed amount: B_n=3^n.
  3. Step 3: Exponential growth such as 3^n eventually outpaces every linear rule an+b.
  4. Answer: The count triples at every step, so it grows exponentially.

Textbook page 189 · solved item 70

NCERT Class 9 Maths Chapter 8, solved question 70
Question from the current NCERT textbook

Explain why the black area at Stage n is (3/4)^n.

Show detailed solution
  1. Step 1: Stage 0 has area 1.
  2. Step 2: At each construction, the central one of four equal parts is removed, leaving 3/4 of the earlier black area.
  3. Step 3: After n stages this multiplier has been applied n times, giving 1\times(3/4)^n.
  4. Answer: The black area is (3/4)^n square units.

Detailed worked answers

Textbook page 193

Textbook page 193 · solved item 71

NCERT Class 9 Maths Chapter 8, solved question 71
Question from the current NCERT textbook

Exercise 8.3, 1: Find the 12th term of a GP with ratio 2 whose 8th term is 192.

Show detailed solution
  1. Step 1: In four steps from the 8th to the 12th term, multiply by 2^4.
  2. Step 2: t_{12}=t_8\times2^4=192\times16.
  3. Step 3: 192\times16=3072.
  4. Answer: The 12th term is 3072.

Textbook page 193 · solved item 72

NCERT Class 9 Maths Chapter 8, solved question 72
Question from the current NCERT textbook

Exercise 8.3, 2: Find the 10th and nth terms of 5, 25, 125, ... .

Show detailed solution
  1. Step 1: The GP has a=5 and r=5.
  2. Step 2: t_n=5\times5^{n-1}=5^n.
  3. Step 3: t_{10}=5^{10}=9765625.
  4. Answer: t_{10}=9,765,625 and t_n=5^n.

Textbook page 193 · solved item 73

NCERT Class 9 Maths Chapter 8, solved question 73
Question from the current NCERT textbook

Exercise 8.3, 3: For t_1=2 and t_{n+1}=3t_n-2, which term is 730?

Show detailed solution
  1. Step 1: Subtract the fixed value 1. If u_n=t_n-1, then u_{n+1}=3u_n and u_1=1.
  2. Step 2: Hence u_n=3^{n-1} and t_n=1+3^{n-1}.
  3. Step 3: Set 1+3^{n-1}=730. Then 3^{n-1}=729=3^6, so n=7.
  4. Answer: 730 is the 7th term.

Textbook page 193 · solved item 74

NCERT Class 9 Maths Chapter 8, solved question 74
Question from the current NCERT textbook

Exercise 8.3, 4: Which term of 2, 6, 18, ... is 4374? Give explicit and recursive rules.

Show detailed solution
  1. Step 1: Here a=2,r=3, so the explicit rule is t_n=2\times3^{n-1}.
  2. Step 2: Solve 2\times3^{n-1}=4374. Dividing by 2 gives 3^{n-1}=2187=3^7, so n=8.
  3. Step 3: The recursive rule is t_1=2,\ t_n=3t_{n-1} for n\ge2.
  4. Answer: 4374 is the 8th term; the rules are as stated above.

Textbook page 193 · solved item 75

NCERT Class 9 Maths Chapter 8, solved question 75
Question from the current NCERT textbook

Exercise 8.3, 5(i): An 80 m ball rebounds to 60% each time. What height does it reach after the 5th bounce?

Show detailed solution
  1. Step 1: After bounce n, the height is h_n=80(0.6)^n.
  2. Step 2: h_5=80(0.6)^5, and (0.6)^5=0.07776.
  3. Step 3: 80\times0.07776=6.2208.
  4. Answer: The ball reaches 6.2208 m after the 5th bounce.

Textbook page 193 · solved item 76

NCERT Class 9 Maths Chapter 8, solved question 76
Question from the current NCERT textbook

Exercise 8.3, 5(ii): Find the total vertical distance by the time the ball hits the ground for the 6th time.

Show detailed solution
  1. Step 1: The first ground hit contributes the initial 80 m fall.
  2. Step 2: Before hits 2 through 6, the ball rises and falls through heights (48,28.8,17.28,10.368,6.2208) m.
  3. Step 3: Total distance is 80+2(48+28.8+17.28+10.368+6.2208).
  4. Step 4: The bracketed sum is 110.6688, so the distance is 80+221.3376=301.3376 m.
  5. Answer: 301.3376 m.

Detailed worked answers

Textbook page 194

Textbook page 194 · solved item 77

NCERT Class 9 Maths Chapter 8, solved question 77
Question from the current NCERT textbook

Exercise 8.3, 6: Which term of 2,2\sqrt2,4,\ldots is 128?

Show detailed solution
  1. Step 1: The common ratio is \sqrt2, so t_n=2(\sqrt2)^{n-1}.
  2. Step 2: Write powers of 2: t_n=2^{1+(n-1)/2} and 128=2^7.
  3. Step 3: Set 1+(n-1)/2=7. Then n-1=12, so n=13.
  4. Answer: 128 is the 13th term.

Textbook page 194 · solved item 78

NCERT Class 9 Maths Chapter 8, solved question 78
Question from the current NCERT textbook

Exercise 8.3, 7(i): How many red squares are in Stages 0 to 3 of the Sierpinski square carpet?

Show detailed solution
  1. Step 1: Stage 0 has one red square.
  2. Step 2: Each red square produces 8 retained red squares at the next stage.
  3. Step 3: The counts are 8^0,8^1,8^2,8^3.
  4. Answer: 1, 8, 64 and 512 red squares.

Textbook page 194 · solved item 79

NCERT Class 9 Maths Chapter 8, solved question 79
Question from the current NCERT textbook

Exercise 8.3, 7(ii): Predict the number of red squares in Stages 4 and 5.

Show detailed solution
  1. Step 1: Continue multiplying the Stage 3 count, 512, by 8.
  2. Step 2: Stage 4 has 512\times8=4096 red squares.
  3. Step 3: Stage 5 has 4096\times8=32768 red squares.
  4. Answer: 4096 and 32768 red squares.

Textbook page 194 · solved item 80

NCERT Class 9 Maths Chapter 8, solved question 80
Question from the current NCERT textbook

Exercise 8.3, 7(iii): Give explicit and recursive rules for the red-square count at Stage n.

Show detailed solution
  1. Step 1: Stage 0 contains one red square and each stage multiplies the count by 8.
  2. Step 2: The explicit rule is R_n=8^n.
  3. Step 3: The recursive rule is R_0=1,\ R_n=8R_{n-1} for n\ge1.
  4. Answer: R_n=8^n; recursively, R_0=1,\ R_n=8R_{n-1}.

Textbook page 194 · solved item 81

NCERT Class 9 Maths Chapter 8, solved question 81
Question from the current NCERT textbook

Exercise 8.3, 7(iv), area question 1: If Stage 0 has area 1, find the red area in Stages 1, 2 and 3.

Show detailed solution
  1. Step 1: Each stage retains 8 of 9 equal subsquares, so the red area is multiplied by 8/9.
  2. Step 2: Stage 1 and 2 areas are 8/9 and (8/9)^2=64/81.
  3. Step 3: Stage 3 area is (8/9)^3=512/729.
  4. Answer: 8/9,\ 64/81,\ 512/729 square units.

Textbook page 194 · solved item 82

NCERT Class 9 Maths Chapter 8, solved question 82
Question from the current NCERT textbook

Exercise 8.3, 7(iv), area question 2: Find the red area in Stages 4 and 5.

Show detailed solution
  1. Step 1: Use A_n=(8/9)^n.
  2. Step 2: A_4=(8/9)^4=4096/6561.
  3. Step 3: A_5=(8/9)^5=32768/59049.
  4. Answer: 4096/6561 and 32768/59049 square units.

Textbook page 194 · solved item 83

NCERT Class 9 Maths Chapter 8, solved question 83
Question from the current NCERT textbook

Exercise 8.3, 7(iv), area question 3: Give explicit and recursive rules for the red area at Stage n.

Show detailed solution
  1. Step 1: The initial red area is 1 square unit.
  2. Step 2: Each stage retains 8/9 of the preceding red area.
  3. Step 3: Explicitly, A_n=(8/9)^n. Recursively, A_0=1,\ A_n=(8/9)A_{n-1}.
  4. Answer: A_n=(8/9)^n; recursively, A_0=1,\ A_n=(8/9)A_{n-1}.

Textbook page 194 · solved item 84

NCERT Class 9 Maths Chapter 8, solved question 84
Question from the current NCERT textbook

Exercise 8.3, 7(iv), area question 4: What happens to the red area as n increases?

Show detailed solution
  1. Step 1: The area follows A_n=(8/9)^n.
  2. Step 2: The ratio 8/9 is positive and less than 1, so every stage has less red area than the preceding stage.
  3. Step 3: \lim_{n\to\infty}(8/9)^n=0.
  4. Answer: The red area approaches 0 while remaining positive at every finite stage.

Textbook page 194 · solved item 85

NCERT Class 9 Maths Chapter 8, solved question 85
Question from the current NCERT textbook

End-of-Chapter Exercise 1: Find the 31st term of an AP with t11 = 38 and t16 = 73.

Show detailed solution
  1. Step 1: a+10d=38 and a+15d=73.
  2. Step 2: Subtract: 5d=35, so d=7. Then a=38-70=-32.
  3. Step 3: t_{31}=a+30d=-32+210=178.
  4. Answer: The 31st term is 178.

Textbook page 194 · solved item 86

NCERT Class 9 Maths Chapter 8, solved question 86
Question from the current NCERT textbook

End-of-Chapter Exercise 2: Determine the AP whose t3 = 16 and whose t7 exceeds t5 by 12.

Show detailed solution
  1. Step 1: t_7-t_5=(a+6d)-(a+4d)=2d=12, so d=6.
  2. Step 2: Since t_3=a+2d=16, a+12=16, giving a=4.
  3. Step 3: The nth term is t_n=4+6(n-1)=6n-2.
  4. Answer: The AP is 4, 10, 16, 22, ... .

Detailed worked answers

Textbook page 195

Textbook page 195 · solved item 87

NCERT Class 9 Maths Chapter 8, solved question 87
Question from the current NCERT textbook

End-of-Chapter Exercise 3: How many three-digit numbers are divisible by 7?

Show detailed solution
  1. Step 1: The first three-digit multiple is 105 = 7 x 15, and the last is 994 = 7 x 142.
  2. Step 2: These form an AP with a=105,d=7,l=994.
  3. Step 3: n=(994-105)/7+1=889/7+1=127+1=128.
  4. Answer: 128 three-digit numbers are divisible by 7.

Textbook page 195 · solved item 88

NCERT Class 9 Maths Chapter 8, solved question 88
Question from the current NCERT textbook

End-of-Chapter Exercise 4: How many multiples of 4 lie between 10 and 250?

Show detailed solution
  1. Step 1: The smallest such multiple is 12 and the largest is 248.
  2. Step 2: They form an AP with common difference 4.
  3. Step 3: n=(248-12)/4+1=236/4+1=59+1=60.
  4. Answer: 60 multiples of 4.

Textbook page 195 · solved item 89

NCERT Class 9 Maths Chapter 8, solved question 89
Question from the current NCERT textbook

End-of-Chapter Exercise 5: Find a GP whose first two terms sum to -4 and whose fifth term is four times its third.

Show detailed solution
  1. Step 1: Let the GP begin a,ar,ar^2,\ldots. From ar^4=4ar^2 and nonzero terms, r^2=4, so r=2 or r=-2.
  2. Step 2: If r=2, then a+ar=3a=-4, so a=-4/3. This gives -4/3,-8/3,-16/3,\ldots.
  3. Step 3: If r=-2, then a+ar=-a=-4, so a=4. This gives 4,-8,16,-32,64,\ldots.
  4. Answer: There are two GPs, with (a,r)=-4/3,2 or 4,-2.

Textbook page 195 · solved item 90

NCERT Class 9 Maths Chapter 8, solved question 90
Question from the current NCERT textbook

End-of-Chapter Exercise 6: Express 100 in all possible ways as a sum of consecutive natural numbers.

Show detailed solution
  1. Step 1: For k terms starting at m, 100=k(2m+k-1)/2, so m=(200/k-k+1)/2 must be a positive integer.
  2. Step 2: Testing positive divisors k of 200 gives k = 1, 5 and 8.
  3. Step 3: These give m=100,18,9, respectively.
  4. Answer: (100), 18+19+20+21+22, and 9+10+11+12+13+14+15+16. If at least two addends are required, omit the one-term form.

Textbook page 195 · solved item 91

NCERT Class 9 Maths Chapter 8, solved question 91
Question from the current NCERT textbook

End-of-Chapter Exercise 7: A culture starts with 30 bacteria and doubles hourly. Find the counts after 2 hours, 4 hours and n hours.

Show detailed solution
  1. Step 1: After h hours, the original count has been doubled h times, so N_h=30\times2^h.
  2. Step 2: N_2=30\times4=120 and N_4=30\times16=480.
  3. Step 3: Keep h = n for the general rule.
  4. Answer: 120, 480 and 30\times2^n bacteria.

Textbook page 195 · solved item 92

NCERT Class 9 Maths Chapter 8, solved question 92
Question from the current NCERT textbook

End-of-Chapter Exercise 8: Given t4 + t8 = 24 and t6 + t10 = 44, find the first three AP terms.

Show detailed solution
  1. Step 1: t_4+t_8=(a+3d)+(a+7d)=2a+10d=24, so a+5d=12.
  2. Step 2: t_6+t_{10}=2a+14d=44, so a+7d=22. Subtracting gives 2d=10, hence d=5.
  3. Step 3: Then a+25=12, so a=-13.
  4. Answer: The first three terms are -13, -8 and -3.

Textbook page 195 · solved item 93

NCERT Class 9 Maths Chapter 8, solved question 93
Question from the current NCERT textbook

End-of-Chapter Exercise 9: Find the least n for which 1 + 2 + ... + n is greater than 1000.

Show detailed solution
  1. Step 1: Require n(n+1)/2>1000.
  2. Step 2: S_{44}=44(45)/2=990, which is not greater than 1000.
  3. Step 3: S_{45}=45(46)/2=1035, which is greater than 1000.
  4. Answer: The smallest value is n = 45.

Textbook page 195 · solved item 94

NCERT Class 9 Maths Chapter 8, solved question 94
Question from the current NCERT textbook

End-of-Chapter Exercise 10: Which term of 2, 8, 32, ... is 131072? Give explicit and recursive rules.

Show detailed solution
  1. Step 1: The GP has a=2,r=4, so t_n=2\times4^{n-1}.
  2. Step 2: Write t_n=2^{2n-1} and 131072=2^{17}. Thus 2n-1=17, so n=9.
  3. Step 3: Recursively, t_1=2,\ t_n=4t_{n-1} for n\ge2.
  4. Answer: 131072 is the 9th term; the explicit and recursive rules are given above.

Textbook page 195 · solved item 95

NCERT Class 9 Maths Chapter 8, solved question 95
Question from the current NCERT textbook

End-of-Chapter Exercise 11: Three GP terms have sum 13/12 and product -1. Find the terms and common ratio.

Show detailed solution
  1. Step 1: Write the terms as x/r,x,xr. Their product is x^3=-1, so x=-1.
  2. Step 2: The sum equation is -1/r-1-r=13/12. Multiplying by 12r gives 12r^2+25r+12=0.
  3. Step 3: Factor: (3r+4)(4r+3)=0, so r=-4/3 or r=-3/4.
  4. Answer: The terms are 3/4,-1,4/3 with ratio -4/3, or the reverse order 4/3,-1,3/4 with ratio -3/4.

Textbook page 195 · solved item 96

NCERT Class 9 Maths Chapter 8, solved question 96
Question from the current NCERT textbook

End-of-Chapter Exercise 12: If the 4th, 10th and 16th GP terms are x, y and z, prove that x, y, z are in GP.

Show detailed solution
  1. Step 1: For a GP with first term a and ratio r, x=ar^3, y=ar^9, and z=ar^{15}.
  2. Step 2: y^2=(ar^9)^2=a^2r^{18}.
  3. Step 3: Also xz=(ar^3)(ar^{15})=a^2r^{18}, so y^2=xz.
  4. Answer: The square of the middle term equals the product of its neighbours, hence x, y, z are in GP.

Textbook page 195 · solved item 97

NCERT Class 9 Maths Chapter 8, solved question 97
Question from the current NCERT textbook

End-of-Chapter Exercise 13: Three GP terms have sum 26 and sum of squares 364. Find the terms.

Show detailed solution
  1. Step 1: Let the terms be x, y, z. From (x+y+z)^2=364+2(xy+yz+zx), obtain xy+yz+zx=156.
  2. Step 2: Since they are in GP, xz=y^2. Then xy+yz+xz=y(x+z)+y^2=y(x+y+z)=26y.
  3. Step 3: Thus 26y=156, so y=6. Now x+z=20 and xz=36, giving (x,z)=(2,18) or (18,2).
  4. Answer: The terms are 2, 6, 18, or the reverse order 18, 6, 2.

Textbook page 195 · solved item 98

NCERT Class 9 Maths Chapter 8, solved question 98
Question from the current NCERT textbook

End-of-Chapter Exercise 14(a): Find P1 through P8 for the given recurrence.

Show detailed solution
  1. Step 1: P_1=1 and P_2=2.
  2. Step 2: P_3=P_1+P_2+1=4, and P_4=1+2+4+1=8.
  3. Step 3: Continuing the same rule doubles each preceding value.
  4. Answer: 1, 2, 4, 8, 16, 32, 64, 128.

Textbook page 195 · solved item 99

NCERT Class 9 Maths Chapter 8, solved question 99
Question from the current NCERT textbook

End-of-Chapter Exercise 14(b): Find a simpler recursive formula for P_n.

Show detailed solution
  1. Step 1: For (n>2), P_n=P_1+\cdots+P_{n-1}+1.
  2. Step 2: The preceding rule gives P_{n-1}=P_1+\cdots+P_{n-2}+1.
  3. Step 3: Subtracting shows P_n-P_{n-1}=P_{n-1}, hence P_n=2P_{n-1}.
  4. Answer: P_1=1,\ P_n=2P_{n-1} for n\ge2.

Textbook page 195 · solved item 100

NCERT Class 9 Maths Chapter 8, solved question 100
Question from the current NCERT textbook

End-of-Chapter Exercise 14(c): Give an explicit formula for P_n.

Show detailed solution
  1. Step 1: The simpler recurrence has first term 1 and common ratio 2.
  2. Step 2: A GP has explicit form P_n=P_1r^{n-1}.
  3. Step 3: Substitute P_1=1,r=2.
  4. Answer: P_n=2^{n-1}.

Textbook page 195 · solved item 101

NCERT Class 9 Maths Chapter 8, solved question 101
Question from the current NCERT textbook

End-of-Chapter Exercise 15(a): Find W1 through W8 for the given recurrence.

Show detailed solution
  1. Step 1: W_1=1, W_2=2, and W_3=W_1+2=3.
  2. Step 2: W_4=W_1+W_2+2=5, then W_5=8 and W_6=13.
  3. Step 3: Continuing gives W_7=21 and W_8=34.
  4. Answer: 1, 2, 3, 5, 8, 13, 21, 34.

Textbook page 195 · solved item 102

NCERT Class 9 Maths Chapter 8, solved question 102
Question from the current NCERT textbook

End-of-Chapter Exercise 15(b): Recognise the W_n sequence and give its simpler recurrence.

Show detailed solution
  1. Step 1: Compare consecutive rules: W_n=W_1+\cdots+W_{n-2}+2 and W_{n-1}=W_1+\cdots+W_{n-3}+2.
  2. Step 2: Subtract to obtain W_n-W_{n-1}=W_{n-2}.
  3. Step 3: Hence W_n=W_{n-1}+W_{n-2} with W_1=1,W_2=2.
  4. Answer: It is the Virahanka-Fibonacci sequence 1, 2, 3, 5, 8, 13, 21, 34, ... .
Practise this chapter Open Mathwise