Class 9 Maths Part 1 Chapter 8 – Predicting What Comes Next: Exploring Sequences and Progressions from Ganita Manjari with easy Notes and complete Book Solutions. Get clear explanations, important concepts, questions and answers, and revision material for Chapter 8 to help Class 9 students learn, practise and prepare effectively for examinations.
100%
Chapter 8: Predicting What Comes Next
Exploring Sequences and Progressions
Easy Notes • Concepts • Book Questions • Easy Solutions
8.1 Introduction to Sequences
What is a Sequence?
A sequence is an ordered list of numbers. Each number in the list is called a term.
A sequence is an ordered list of numbers. Each number in the list is called a term.
Example:
1, 3, 5, 7, 9, …
Here the terms are 1, 3, 5, 7, 9, …
1, 3, 5, 7, 9, …
Here the terms are 1, 3, 5, 7, 9, …
Remember: The dots (…) show that the sequence continues.
Think and Reflect: Can you describe the pattern in each of the above
sequences? Can you predict the next few numbers in these sequences?
1, 2, 3, 4, 5, 6, …
1, 3, 5, 7, 9, 11, …
1, 3, 6, 10, 15, 21, …
1, 4, 9, 16, 25, 36, …
1, 2, 3, 4, 5, 6, …
1, 3, 5, 7, 9, 11, …
1, 3, 6, 10, 15, 21, …
1, 4, 9, 16, 25, 36, …
Solution:
• 1, 2, 3, 4, 5, 6, … increases by 1.
Next numbers: 7, 8, 9.
• 1, 3, 5, 7, 9, 11, … increases by 2.
Next numbers: 13, 15, 17.
• 1, 3, 6, 10, 15, 21, … has differences 2, 3, 4, 5, 6, …
Next numbers: 28, 36, 45.
• 1, 4, 9, 16, 25, 36, … are square numbers.
Next numbers: 49, 64, 81.
• 1, 2, 3, 4, 5, 6, … increases by 1.
Next numbers: 7, 8, 9.
• 1, 3, 5, 7, 9, 11, … increases by 2.
Next numbers: 13, 15, 17.
• 1, 3, 6, 10, 15, 21, … has differences 2, 3, 4, 5, 6, …
Next numbers: 28, 36, 45.
• 1, 4, 9, 16, 25, 36, … are square numbers.
Next numbers: 49, 64, 81.
Exercise: Consider the sequence 1, 4, 7, 10, 13, … Can you predict
the next four terms? Can you derive the first 10 terms of the sequence
obtained by adding all the terms up to a given term of this sequence?
The sequence increases by 3 each time.So the next four terms are:16, 19, 22, 25Adding the terms successively:1 = 1
1 + 4 = 5
1 + 4 + 7 = 12
1 + 4 + 7 + 10 = 22
1 + 4 + 7 + 10 + 13 = 35
1 + … + 16 = 51
1 + … + 19 = 70
1 + … + 22 = 92
1 + … + 25 = 117
1 + … + 28 = 145
1 + 4 = 5
1 + 4 + 7 = 12
1 + 4 + 7 + 10 = 22
1 + 4 + 7 + 10 + 13 = 35
1 + … + 16 = 51
1 + … + 19 = 70
1 + … + 22 = 92
1 + … + 25 = 117
1 + … + 28 = 145
First 10 cumulative terms:
1, 5, 12, 22, 35, 51, 70, 92, 117, 145
Exercise: Can you write t₅, t₆, t₇ and t₈ for the sequence of
triangular numbers?
Triangular numbers are:1, 3, 6, 10, 15, 21, 28, 36, …Therefore:t₅ = 15
t₆ = 21
t₇ = 28
t₈ = 36
t₆ = 21
t₇ = 28
t₈ = 36
Think and Reflect: Can you think of any other kinds of sequences?
List out five different types of sequences and discuss their properties with
your friends.
Possible examples are:1. Natural numbers: 1, 2, 3, 4, …
2. Even numbers: 2, 4, 6, 8, …
3. Odd numbers: 1, 3, 5, 7, …
4. Square numbers: 1, 4, 9, 16, …
5. Triangular numbers: 1, 3, 6, 10, …Other correct examples are also possible.
2. Even numbers: 2, 4, 6, 8, …
3. Odd numbers: 1, 3, 5, 7, …
4. Square numbers: 1, 4, 9, 16, …
5. Triangular numbers: 1, 3, 6, 10, …Other correct examples are also possible.
8.2 Explicit Rule for a Sequence
Explicit Rule
An explicit rule gives the nth term directly.
An explicit rule gives the nth term directly.
uₙ = 2n − 1
For example, putting n = 5 gives:
u₅ = 2(5) − 1 = 9.
Think and Reflect: Why is it useful to have an explicit formula for
the nth term of a sequence?
An explicit formula helps us find any term directly.We do not have to calculate all the terms before it.For example, if we know the formula for tₙ, we can directly put
n = 100 and find the 100th term.
Exercise: Using the explicit rule uₙ = 2n − 1, find the 53rd term,
the 108th term, and the 1170th term of the odd number sequence.
For the 53rd term:u₅₃ = 2(53) − 1 = 106 − 1 = 105For the 108th term:u₁₀₈ = 2(108) − 1 = 216 − 1 = 215For the 1170th term:u₁₁₇₀ = 2(1170) − 1
= 2340 − 1
= 2339
Answers: 105, 215 and 2339
Think and Reflect: Can you find the rule describing the nth term of
the sequence of square numbers?
The square numbers are:1², 2², 3², 4², …Therefore the nth term is:
tₙ = n²
Think and Reflect: Here is the sequence of the first ten prime
numbers: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29. Do you see any pattern in
this sequence? Can you think of a rule that can predict the next few prime
numbers?
These are prime numbers.Unlike an arithmetic progression, their differences are not constant.The next few prime numbers are:31, 37, 41, 43, …There is no simple constant-difference rule like an AP.
Exercise: Consider the expression tₙ = 3n − 7.
(i) Find its first, second, third, 12th, 18th and 50th terms.
(ii) Which term of the sequence is 332?
(iii) Is 557 a term of this sequence? Why or why not?
(i) Find its first, second, third, 12th, 18th and 50th terms.
(ii) Which term of the sequence is 332?
(iii) Is 557 a term of this sequence? Why or why not?
(i)t₁ = 3(1) − 7 = −4
t₂ = 3(2) − 7 = −1
t₃ = 3(3) − 7 = 2
t₁₂ = 36 − 7 = 29
t₁₈ = 54 − 7 = 47
t₅₀ = 150 − 7 = 143(ii)3n − 7 = 332
3n = 339
n = 113So 332 is the 113th term.(iii)3n − 7 = 557
3n = 564
n = 188Since 188 is a whole-number position, 557 is a term.
t₂ = 3(2) − 7 = −1
t₃ = 3(3) − 7 = 2
t₁₂ = 36 − 7 = 29
t₁₈ = 54 − 7 = 47
t₅₀ = 150 − 7 = 143(ii)3n − 7 = 332
3n = 339
n = 113So 332 is the 113th term.(iii)3n − 7 = 557
3n = 564
n = 188Since 188 is a whole-number position, 557 is a term.
8.3 Recursive Rule for a Sequence
Recursive Rule: A recursive rule tells us how to obtain a term from
the previous term or terms.
t₁ = 1, tₙ = tₙ₋₁ + 3
Exercise: Can you write the next two terms of this sequence?
1, 2, 3, 5, 8, 13, 21, 34, …
1, 2, 3, 5, 8, 13, 21, 34, …
Each term is obtained by adding the previous two terms.21 + 34 = 5534 + 55 = 89
Next two terms: 55, 89
Exercise Set 8.1
1. Find the first five terms of the sequence in which the nth term is
given by (i) tₙ = 3n − 4, (ii) tₙ = 2 − 5n, and (iii) tₙ = n² − 2n + 3
for n ≥ 1.
(i) tₙ = 3n − 4n = 1,2,3,4,5 gives:−1, 2, 5, 8, 11
(ii) tₙ = 2 − 5nFirst five terms:−3, −8, −13, −18, −23
(iii) tₙ = n² − 2n + 3First five terms:2, 3, 5, 9, 15
(ii) tₙ = 2 − 5nFirst five terms:−3, −8, −13, −18, −23
(iii) tₙ = n² − 2n + 3First five terms:2, 3, 5, 9, 15
2. Find the 10th and 15th terms of the sequence tₙ = 5n − 3
for n ≥ 1.
t₁₀ = 5(10) − 3 = 47t₁₅ = 5(15) − 3 = 72
3. Determine whether 97 and 172 are terms of the sequence
tₙ = 5n − 3 for n ≥ 1.
For 97:5n − 3 = 97
5n = 100
n = 20So 97 is a term.For 172:5n − 3 = 172
5n = 175
n = 35So 172 is also a term.
5n = 100
n = 20So 97 is a term.For 172:5n − 3 = 172
5n = 175
n = 35So 172 is also a term.
4. Which term of the sequence tₙ = 5n − 3 for n ≥ 1 is 607?
5n − 3 = 607
5n = 610
n = 122Therefore 607 is the 122nd term.
5n = 610
n = 122Therefore 607 is the 122nd term.
5. A sequence is given by the recursive rule t₁ = − 5,
tₙ₊₁ = tₙ + 3 for n ≥ 1. Find the first five terms of the sequence.
Is 52 a term of this sequence? If so, which term is it?
Starting with −5 and adding 3:−5, −2, 1, 4, 7So the first five terms are:−5, −2, 1, 4, 7For 52:tₙ = −5 + 3(n−1)52 = −5 + 3(n−1)57 = 3(n−1)19 = n−1n = 20Therefore 52 is the 20th term.
6. Let T₁ = 1, T₂ = 2, T₃ = 4, and Tₙ = Tₙ₋₁ + Tₙ₋₂ + Tₙ₋₃
for n ≥ 4. Find T₄, T₅, T₆, T₇, and T₈.
T₄ = 4 + 2 + 1 = 7T₅ = 7 + 4 + 2 = 13T₆ = 13 + 7 + 4 = 24T₇ = 24 + 13 + 7 = 44T₈ = 44 + 24 + 13 = 81
7, 13, 24, 44, 81
8.4 Arithmetic Progressions
Arithmetic Progression (AP)
An AP is a sequence in which the difference between consecutive terms is always the same.
An AP is a sequence in which the difference between consecutive terms is always the same.
tₙ = a + (n − 1)d
a = first term
d = common difference
d = common difference
Think and Reflect: Can you predict the number of squares in Stages 5
and 6 of the sequence? In Stages 10, 11 and 12? In Stage 20? At any stage?
The sequence is:1, 5, 9, 13, …The common difference is 4.So:tₙ = 1 + (n−1)4tₙ = 4n − 3Stage 5 = 17
Stage 6 = 21
Stage 10 = 37
Stage 11 = 41
Stage 12 = 45
Stage 20 = 77
Stage 6 = 21
Stage 10 = 37
Stage 11 = 41
Stage 12 = 45
Stage 20 = 77
Think and Reflect: Consider all the sequences we have discussed so
far in this chapter. Which ones are arithmetic progressions and which ones
are not? Can you justify your claim?
An AP has a constant difference.For example:1, 2, 3, 4, … → difference = 11, 3, 5, 7, … → difference = 21, 4, 7, 10, … → difference = 3Therefore these are APs.Square numbers and triangular numbers are not APs because their differences
change.
Exercise: Verify that the following sequences are arithmetic
progressions and write their nth terms. What do you observe when you plot
the ordered pairs emerging from them?
(i) 2, 5, 8, 11, …
(ii) –5, –1, 3, 7, …
(i) 2, 5, 8, 11, …
(ii) –5, –1, 3, 7, …
(i)d = 5−2 = 3Therefore:tₙ = 2 + (n−1)3tₙ = 3n − 1
(ii)d = −1−(−5) = 4Therefore:tₙ = −5 + (n−1)4tₙ = 4n − 9
When the ordered pairs are plotted, they lie on a straight line.
(ii)d = −1−(−5) = 4Therefore:tₙ = −5 + (n−1)4tₙ = 4n − 9
When the ordered pairs are plotted, they lie on a straight line.
Exercise: Using the formula tₙ = a + (n − 1) × d, find the nth term
of the following arithmetic progressions.
(i) 1/2, 5/2, 9/2, 13/2, …
(ii) 1.5, 3.5, 5.5, 7.5, …
(i) 1/2, 5/2, 9/2, 13/2, …
(ii) 1.5, 3.5, 5.5, 7.5, …
(i)a = 1/2, d = 2tₙ = 1/2 + (n−1)2Therefore:tₙ = 2n − 3/2
(ii)a = 1.5, d = 2tₙ = 1.5 + (n−1)2Therefore:tₙ = 2n − 0.5
(ii)a = 1.5, d = 2tₙ = 1.5 + (n−1)2Therefore:tₙ = 2n − 0.5
Exercise: Find recursive rules for the APs in the previous exercises.
For 2, 5, 8, 11, …:t₁ = 2, tₙ = tₙ₋₁ + 3For −5, −1, 3, 7, …:t₁ = −5, tₙ = tₙ₋₁ + 4For 1/2, 5/2, 9/2, …:t₁ = 1/2, tₙ = tₙ₋₁ + 2For 1.5, 3.5, 5.5, …:t₁ = 1.5, tₙ = tₙ₋₁ + 2
Example 5: A person books a taxi to travel in the city. The taxi
company charges a fixed booking fee of `200 plus `40 per kilometre travelled.
Let us write the sequence representing the total fare after travelling 1 km,
2 km, 3 km, and so on. If the person travels 10 km, what will be the total
fare?
For 1 km:₹200 + ₹40 = ₹240The sequence is:240, 280, 320, …Here:a = 240d = 40For 10 km:t₁₀ = 240 + 9(40)= 240 + 360= ₹600
8.5 Sum of the First n Natural Numbers
Sₙ = 1 + 2 + 3 + … + n
Sₙ = n(n + 1) / 2
Sₙ = n(n + 1) / 2
Easy trick: Multiply the number of terms by the next number and divide
by 2.
Think and Reflect: Can the same approach be used to find the sum of
1 + 2 + 3 + … + 100?
S₁₀₀ = 100 × 101 / 2= 5050
Think and Reflect: Can you use this formula to find S₂₀, S₅₀ or
S₁₀₀₀?
S₂₀ = 20 × 21 / 2 = 210
S₅₀ = 50 × 51 / 2 = 1275
S₁₀₀₀ = 1000 × 1001 / 2 = 500500
S₅₀ = 50 × 51 / 2 = 1275
S₁₀₀₀ = 1000 × 1001 / 2 = 500500
Think and Reflect: Also, this formula can be used to find the sum of
consecutive numbers such as 25 + 26 + 27 + … + 58.
25 + 26 + … + 58= S₅₈ − S₂₄= (58×59)/2 − (24×25)/2= 1711 − 300= 1411
Think and Reflect: Can you use this to find the 10th, 17th and 80th
triangular numbers?
The nth triangular number is:n(n+1)/210th = 10×11/2 = 5517th = 17×18/2 = 15380th = 80×81/2 = 3240
Exercise Set 8.2
1. Find the 10th and 26th terms of the AP: 3, 8, 13, 18, ….
a = 3, d = 5t₁₀ = 3 + 9×5 = 48t₂₆ = 3 + 25×5 = 128
2. Which term of the AP : 21, 18, 15, … is – 81? Also, is 0 a term
of this AP? Give reasons for your answer.
a = 21, d = −3For −81:21 − 3(n−1) = −81n = 35So −81 is the 35th term.For 0:21 − 3(n−1) = 0n = 8So 0 is the 8th term.
3. Find the nth term of the AP: 11, 8, 5, 2 … Write the recursive
rule for this AP.
a = 11, d = −3tₙ = 11 + (n−1)(−3)Therefore:tₙ = 14 − 3nRecursive rule:t₁ = 11tₙ = tₙ₋₁ − 3
4. An AP consists of 50 terms in which the 3rd term is 12 and the last
term is 106. Find the 29th term. (Hint: If ‘a’ is the first term and ‘d’ the
common difference, then we arrive at the equations a + 2d = 12 and
a + 49d = 106. Solve this pair of linear equations for ‘a’ and ‘d’.)
a + 2d = 12a + 49d = 106Subtracting:47d = 94d = 2Now:a + 4 = 12a = 8Therefore:t₂₉ = 8 + 28×2= 64
5. How many 2-digit numbers are divisible by 3? What is the sum of all
these 2-digit numbers?
The numbers are:12, 15, 18, …, 99This is an AP.Number of terms:(99−12)/3 + 1 = 30Sum:30(12+99)/2= 1665
There are 30 numbers 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?
5,00,000 + (n−1)20,000 = 7,00,000(n−1)20,000 = 2,00,000n−1 = 10n = 11So the salary reached ₹7,00,000 in the 11th year.
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?
Total marbles:1 + 2 + 3 + … + 25= 25×26/2= 325
The child uses 325 marbles.
8.6 Geometric Progressions
Geometric Progression (GP)
A GP is a sequence in which every term is obtained by multiplying the previous term by the same number.
A GP is a sequence in which every term is obtained by multiplying the previous term by the same number.
tₙ = arⁿ⁻¹
r is called the common ratio.
Example 6: Is 1, 2, 4, 8, 16, … a geometric progression? If so,
what is the common ratio?
Each term is obtained by multiplying the previous term by 2.Therefore it is a GP.Common ratio = 2
Example 7: Is 1, 3, 9, 27, 81, … a geometric progression? If so,
what is the common ratio?
Each term is multiplied by 3.Therefore it is a GP.Common ratio = 3
Example 8: Is 1, –1, 1, –1, 1, … a geometric progression? If so,
what is the common ratio?
Each term is multiplied by −1.Therefore it is a GP.Common ratio = −1
Example 9: Check whether the sequence 5, 15/4, 45/16, 135/64, …
is a geometric progression and find its nth term.
15/4 ÷ 5 = 3/445/16 ÷ 15/4 = 3/4135/64 ÷ 45/16 = 3/4The ratio is constant.Therefore it is a GP.a = 5, r = 3/4
tₙ = 5(3/4)ⁿ⁻¹
Exercise: Check whether the following sequences are geometric
progressions and find their nth terms.
(i) 2, 10, 50, 250, …
(ii) 4, 8/3, 16/9, 32/27, …
(iii) 3, −3/2, 3/4, −3/8, …
(i) 2, 10, 50, 250, …
(ii) 4, 8/3, 16/9, 32/27, …
(iii) 3, −3/2, 3/4, −3/8, …
(i)Common ratio = 5tₙ = 2×5ⁿ⁻¹
(ii)Common ratio = 2/3tₙ = 4(2/3)ⁿ⁻¹
(iii)Common ratio = −1/2tₙ = 3(−1/2)ⁿ⁻¹All three are GPs.
(ii)Common ratio = 2/3tₙ = 4(2/3)ⁿ⁻¹
(iii)Common ratio = −1/2tₙ = 3(−1/2)ⁿ⁻¹All three are GPs.
Exercise: Can you find a recursive rule for the formula
tₙ = 3 × 10ⁿ⁻¹ that generates the geometric progression
3, 30, 300, 3000, …?
The first term is 3.Each term is multiplied by 10.Therefore:t₁ = 3tₙ = 10tₙ₋₁
Fun with Fractals – Sierpiński Triangle
Think and Reflect: Observe the Sierpiński triangle and try to answer
the following questions
(a) How many black triangles are there in Stages 0 to 3 of Fig. 8.7?
(b) Can you predict the number of black triangles at Stages 4 and 5?
(c) Can you find a rule for the number of black triangles at the nth stage?
(d) Suppose the area of the triangle (that is, the black region) in Stage 0 is 1 square unit. What is the area of the black region in Stages 1, 2 and 3? What will be the area of the black region in Stages 4 and 5? Find a rule for the area of the black region at the nth stage. What happens to this area as n, the number of stages, goes on increasing?
(a) How many black triangles are there in Stages 0 to 3 of Fig. 8.7?
(b) Can you predict the number of black triangles at Stages 4 and 5?
(c) Can you find a rule for the number of black triangles at the nth stage?
(d) Suppose the area of the triangle (that is, the black region) in Stage 0 is 1 square unit. What is the area of the black region in Stages 1, 2 and 3? What will be the area of the black region in Stages 4 and 5? Find a rule for the area of the black region at the nth stage. What happens to this area as n, the number of stages, goes on increasing?
(a)Stages 0, 1, 2, 3:1, 3, 9, 27
(b)Stage 4 = 81Stage 5 = 243
(c)Every stage has 3 times as many black triangles.Therefore:
(d)The area is multiplied by 3/4 at every stage.Stage 1 = 3/4Stage 2 = 9/16Stage 3 = 27/64Stage 4 = 81/256Stage 5 = 243/1024Rule:
(b)Stage 4 = 81Stage 5 = 243
(c)Every stage has 3 times as many black triangles.Therefore:
tₙ = 3ⁿ
(d)The area is multiplied by 3/4 at every stage.Stage 1 = 3/4Stage 2 = 9/16Stage 3 = 27/64Stage 4 = 81/256Stage 5 = 243/1024Rule:
Area = (3/4)ⁿ
As n increases, the area gets closer and closer to 0.Exercise Set 8.3
1. Find the 12th term of a GP with common ratio 2, whose 8th term
is 192.
From the 8th term to the 12th term there are 4 steps.t₁₂ = 192 × 2⁴= 192 × 16= 3072
2. Find the 10th and nth terms of the GP: 5, 25, 125, ….
a = 5r = 5Therefore:
tₙ = 5×5ⁿ⁻¹ = 5ⁿ
10th term:t₁₀ = 5¹⁰= 9,765,625
*3. A sequence is given by the recursive rule t₁ = 2,
tₙ₊₁ = 3tₙ – 2 for n ≥ 1. Which term of the sequence is 730?
Start with t₁ = 2.t₂ = 3(2)−2 = 4t₃ = 3(4)−2 = 10t₄ = 28t₅ = 82t₆ = 244t₇ = 730Therefore:
730 is the 7th term.
4. Which term of the GP: 2, 6, 18, … is 4374? Write the explicit
formula as well as the recursive formula for the nth term.
a = 2, r = 32×3ⁿ⁻¹ = 43744374 ÷ 2 = 21872187 = 3⁷Therefore n−1 = 7n = 8Explicit formula:tₙ = 2×3ⁿ⁻¹Recursive formula:t₁ = 2tₙ = 3tₙ₋₁
5. A ball is dropped from a height of 80 metres. After hitting the
ground, it bounces back to 60% of the height from which it fell. It continues
bouncing in this way — each time rising to 60% of the previous height.
(i) What height does the ball reach after the 5th bounce?
(ii) What is the total vertical distance the ball has travelled by the time it hits the ground for the 6th time?
(i) What height does the ball reach after the 5th bounce?
(ii) What is the total vertical distance the ball has travelled by the time it hits the ground for the 6th time?
The bounce heights form a GP:48, 28.8, 17.28, 10.368, …(i)5th bounce:80 × (0.6)⁵= 6.2208 m
(ii)Add the downward and upward distances carefully.The total distance by the 6th ground hit is:305.07008 m
(ii)Add the downward and upward distances carefully.The total distance by the 6th ground hit is:305.07008 m
5th bounce = 6.2208 m
Total distance = 305.07008 m
Total distance = 305.07008 m
6. Which term of the sequence 2, 2√2, 4, … is 128?
This is a GP.a = 2r = √2So:tₙ = 2(√2)ⁿ⁻¹Put tₙ = 128.Solving gives:n = 13Therefore 128 is the 13th term.
7. Fig. 8.12 shows Stages 0 to 3 of the Sierpiński 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.
Look at Fig. 8.12 and try to answer the following questions.
(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 nth 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 nth stage. What happens to this area as n, the number of stages, goes on increasing?
Look at Fig. 8.12 and try to answer the following questions.
(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 nth 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 nth stage. What happens to this area as n, the number of stages, goes on increasing?
(i)Stages 0 to 3:1, 8, 64, 512
(ii)Stage 4:8⁴ = 4096Stage 5:8⁵ = 32768
(iii)Each stage has 8 times as many squares.Explicit formula:
(iv)At each stage the area is multiplied by 8/9.Therefore:Stage 1 = 8/9Stage 2 = 64/81Stage 3 = 512/729Stage 4 = 4096/6561Stage 5 = 32768/59049Explicit formula:
(ii)Stage 4:8⁴ = 4096Stage 5:8⁵ = 32768
(iii)Each stage has 8 times as many squares.Explicit formula:
tₙ = 8ⁿ
Recursive formula:t₀ = 1tₙ = 8tₙ₋₁(iv)At each stage the area is multiplied by 8/9.Therefore:Stage 1 = 8/9Stage 2 = 64/81Stage 3 = 512/729Stage 4 = 4096/6561Stage 5 = 32768/59049Explicit formula:
Area = (8/9)ⁿ
As n becomes larger, the area gets closer and closer to 0.End-of-Chapter Exercises
1. Find the 31st term of an AP whose 11th term is 38 and 16th
term is 73.
a + 10d = 38a + 15d = 73Subtract:5d = 35d = 7Then:a + 70 = 38a = −32Therefore:t₃₁ = −32 + 30×7= 178
2. Determine the AP whose third term is 16 and whose 7th term exceeds
the 5th term by 12.
Third term:a + 2d = 16Also:t₇ − t₅ = 12(a+6d) − (a+4d) = 122d = 12d = 6Therefore:a + 12 = 16a = 4
AP = 4, 10, 16, 22, …
*3. How many three-digit numbers are divisible by 7? (Hint: All
three-digit numbers divisible by 7 form an AP. Find the smallest and largest
such three-digit numbers.)
Smallest three-digit multiple of 7 = 105.Largest three-digit multiple of 7 = 994.Number of terms:(994−105)/7 + 1= 127 + 1= 128
*4. How many multiples of 4 lie between 10 and 250? (Hint: All
multiples of 4 form an AP. Find the smallest and largest multiples of 4
between 10 and 250.)
Smallest = 12Largest = 248Number:(248−12)/4 + 1= 59 + 1= 60
*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.
Let the terms be:a, ar, ar², ar³, ar⁴Given:a + ar = −4andar⁴ = 4ar²So:r² = 4Therefore:r = 2 or −2For r = 2:3a = −4a = −4/3For r = −2:−a = −4a = 4Hence two possible GPs are:−4/3, −8/3, −16/3, …and4, −8, 16, −32, …
*6. Find all possible ways of expressing 100 as the sum of consecutive
natural numbers.
The possible expressions are:1009 + 10 + 11 + 12 + 13 + 14 + 15 + 16 = 10018 + 19 + 20 + 21 + 22 = 100These are the possible ways using consecutive natural numbers.
*7. The number of bacteria in a certain culture doubles every hour.
If there were 30 bacteria present in the culture originally, how many
bacteria will be present at the end of the 2nd hour, 4th hour and nth hour?
After h hours:30 × 2ʰ2nd hour:30×2² = 1204th hour:30×2⁴ = 480nth hour:
30×2ⁿ
*8. The sum of the 4th and 8th terms of an AP is 24 and the sum of
the 6th and 10th terms is 44. Find the first three terms of the AP.
t₄ + t₈ = 242a + 10d = 24a + 5d = 12Also:t₆ + t₁₀ = 442a + 14d = 44a + 7d = 22Subtracting:2d = 10d = 5Then:a + 25 = 12a = −13Therefore first three terms are:
−13, −8, −3
*9. Find the smallest value of n such that the sum of the first n
natural numbers is greater than 1,000.
S₄₄ = 44×45/2 = 990This is less than 1000.S₄₅ = 45×46/2 = 1035This is greater than 1000.
Smallest value of n = 45
*10. Which term of the GP: 2, 8, 32, … is 131072? Write the
explicit formula as well as the recursive formula for the nth term.
a = 2r = 4tₙ = 2×4ⁿ⁻¹Put tₙ = 131072.This gives:n = 9Explicit formula:tₙ = 2×4ⁿ⁻¹Recursive formula:t₁ = 2tₙ = 4tₙ₋₁
*11. The sum of the first three terms of a GP is 13/12 and their
product is –1. Find the common ratio and the terms.
The three terms are:3/4, −1, 4/3Their sum:3/4 − 1 + 4/3 = 13/12Their product:(3/4)(−1)(4/3) = −1The common ratio is:−4/3The reverse order is also possible, with common ratio −3/4.
*12. If the 4th, 10th and 16th terms of a GP are x, y and z
respectively, prove that x, y, z are in GP.
Let the common ratio be r.x = ar³y = ar⁹z = ar¹⁵Then:y/x = r⁶z/y = r⁶Therefore:y/x = z/ySo:y² = xzHence x, y, z are in GP.
*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.
The terms are:2, 6, 18Check:2 + 6 + 18 = 26and2² + 6² + 18²= 4 + 36 + 324= 364Therefore the terms are:
2, 6, 18
Their reverse order also forms a GP.
*14. Suppose P₁ = 1, P₂ = 2 and for n > 2,
Pₙ = P₁ + P₂ + … + Pₙ₋₁ + 1. Find the values of
P₁, P₂, …, P₈. Can you find a simpler recursive formula for Pₙ?
Can you give an explicit formula?
The terms are:P₁ = 1P₂ = 2P₃ = 1+2+1 = 4P₄ = 1+2+4+1 = 8Continuing:1, 2, 4, 8, 16, 32, 64, 128The simpler recursive rule is:Pₙ = 2Pₙ₋₁Explicit formula:
Pₙ = 2ⁿ⁻¹
*15. Suppose W₁ = 1, W₂ = 2 and for n > 2,
Wₙ = W₁ + W₂ + … + Wₙ₋₂ + 2. Find the values of
W₁, W₂, …, W₈. Do you recognise this sequence?
W₁ = 1W₂ = 2W₃ = 1+2+2 = 5W₄ = 1+2+5+2 = 10Continuing:1, 2, 5, 10, 20, 40, 80, 160From W₄ onward the terms double.For n ≥ 3:
Wₙ = 5×2ⁿ⁻³
Chapter 8 Quick Revision
Sequence: An ordered list of numbers.
Explicit rule: Gives the nth term directly.
Recursive rule: Gives a term using earlier term(s).
AP: Constant difference.
tₙ = a + (n−1)d
Sum of first n natural numbers:
Sₙ = n(n+1)/2
GP: Constant ratio.
tₙ = arⁿ⁻¹
Explicit rule: Gives the nth term directly.
Recursive rule: Gives a term using earlier term(s).
AP: Constant difference.
tₙ = a + (n−1)d
Sum of first n natural numbers:
Sₙ = n(n+1)/2
GP: Constant ratio.
tₙ = arⁿ⁻¹
prepare more with the given Worksheet of chapte8 , which will help you to improve your knowledge .
Class 9 Maths Part 1 Chapter 8 Notes, Class 9 Maths Chapter 8 Notes, Class 9 Ganita Manjari Chapter 8 Notes, Class 9 Ganita Manjari Part 1 Chapter 8 Notes, Class 9 Maths Part 1 Chapter 8 Book Solutions, Class 9 Maths Chapter 8 Book Solutions, Class 9 Ganita Manjari Chapter 8 Book Solutions, Class 9 Ganita Manjari Chapter 8 Notes and Solutions, Class 9 Maths Chapter 8 Questions and Answers, Class 9 Maths Part 1 Chapter 8 Questions and Answers, Class 9 Ganita Manjari Chapter 8 Questions and Answers, NCERT Class 9 Maths Chapter 8 Notes, NCERT Class 9 Mathematics Part 1 Chapter 8, CBSE Class 9 Maths Chapter 8 Notes, Class 9 Maths Chapter 8 Study Material, Class 9 Maths Chapter 8 Revision Notes, Class 9 Ganita Manjari Chapter 8 Revision Notes