Class 9 Maths Chapter 8 NCERT Solutions: Predicting What Comes Next: Exploring Sequences and Progressions

Section 8.1: Introduction to Sequences

Think and Reflect (Page 174)

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, ... (Natural Numbers)
1, 3, 5, 7, 9, 11, ... (Odd Numbers)
1, 3, 6, 10, 15, 21, ... (Triangular Numbers)
1, 4, 9, 16, 25, 36, ... (Square Numbers)

Solution

Natural Numbers: The pattern is to add 1 to the previous number. The next three numbers are 7, 8, and 9.
Odd Numbers: The pattern is to add 2 to the previous number. The next three numbers are 13, 15, and 17.
Triangular Numbers: The pattern involves adding the next counting number to the previous term. To get the 2nd term, you add 2. To get the 3rd term, you add 3. Following this, the next gap after 21 (which is the 6th term) will be adding 7, then 8, then 9. The next three numbers are 28, 36, and 45.
Square Numbers: The pattern is multiplying a natural number by itself ( $1 \times 1$ , $2 \times 2$ , $3 \times 3$ , etc.). The next three numbers are $7 \times 7 = 49$ , $8 \times 8 = 64$ , and $9 \times 9 = 81$ .

In-Text Question (Page 174)

But the sequence 6, 12, 24, 48, 96 is a finite sequence of five terms. Can you think of other finite sequences that you see in your daily life?

Solution

(Answers may vary, but here are some logical examples)
The days of the week: Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, Sunday.
The periods in a school timetable: 1st period, 2nd period, 3rd period, etc., up to the final period.
The months of the year: January, February, March... up to December.

In-Text Question (Page 175)

This is represented by the diagram in Fig. 8.1, where each triangular number is represented by a triangular array of dots. Can you draw the patterns for the next two terms of the sequence?

Fig. 8.1: Triangular numbers as triangular arrays of dots (terms 1–7)

Solution

The next two patterns follow the arrangement shown in Fig. 8.1:
For the 6th term (21): Draw a horizontal row of 6 dots at the bottom. Above that, center a row of 5 dots, then 4, 3, 2, and finally 1 dot at the top to form a triangle.
For the 7th term (28): Draw a horizontal row of 7 dots at the bottom. Stack rows of 6, 5, 4, 3, 2, and 1 dot on top of each other, forming an even larger triangle.

In-Text Question (Page 175)

This interesting relationship between the odd numbers and square numbers can be represented by the diagram in Fig. 8.2. Can you explain the relationship?

Fig. 8.2: Square numbers built by adding consecutive odd numbers

Solution

Looking at Fig. 8.2, if you start with 1 dot (the first square number, 1) and add an "L" shape of 3 new dots around it (the next odd number), you get a $2 \times 2$ square (the square number 4). If you wrap another "L" shape of 5 new dots around that, you get a $3 \times 3$ square (the square number 9). In short, adding consecutive odd numbers starting from 1 will always build the next perfect square number.

Exercise (Page 176)

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? (Hint: The first term is 1. The second term is 1 + 4 = 5, the third term is 1 + 4 + 7 = 12, and so on.)

Solution

Next four terms: The rule for this sequence is to add 3 to the previous term. $13 + 3 = 16$ $16 + 3 = 19$ $19 + 3 = 22$ $22 + 3 = 25$ The next four terms are 16, 19, 22, and 25.
First 10 terms of the sum sequence: We need to keep a running total of the sequence 1, 4, 7, 10, 13, 16, 19, 22, 25, 28. Term 1: 1 Term 2: $1 + 4 = 5$ Term 3: $5 + 7 = 12$ Term 4: $12 + 10 = 22$ Term 5: $22 + 13 = 35$ Term 6: $35 + 16 = 51$ Term 7: $51 + 19 = 70$ Term 8: $70 + 22 = 92$ Term 9: $92 + 25 = 117$ Term 10: $117 + 28 = 145$ The sequence of sums is: 1, 5, 12, 22, 35, 51, 70, 92, 117, 145.

Exercise (Page 176)

Can you write $𝑡 5$ , $𝑡 6$ , $𝑡 7$ and $𝑡 8$ for the sequence of triangular numbers?

Solution

The sequence of triangular numbers is built by adding the position number to the previous term.
We know from the text that $𝑡 4 = 10$ .
$𝑡 5 = 𝑡 4 + 5 = 10 + 5 = 15$
$𝑡 6 = 𝑡 5 + 6 = 15 + 6 = 21$
$𝑡 7 = 𝑡 6 + 7 = 21 + 7 = 28$
$𝑡 8 = 𝑡 7 + 8 = 28 + 8 = 36$

Think and Reflect (Page 176)

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

Solution

(Answers may vary, but here are five common types of sequences)
Sequences with negative numbers: (e.g., -5, -10, -15, -20...) This sequence decreases by a constant value.
Fractional sequences: (e.g., 1/2, 1/3, 1/4, 1/5...) As shown in the text, the terms get smaller and approach zero.
Alternating sequences: (e.g., 1, -1, 1, -1, 1...) The numbers flip back and forth between positive and negative.
Prime number sequence: (e.g., 2, 3, 5, 7, 11...) A sequence of numbers greater than 1 that only have two factors: 1 and themselves.
Doubling sequences: (e.g., 2, 4, 8, 16, 32...) Each term is multiplied by a constant number (in this case, 2) to get the next term.

Section 8.2: Explicit Rule for a Sequence

Think and Reflect (Page 177)

Why is it useful to have an explicit formula for the $𝑛 𝑡 ℎ$ term of a sequence?

Solution

An explicit formula is useful because it allows us to find the value of any term in the sequence directly (like the 100th term or the 1000th term) just by plugging in its position number ( $𝑛$ ). We don't have to calculate all the preceding terms to get there. It also helps us solve backwards to check if a specific number is part of the sequence.

Exercise (Page 177)

Using the explicit rule $𝑢 𝑛 = 2 𝑛 - 1$ , find the 53rd term, the 108th term, and the 1170th term of the odd number sequence.

Solution

Substitute the given position numbers for $𝑛$ into the formula:
53rd term: $𝑢 53 = 2 (53) - 1 = 106 - 1 = 105$
108th term: $𝑢 108 = 2 (108) - 1 = 216 - 1 = 215$
1170th term: $𝑢 1170 = 2 (1170) - 1 = 2340 - 1 = 2339$

Example 2 (Page 177)

Consider the sequence that is generated by the explicit formula $𝑠 𝑛 = 5 𝑛 - 2$ . Can you write the first 6 terms of this sequence? What is the 100th term? The 1000th term?... Can you explain why we need $𝑛$ to be a natural number?

Solution

First 6 terms: Substitute $𝑛$ = 1, 2, 3, 4, 5, 6. $𝑠 1 = 5 (1) - 2 = 3$ $𝑠 2 = 5 (2) - 2 = 8$ $𝑠 3 = 5 (3) - 2 = 13$ $𝑠 4 = 5 (4) - 2 = 18$ $𝑠 5 = 5 (5) - 2 = 23$ $𝑠 6 = 5 (6) - 2 = 28$ The first 6 terms are 3, 8, 13, 18, 23, and 28.
100th term: $𝑠 100 = 5 (100) - 2 = 500 - 2 = 498$
1000th term: $𝑠 1000 = 5 (1000) - 2 = 5000 - 2 = 4998$
Why $𝑛$ needs to be a natural number: In sequences, $𝑛$ represents the position of the term (1st, 2nd, 3rd, etc.). Because we count positions using counting numbers, $𝑛$ must be a natural number (a positive integer). We cannot have a "94.6th" position.

Think and Reflect (Page 177)

Can you find the rule describing the $𝑛 𝑡 ℎ$ term of the sequence of square numbers?

Solution

The sequence of square numbers is 1, 4, 9, 16, 25... These are found by squaring their position number (1 squared, 2 squared, 3 squared, etc.). Therefore, the explicit rule is $𝑡 𝑛 = 𝑛 2$ .

Think and Reflect (Page 177 - 178)

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?

Solution

Unlike sequences that grow by a constant difference or multiplier, there is no simple mathematical formula or explicit algebraic rule that generates the prime numbers. The only "pattern" is their definition: each number is a natural number greater than 1 that is only divisible by 1 and itself.

Exercise (Page 178)

Consider the expression $𝑡 𝑛 = 3 𝑛 - 7$ . (i) Find its first, second, third, 12th, 18th and $50 𝑡 ℎ$ terms. (ii) Which term of the sequence is 332? (iii) Is 557 a term of this sequence? Why or why not?

Solution

(i) Plug the positions into $𝑛$ : $𝑡 1 = 3 (1) - 7 = - 4$ $𝑡 2 = 3 (2) - 7 = - 1$ $𝑡 3 = 3 (3) - 7 = 2$ $𝑡 12 = 3 (12) - 7 = 36 - 7 = 29$ $𝑡 18 = 3 (18) - 7 = 54 - 7 = 47$ $𝑡 50 = 3 (50) - 7 = 150 - 7 = 143$
(ii) Set the formula equal to 332 and solve for $𝑛$ : $3 𝑛 - 7 = 332$ $3 𝑛 = 339$ $𝑛 = 113$ 332 is the 113th term.
(iii) Set the formula equal to 557 and solve for $𝑛$ : $3 𝑛 - 7 = 557$ $3 𝑛 = 564$ $𝑛 = 188$ Because 188 is a natural number, yes, 557 is the 188th term of this sequence.

Section 8.3: Recursive Rule for a Sequence

Example 3 (Page 178)

Find the first four terms of the sequence given by the recursive rule $𝑢 1 = 1$ , $𝑢 𝑛 = 2 𝑢 𝑛 - 1 + 3$ for $𝑛 \geq 2$ . Is 133 a term of this sequence?

Solution

First, we find the first four terms by plugging the previous term into the formula:
$𝑢 1 = 1$ (Given)
$𝑢 2 = 2 (1) + 3 = 5$
$𝑢 3 = 2 (5) + 3 = 13$
$𝑢 4 = 2 (13) + 3 = 29$ The first four terms are 1, 5, 13, 29. To check if 133 is a term, we must keep generating terms:
$𝑢 5 = 2 (29) + 3 = 58 + 3 = 61$
$𝑢 6 = 2 (61) + 3 = 122 + 3 = 125$
$𝑢 7 = 2 (125) + 3 = 250 + 3 = 253$ Since the sequence jumps from 125 to 253, skipping 133 entirely, 133 is not a term in this sequence.

Example 4 (Page 178)

Find the first four terms of the sequence given by the recursive rule $𝑠 1 = 3$ , $𝑠 𝑛 = 𝑠 𝑛 - 1 (𝑠 𝑛 - 1 - 1)$ for $𝑛 \geq 2$ .

Solution

Substitute the previous term into the formula for each new term:
$𝑠 1 = 3$ (Given)
$𝑠 2 = 𝑠 1 (𝑠 1 - 1) = 3 (3 - 1) = 3 (2) = 6$
$𝑠 3 = 𝑠 2 (𝑠 2 - 1) = 6 (6 - 1) = 6 (5) = 30$
$𝑠 4 = 𝑠 3 (𝑠 3 - 1) = 30 (30 - 1) = 30 (29) = 870$ The first four terms are 3, 6, 30, and 870.

In-Text Question (Page 179)

The most famous example of such a sequence is $𝑉 1 = 1, 𝑉 2 = 2$ and $𝑉 𝑛 = 𝑉 𝑛 - 1 + 𝑉 𝑛 - 2$ for $𝑛 \geq 3$ ... So we get the sequence 1, 2, 3, 5, 8, 13, 21, 34, ... Can you write the next two terms of this sequence?

Solution

The recursive rule states that each term is the sum of the two preceding terms.
To find the next term ( $𝑉 9$ ), add the 7th and 8th terms: $21 + 34 = 55$ .
To find the following term ( $𝑉 10$ ), add the 8th and 9th terms: $34 + 55 = 89$ . The next two terms are 55 and 89.

EXERCISE SET 8.1 (Pages 179 - 180)

Question 1

Find the first five terms of the sequence in which the $𝑛 𝑡 ℎ$ term is given by (i) $𝑡 𝑛 = 3 𝑛 - 4$ , (ii) $𝑡 𝑛 = 2 - 5 𝑛$ , and (iii) $𝑡 𝑛 = 𝑛 2 - 2 𝑛 + 3$ for $𝑛 \geq 1$ .

Solution

Substitute $𝑛$ = 1, 2, 3, 4, 5 for each rule.
(i) $𝑡 𝑛 = 3 𝑛 - 4$ $𝑡 1 = 3 (1) - 4 = - 1$ $𝑡 2 = 3 (2) - 4 = 2$ $𝑡 3 = 3 (3) - 4 = 5$ $𝑡 4 = 3 (4) - 4 = 8$ * $𝑡 5 = 3 (5) - 4 = 11$ First five terms: -1, 2, 5, 8, 11
(ii) $𝑡 𝑛 = 2 - 5 𝑛$ $𝑡 1 = 2 - 5 (1) = - 3$ $𝑡 2 = 2 - 5 (2) = - 8$ $𝑡 3 = 2 - 5 (3) = - 13$ $𝑡 4 = 2 - 5 (4) = - 18$ * $𝑡 5 = 2 - 5 (5) = - 23$ First five terms: -3, -8, -13, -18, -23
(iii) $𝑡 𝑛 = 𝑛 2 - 2 𝑛 + 3$ $𝑡 1 = (1) 2 - 2 (1) + 3 = 1 - 2 + 3 = 2$ $𝑡 2 = (2) 2 - 2 (2) + 3 = 4 - 4 + 3 = 3$ $𝑡 3 = (3) 2 - 2 (3) + 3 = 9 - 6 + 3 = 6$ $𝑡 4 = (4) 2 - 2 (4) + 3 = 16 - 8 + 3 = 11$ * $𝑡 5 = (5) 2 - 2 (5) + 3 = 25 - 10 + 3 = 18$ First five terms: 2, 3, 6, 11, 18

Question 2

Find the $10 𝑡 ℎ$ and 15th terms of the sequence $𝑡 𝑛 = 5 𝑛 - 3$ for $𝑛 \geq 1$ .

Solution

$𝑡 10 = 5 (10) - 3 = 50 - 3 = 47$
$𝑡 15 = 5 (15) - 3 = 75 - 3 = 72$

Question 3

Determine whether 97 and 172 are terms of the sequence $𝑡 𝑛 = 5 𝑛 - 3$ for $𝑛 \geq 1$ .

Solution

Set the explicit formula equal to the number and solve for $𝑛$ .
For 97: $5 𝑛 - 3 = 97$ $5 𝑛 = 100$ $𝑛 = 20$ Yes, 97 is the 20th term.
For 172: $5 𝑛 - 3 = 172$ $5 𝑛 = 175$ $𝑛 = 35$ Yes, 172 is the 35th term.

Question 4

Which term of the sequence $𝑡 𝑛 = 5 𝑛 - 3$ for $𝑛 \geq 1$ is 607?

Solution

Set the explicit formula equal to 607 and solve for $𝑛$ . $5 𝑛 - 3 = 607$ $5 𝑛 = 610$ $𝑛 = 122$ 607 is the 122nd term.

Question 5

A sequence is given by the recursive rule $𝑡 1 = - 5$ , $𝑡 𝑛 + 1 = 𝑡 𝑛 + 3$ for $𝑛 \geq 1$ . Find the first five terms of the sequence. Is 52 a term of this sequence? If so, which term is it?

Solution

Use the recursive rule to find the terms by adding 3 to the previous term:
$𝑡 1 = - 5$
$𝑡 2 = - 5 + 3 = - 2$
$𝑡 3 = - 2 + 3 = 1$
$𝑡 4 = 1 + 3 = 4$
$𝑡 5 = 4 + 3 = 7$ The first five terms are -5, -2, 1, 4, 7. To check if 52 is a term, we can find the explicit formula. The sequence starts at -5 and increases by 3 every time. This is an arithmetic progression. The explicit rule is $𝑡 𝑛 = 𝑎 + (𝑛 - 1) 𝑑 = - 5 + (𝑛 - 1) 3 = - 5 + 3 𝑛 - 3 = 3 𝑛 - 8$ . Now, set it equal to 52: $3 𝑛 - 8 = 52$ $3 𝑛 = 60$ $𝑛 = 20$ Because 20 is a natural number, yes, 52 is the 20th term of this sequence.

Question 6

Let $𝑇 1 = 1, 𝑇 2 = 2, 𝑇 3 = 4$ , and $𝑇 𝑛 = 𝑇 𝑛 - 1 + 𝑇 𝑛 - 2 + 𝑇 𝑛 - 3$ for $𝑛 \geq 4$ . Find $𝑇 4$ , $𝑇 5$ , $𝑇 6$ , $𝑇 7$ and $𝑇 8$ .

Solution

The recursive rule tells us that any term starting from the 4th term is the sum of the three previous terms.
$𝑇 4 = 𝑇 3 + 𝑇 2 + 𝑇 1 = 4 + 2 + 1 = 7$
$𝑇 5 = 𝑇 4 + 𝑇 3 + 𝑇 2 = 7 + 4 + 2 = 13$
$𝑇 6 = 𝑇 5 + 𝑇 4 + 𝑇 3 = 13 + 7 + 4 = 24$
$𝑇 7 = 𝑇 6 + 𝑇 5 + 𝑇 4 = 24 + 13 + 7 = 44$
$𝑇 8 = 𝑇 7 + 𝑇 6 + 𝑇 5 = 44 + 24 + 13 = 81$

Section 8.4: Arithmetic Progressions

Think and Reflect (Page 180)

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?

Solution

The pattern of squares shown in Fig. 8.3 grows by adding 4 squares to the corners at each stage. The sequence is 1, 5, 9, 13... which means the common difference is 4. The explicit formula for the $𝑛 𝑡 ℎ$ stage is $𝑡 𝑛 = 4 𝑛 - 3$ . Using this rule:
Stage 5: $4 (5) - 3 = 20 - 3 = 17$ squares.
Stage 6: $4 (6) - 3 = 24 - 3 = 21$ squares.
Stage 10: $4 (10) - 3 = 40 - 3 = 37$ squares.
Stage 11: $4 (11) - 3 = 44 - 3 = 41$ squares.
Stage 12: $4 (12) - 3 = 48 - 3 = 45$ squares.
Stage 20: $4 (20) - 3 = 80 - 3 = 77$ squares.
At any stage $𝑛$ : The number of squares will be $4 𝑛 - 3$ .

Think and Reflect (Page 181)

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?

Solution

An Arithmetic Progression (AP) requires a constant difference between any two consecutive terms.
Natural Numbers (1, 2, 3, 4...): Yes, it is an AP. The common difference is exactly 1.
Odd Numbers (1, 3, 5, 7...): Yes, it is an AP. The common difference is exactly 2.
Triangular Numbers (1, 3, 6, 10...): No, it is not an AP. The differences between consecutive terms are changing (2, 3, 4, etc.).
Square Numbers (1, 4, 9, 16...): No. The differences between consecutive terms are changing (3, 5, 7, etc.).
Prime Numbers (2, 3, 5, 7...): No. The differences fluctuate (1, 2, 2, 4, etc.) without a constant pattern.
Virahānka-Fibonacci sequence (1, 2, 3, 5, 8...): No. The differences increase (1, 1, 2, 3, etc.).

Exercise (Page 182)

Verify that the following sequences are arithmetic progressions and write their $𝑛 𝑡 ℎ$ terms. What do you observe when you plot the ordered pairs emerging from them? (i) 2, 5, 8, 11, ... (ii) -5, -1, 3, 7, ...

Solution

(i) 2, 5, 8, 11, ... Verification: Find the difference between consecutive terms: $5 - 2 = 3$ , $8 - 5 = 3$ , $11 - 8 = 3$ . Because the difference is constantly 3, it is an AP. $𝑛 𝑡 ℎ$ term: The first term ( $𝑎$ ) is 2, and the common difference ( $𝑑$ ) is 3. $𝑡 𝑛 = 𝑎 + (𝑛 - 1) 𝑑$ $𝑡 𝑛 = 2 + (𝑛 - 1) 3$ $𝑡 𝑛 = 2 + 3 𝑛 - 3$ $𝑡 𝑛 = 3 𝑛 - 1$
(ii) -5, -1, 3, 7, ... Verification: Find the difference: $- 1 - (- 5) = 4$ , $3 - (- 1) = 4$ , $7 - 3 = 4$ . Because the difference is constantly 4, it is an AP. $𝑛 𝑡 ℎ$ term: The first term ( $𝑎$ ) is -5, and the common difference ( $𝑑$ ) is 4. $𝑡 𝑛 = - 5 + (𝑛 - 1) 4$ $𝑡 𝑛 = - 5 + 4 𝑛 - 4$ $𝑡 𝑛 = 4 𝑛 - 9$
Observation upon plotting: If you take the position number as the x-coordinate and the sequence term as the y-coordinate (e.g., (1, 2), (2, 5), (3, 8)), and plot these ordered pairs on a graph, you will observe that they form a perfectly straight line. This is characteristic of all arithmetic progressions.

Exercise (Page 182)

Using the formula $𝑡 𝑛 = 𝑎 + (𝑛 - 1) \times 𝑑$ find the $𝑛 𝑡 ℎ$ term of the following arithmetic progressions. (i) 1/2, 5/2, 9/2, 13/2, ... (ii) 1.5, 3.5, 5.5, 7.5, ...

Solution

(i) $𝑎 = 1 / 2$ . To find $𝑑$ , subtract the first term from the second: $5 / 2 - 1 / 2 = 4 / 2 = 2$ . $𝑡 𝑛 = 1 / 2 + (𝑛 - 1) 2$ $𝑡 𝑛 = 1 / 2 + 2 𝑛 - 2$ $𝑡 𝑛 = 2 𝑛 - 3 / 2$
(ii) $𝑎 = 1.5$ . To find $𝑑$ , subtract the first term from the second: $3.5 - 1.5 = 2.0$ . $𝑡 𝑛 = 1.5 + (𝑛 - 1) 2$ $𝑡 𝑛 = 1.5 + 2 𝑛 - 2$ $𝑡 𝑛 = 2 𝑛 - 0.5$

Exercise (Page 183)

Find recursive rules for the APs in the previous exercises.

Solution

A recursive rule defines the first term and then provides a formula to find the next term by adding the common difference ( $𝑑$ ) to the previous term.
For 2, 5, 8, 11... $𝑡 1 = 2$ $𝑡 𝑛 = 𝑡 𝑛 - 1 + 3 f o r 𝑛 \geq 2$
For -5, -1, 3, 7... $𝑡 1 = - 5$ $𝑡 𝑛 = 𝑡 𝑛 - 1 + 4 f o r 𝑛 \geq 2$
For 1/2, 5/2, 9/2, 13/2... $𝑡 1 = 1 / 2$ $𝑡 𝑛 = 𝑡 𝑛 - 1 + 2 f o r 𝑛 \geq 2$
For 1.5, 3.5, 5.5, 7.5... $𝑡 1 = 1.5$ $𝑡 𝑛 = 𝑡 𝑛 - 1 + 2 f o r 𝑛 \geq 2$

Example 5 (Page 183)

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?

Solution

Let's build the sequence of the total fare based on the distance travelled:
After 1 km: Base fee of ₹200 + (1 km × ₹40) = ₹240
After 2 km: Base fee of ₹200 + (2 km × ₹40) = ₹280
After 3 km: Base fee of ₹200 + (3 km × ₹40) = ₹320 The sequence is 240, 280, 320... This is an AP where the first term ( $𝑎$ ) is 240 and the common difference ( $𝑑$ ) is 40. To find the fare after traveling 10 km, we need to find the 10th term in this sequence (where $𝑛 = 10$ ). We can use the formula $𝑡 𝑛 = 𝑎 + (𝑛 - 1) 𝑑$ : $𝑡 10 = 240 + (10 - 1) 40$ $𝑡 10 = 240 + (9) 40$ $𝑡 10 = 240 + 360 = 600$ The total fare for 10 km will be ₹600.

Section 8.5: Sum of the First $𝑛$ Natural Numbers

Think and Reflect (Page 184)

Can the same approach be used to find the sum of $1 + 2 + 3 + . . . + 100$ ?

Solution

Yes, it can. The approach is to write the sum forwards and backwards, then add the two equations together. Let $𝑆 = 1 + 2 + 3 + . . . + 100$ Write it backwards: $𝑆 = 100 + 99 + 98 + . . . + 1$ Add them together pair by pair: $2 𝑆 = (1 + 100) + (2 + 99) + (3 + 98) + . . . + (100 + 1)$ $2 𝑆 = 101 + 101 + 101 + . . . + 101$ (This happens 100 times) $2 𝑆 = 100 \times 101$ $2 𝑆 = 10100$ $𝑆 = 5050$ The sum of the first 100 natural numbers is 5050.

Think and Reflect (Page 185)

Can you use this formula to find $𝑆 20$ , $𝑆 50$ or $𝑆 1000$ ?

Solution

Yes, the formula derived in the text is $𝑆 𝑛 = 𝑛 (𝑛 + 1) 2$ , where $𝑛$ is the number of terms.
For $𝑆 20$ : Substitute $𝑛 = 20$ . $𝑆 20 = 20 (20 + 1) 2 = 20 \times 21 2 = 10 \times 21 = 210$
For $𝑆 50$ : Substitute $𝑛 = 50$ . $𝑆 50 = 50 (50 + 1) 2 = 50 \times 51 2 = 25 \times 51 = 1275$
For $𝑆 1000$ : Substitute $𝑛 = 1000$ . $𝑆 1000 = 1000 (1000 + 1) 2 = 1000 \times 1001 2 = 500 \times 1001 = 500500$

EXERCISE SET 8.2 (Pages 185 - 186)

Question 1

Find the 10th and 26th terms of the AP: 3, 8, 13, 18, ...

Solution

First, identify the starting term ( $𝑎$ ) and the common difference ( $𝑑$ ).
$𝑎 = 3$
$𝑑 = 8 - 3 = 5$ Use the AP formula $𝑡 𝑛 = 𝑎 + (𝑛 - 1) 𝑑$ :
10th term ( $𝑛 = 10$ ): $𝑡 10 = 3 + (10 - 1) 5 = 3 + (9) 5 = 3 + 45 = 48$
26th term ( $𝑛 = 26$ ): $𝑡 26 = 3 + (26 - 1) 5 = 3 + (25) 5 = 3 + 125 = 128$

Question 2

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

Solution

Identify the parameters: $𝑎 = 21$ , and $𝑑 = 18 - 21 = - 3$ . The explicit formula is $𝑡 𝑛 = 21 + (𝑛 - 1) (- 3)$ .
To find which term is -81: Set $𝑡 𝑛 = - 81$ and solve for $𝑛$ . $- 81 = 21 - 3 (𝑛 - 1)$ $- 102 = - 3 (𝑛 - 1)$ $34 = 𝑛 - 1$ $𝑛 = 35$ Since $𝑛$ is a natural number, -81 is the 35th term.
To check if 0 is a term: Set $𝑡 𝑛 = 0$ and solve for $𝑛$ . $0 = 21 - 3 (𝑛 - 1)$ $- 21 = - 3 (𝑛 - 1)$ $7 = 𝑛 - 1$ $𝑛 = 8$ Because $𝑛 = 8$ is a natural number, yes, 0 is a term of this AP. It is the 8th term.

Question 3

Find the $𝑛 𝑡 ℎ$ term of the AP: 11, 8, 5, 2... Write the recursive rule for this AP.

Solution

Identify the parameters: $𝑎 = 11$ , and $𝑑 = 8 - 11 = - 3$ .
Explicit $𝑛 𝑡 ℎ$ term: $𝑡 𝑛 = 𝑎 + (𝑛 - 1) 𝑑$ $𝑡 𝑛 = 11 + (𝑛 - 1) (- 3)$ $𝑡 𝑛 = 11 - 3 𝑛 + 3$ $𝑡 𝑛 = 14 - 3 𝑛$
Recursive rule: The first term is 11, and every subsequent term is found by subtracting 3 from the previous term. $𝑡 1 = 11$ $𝑡 𝑛 = 𝑡 𝑛 - 1 - 3 f o r 𝑛 \geq 2$

Question 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 $𝑎 + 2 𝑑 = 12$ and $𝑎 + 49 𝑑 = 106$ . Solve this pair of linear equations for 'a' and 'd'.)

Solution

Using the hint provided, we have a system of two linear equations:
$𝑎 + 2 𝑑 = 12$
$𝑎 + 49 𝑑 = 106$ Subtract the first equation from the second to eliminate $𝑎$ : $(𝑎 + 49 𝑑) - (𝑎 + 2 𝑑) = 106 - 12$ $47 𝑑 = 94$ $𝑑 = 2$ Now, substitute $𝑑 = 2$ back into the first equation to find $𝑎$ : $𝑎 + 2 (2) = 12$ $𝑎 + 4 = 12$ $𝑎 = 8$ We have $𝑎 = 8$ and $𝑑 = 2$ . Now find the 29th term ( $𝑛 = 29$ ): $𝑡 29 = 𝑎 + (29 - 1) 𝑑$ $𝑡 29 = 8 + 28 (2)$ $𝑡 29 = 8 + 56 = 64$ The 29th term is 64.

Question 5

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

Solution

First, list the 2-digit numbers divisible by 3. The smallest is 12, and the largest is 99. The sequence is 12, 15, 18, ..., 99. This is an AP where $𝑎 = 12$ and $𝑑 = 3$ .
How many numbers? Set the $𝑛 𝑡 ℎ$ term formula equal to 99 to find the number of terms. $99 = 12 + (𝑛 - 1) 3$ $87 = 3 (𝑛 - 1)$ $29 = 𝑛 - 1$ $𝑛 = 30$ There are thirty 2-digit numbers divisible by 3.
What is the sum? We want the sum $12 + 15 + 18 + . . . + 99$ . We can factor out a 3 to use the natural number sum formula taught in the chapter: $3 \times (4 + 5 + 6 + . . . + 33)$ To find the sum of $4$ to $33$ , we find the sum of $1$ to $33$ and subtract the sum of $1$ to $3$ : Sum of 1 to 33 = $33 \times 34 2 = 561$ Sum of 1 to 3 = $3 \times 4 2 = 6$ Sum of 4 to 33 = $561 - 6 = 555$ Finally, multiply back by the 3 we factored out: $3 \times 555 = 1665$ The sum of all 2-digit numbers divisible by 3 is 1665.

Question 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?

Solution

This forms an AP where the first year's salary ( $𝑎$ ) is 5,00,000 and the yearly increment ( $𝑑$ ) is 20,000. We want to find the term ( $𝑛$ ) that equals 7,00,000. $𝑡 𝑛 = 𝑎 + (𝑛 - 1) 𝑑$ $7, 00, 000 = 5, 00, 000 + (𝑛 - 1) 20, 000$ $2, 00, 000 = (𝑛 - 1) 20, 000$ Divide both sides by 20,000: $10 = 𝑛 - 1$ $𝑛 = 11$ His income reaches ₹7,00,000 in his 11th year of working, which means it happened after 10 full years of working (and receiving 10 increments).

Question 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?

Solution

The sequence of marbles per row is 1, 2, 3, 4, ..., 25. To find the total number of marbles, we need to find the sum of the first 25 natural numbers. We use the formula $𝑆 𝑛 = 𝑛 (𝑛 + 1) 2$ where $𝑛 = 25$ : $𝑆 25 = 25 (25 + 1) 2$ $𝑆 25 = 25 \times 26 2$ $𝑆 25 = 25 \times 13 = 325$ The child uses 325 marbles in all.

Section 8.6: Geometric Progressions

Think and Reflect (Page 186)

Can you predict the number of squares in Stages 5 and 6 of the pattern? In Stages 10, 11 and 12? In Stage 20? At any stage? How is this different from the growing pattern in Fig. 8.3?

Fig. 8.3: Growing pattern of squares (AP with d = 4)

Solution

The sequence of green squares in Fig. 8.6 is 3, 6, 12, 24. This sequence grows by multiplying the previous term by 2 (the common ratio is 2). The explicit formula for the $𝑛 𝑡 ℎ$ stage is $𝑡 𝑛 = 3 \times 2 𝑛 - 1$ .
Stage 5: $3 \times 2 5 - 1 = 3 \times 24 = 3 \times 16 = 48$ squares.
Stage 6: $3 \times 2 6 - 1 = 3 \times 25 = 3 \times 32 = 96$ squares.
Stage 10: $3 \times 2 10 - 1 = 3 \times 29 = 3 \times 512 = 1536$ squares.
Stage 11: $3 \times 2 11 - 1 = 3 \times 210 = 3 \times 1024 = 3072$ squares.
Stage 12: $3 \times 2 12 - 1 = 3 \times 211 = 3 \times 2048 = 6144$ squares.
Stage 20: $3 \times 2 20 - 1 = 3 \times 219 = 3 \times 524288 = 1572864$ squares.
At any stage $𝑛$ : $3 \times 2 𝑛 - 1$ squares.
Difference from Fig. 8.3: The pattern in Fig. 8.3 was an Arithmetic Progression (AP) where a constant number (4) was added at each step. The pattern here is a Geometric Progression (GP) where the previous number is multiplied by a constant factor (2) at each step.

Example 6 (Page 187)

Is 1, 2, 4, 8, 16, ... a geometric progression? If so, what is the common ratio?

Solution

To check if it is a geometric progression, find the ratio of consecutive terms by dividing a term by the one before it:
$2 \div 1 = 2$
$4 \div 2 = 2$
$8 \div 4 = 2$
$16 \div 8 = 2$ Since the ratio is constantly 2, yes, it is a geometric progression. The common ratio ( $𝑟$ ) is 2.

Example 7 (Page 187)

Is 1, 3, 9, 27, 81, ... a geometric progression? If so, what is the common ratio?

Solution

Check the ratio between consecutive terms:
$3 \div 1 = 3$
$9 \div 3 = 3$
$27 \div 9 = 3$ Since the ratio is constantly 3, yes, it is a geometric progression. The common ratio ( $𝑟$ ) is 3.

Example 8 (Page 187)

Is 1, -1, 1, -1, 1, ... a geometric progression? If so, what is the common ratio?

Solution

Check the ratio between consecutive terms:
$- 1 \div 1 = - 1$
$1 \div - 1 = - 1$ Since the ratio is constantly -1, yes, it is a geometric progression. The common ratio ( $𝑟$ ) is -1.

Example 9 (Page 187)

Check whether the sequence 5, 15/4, 45/16, 135/64, ... is a geometric progression and find its $𝑛 𝑡 ℎ$ term.

Solution

First, calculate the ratio between consecutive pairs to check if it's a GP:
$154 \div 5 = 154 \times 15 = 34$
$4516 \div 154 = 4516 \times 415 = 34$
$13564 \div 4516 = 13564 \times 1645 = 34$ The ratio is constant, so it is a geometric progression with first term $𝑎 = 5$ and common ratio $𝑟 = 34$ . The explicit formula for the $𝑛 𝑡 ℎ$ term is $𝑡 𝑛 = 𝑎 \times 𝑟 𝑛 - 1$ . Substituting our values: $𝑡 𝑛 = 5 \times (34) 𝑛 - 1$ .

Exercise (Page 188)

Check whether the following sequences are geometric progressions and find their $𝑛 𝑡 ℎ$ terms. (i) 2, 10, 50, 250, ... (ii) 4, 8/3, 16/9, 32/27, ... (iii) 3, -3/2, 3/4, -3/8, ...

Solution

(i) 2, 10, 50, 250, ... Check: $10 \div 2 = 5$ ; $50 \div 10 = 5$ ; $250 \div 50 = 5$ . Since the ratio is constant, yes, it is a GP. $𝑛 𝑡 ℎ$ term: $𝑎 = 2, 𝑟 = 5$ . The formula is $𝑡 𝑛 = 2 \times 5 𝑛 - 1$ .
(ii) 4, 8/3, 16/9, 32/27, ... Check: $83 \div 4 = 812 = 23$ ; $169 \div 83 = 169 \times 38 = 23$ . Yes, it is a GP. $𝑛 𝑡 ℎ$ term: $𝑎 = 4, 𝑟 = 23$ . The formula is $𝑡 𝑛 = 4 \times (23) 𝑛 - 1$ .
(iii) 3, -3/2, 3/4, -3/8, ... Check: $- 3 2 \div 3 = - 12$ ; $34 \div - 3 2 = 34 \times - 2 3 = - 12$ . Yes, it is a GP. $𝑛 𝑡 ℎ$ term: $𝑎 = 3, 𝑟 = - 12$ . The formula is $𝑡 𝑛 = 3 \times (- 12) 𝑛 - 1$ .

Exercise (Page 188)

Can you find a recursive rule for the formula $𝑡 𝑛 = 3 \times 10 𝑛 - 1$ that generates the geometric progression 3, 30, 300, 3000, ...?

Solution

A recursive rule defines the first term and then gives the rule to get to the next term from the previous one. The first term ( $𝑡 1$ ) is 3. The sequence multiplies by 10 each time (the common ratio is 10). The recursive rule is: $𝑡 1 = 3$ , $𝑡 𝑛 = 10 \times 𝑡 𝑛 - 1$ for $𝑛 \geq 2$ .

Think and Reflect (Page 189)

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 $𝑛 𝑡 ℎ$ 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 $𝑛 𝑡 ℎ$ stage. What happens to this area as $𝑛$ , the number of stages, goes on increasing?

Fig. 8.7: Stages of the Sierpiński triangle (black triangles)

Solution

(a) Counting directly from Fig. 8.7: Stage 0: 1 black triangle Stage 1: 3 black triangles Stage 2: 9 black triangles Stage 3: 27 black triangles
(b) Each stage multiplies the number of black triangles by 3. Stage 4: $27 \times 3 = 81$ black triangles Stage 5: $81 \times 3 = 243$ black triangles
(c) Notice that the sequence starts at $𝑛 = 0$ (Stage 0). The terms are $30, 31, 32$ , etc. The explicit rule for the number of black triangles at stage $𝑛$ is $𝑡 𝑛 = 3 𝑛$ .
(d) At each stage, the central triangle is removed, leaving 3 of the 4 equal parts, which means the remaining area is multiplied by $34$ each time. Stage 1 area: $1 \times 34 = 34$ Stage 2 area: $34 \times 34 = (34) 2 = 916$ Stage 3 area: $916 \times 34 = (34) 3 = 2764$ Stage 4 area: $2764 \times 34 = (34) 4 = 81256$ Stage 5 area: $81256 \times 34 = (34) 5 = 2431024$ Rule for the $𝑛 𝑡 ℎ$ stage: $𝑠 𝑛 = (34) 𝑛$ * What happens as $𝑛$ increases: The fraction is raised to higher and higher powers. Because the base ( $34$ ) is less than 1, the total area of the black region continually decreases and gets closer and closer to 0.

Example 10 (Page 192)

A ball is dropped from a height of 24 feet above the ground. Each time the ball bounces up to $34$ of its previous height. (a) Can you write the sequence of numbers obtained from the heights attained by the ball in five successive bounces? (b) How many bounces are required for the ball to remain below a height of $16$ of the original height from which it was dropped?

Solution

(a) The sequence is formed by multiplying the previous height by $34$ (or 0.75). 1st bounce: $24 \times 0.75 = 18$ feet 2nd bounce: $18 \times 0.75 = 13.5$ feet 3rd bounce: $13.5 \times 0.75 = 10.125$ feet 4th bounce: $10.125 \times 0.75 = 7.59375$ feet * 5th bounce: $7.59375 \times 0.75 = 5.6953125$ feet (The book truncates this to 5.695)
(b) First, calculate the target height: $16$ of 24 feet is $24 \times 16 = 4$ feet. We need to find when the sequence drops below 4 feet. We left off at the 5th bounce (approx 5.695 ft). Let's continue: 6th bounce: $5.6953125 \times 0.75 \approx 4.271$ feet 7th bounce: $4.27148 . . . \times 0.75 \approx 3.203$ feet After the 7th bounce, the ball reaches a maximum height of ~3.2 feet, which is below 4 feet. Therefore, 7 bounces are required.

EXERCISE SET 8.3 (Pages 193 - 194)

Question 1

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

Solution

We know the common ratio ( $𝑟$ ) is 2 and the 8th term ( $𝑡 8$ ) is 192. We can find the 12th term ( $𝑡 12$ ) without finding the first term by simply multiplying the 8th term by the common ratio 4 more times (since $12 - 8 = 4$ ). $𝑡 12 = 𝑡 8 \times 𝑟 4$ $𝑡 12 = 192 \times 24$ $𝑡 12 = 192 \times 16 = 3072$ The 12th term is 3072.

Question 2

Find the 10th and $𝑛 𝑡 ℎ$ terms of the GP: 5, 25, 125, ...

Solution

The first term ( $𝑎$ ) is 5. The common ratio ( $𝑟$ ) is $25 \div 5 = 5$ .
$𝑛 𝑡 ℎ$ term: The explicit formula is $𝑡 𝑛 = 𝑎 \times 𝑟 𝑛 - 1$ . $𝑡 𝑛 = 5 \times 5 𝑛 - 1$ By exponent rules, adding the powers of 5 gives $𝑡 𝑛 = 5 𝑛$ .
10th term: Substitute $𝑛 = 10$ into our simplified formula. $𝑡 10 = 510$ (Which calculates out to 9,765,625).

Question 3

A sequence is given by the recursive rule $𝑡 1 = 2$ , $𝑡 𝑛 + 1 = 3 𝑡 𝑛 - 2$ for $𝑛 \geq 1$ . Which term of the sequence is 730?

Solution

We use the recursive rule to calculate terms one by one until we hit 730:
$𝑡 1 = 2$
$𝑡 2 = 3 (2) - 2 = 6 - 2 = 4$
$𝑡 3 = 3 (4) - 2 = 12 - 2 = 10$
$𝑡 4 = 3 (10) - 2 = 30 - 2 = 28$
$𝑡 5 = 3 (28) - 2 = 84 - 2 = 82$
$𝑡 6 = 3 (82) - 2 = 246 - 2 = 244$
$𝑡 7 = 3 (244) - 2 = 732 - 2 = 730$ 730 is the 7th term of the sequence.

Question 4

Which term of the GP: 2, 6, 18, ... is 4374? Write the explicit formula as well as the recursive formula for the $𝑛 𝑡 ℎ$ term.

Solution

The first term ( $𝑎$ ) is 2, and the common ratio ( $𝑟$ ) is $6 \div 2 = 3$ .
Explicit formula: $𝑡 𝑛 = 2 \times 3 𝑛 - 1$
Recursive formula: The first term is 2, and each subsequent term multiplies the previous term by 3. $𝑡 1 = 2$ , and $𝑡 𝑛 = 3 \times 𝑡 𝑛 - 1$ for $𝑛 \geq 2$ .
Which term is 4374? Set the explicit formula equal to 4374. $2 \times 3 𝑛 - 1 = 4374$ $3 𝑛 - 1 = 2187$ We know that $37 = 2187$ , so: $𝑛 - 1 = 7$ $𝑛 = 8$ 4374 is the 8th term.

Question 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?

Solution

The bounce heights form a geometric sequence where you multiply by 0.6 (or 60%).
Initial drop: 80 m
1st bounce height: $80 \times 0.6 = 48$ m
2nd bounce height: $48 \times 0.6 = 28.8$ m
3rd bounce height: $28.8 \times 0.6 = 17.28$ m
4th bounce height: $17.28 \times 0.6 = 10.368$ m
5th bounce height: $10.368 \times 0.6 = 6.2208$ m (i) After the 5th bounce, the ball reaches a height of 6.2208 metres.
(ii) To find total distance traveled by the time it hits the ground for the 6th time, we must add the initial drop plus the "up and down" distance of the first 5 bounces:
Initial drop: 80 m
1st bounce (up + down): $48 \times 2 = 96$ m
2nd bounce (up + down): $28.8 \times 2 = 57.6$ m
3rd bounce (up + down): $17.28 \times 2 = 34.56$ m
4th bounce (up + down): $10.368 \times 2 = 20.736$ m
5th bounce (up + down): $6.2208 \times 2 = 12.4416$ m Total distance = $80 + 96 + 57.6 + 34.56 + 20.736 + 12.4416 = 301.3376$ metres.

Question 6

Which term of the sequence $2 \sqrt 2$ , is 128?

Solution

(Note: The textbook has a typo here and cut off parts of the sequence. Based on standard geometric progressions that result in 128, it is safe to assume the sequence intended is $2, 2 \sqrt 2, 4, . . .$ where the first term $𝑎 = 2$ and the common ratio $𝑟 = \sqrt 2$ .) Let $𝑎 = 2$ and $𝑟 = \sqrt 2$ . Set the explicit formula equal to 128: $𝑡 𝑛 = 𝑎 \times 𝑟 𝑛 - 1$ $128 = 2 \times (\sqrt 2) 𝑛 - 1$ Divide by 2: $64 = (\sqrt 2) 𝑛 - 1$ We know that $64$ is $26$ . We also know that $2$ is $(\sqrt 2) 2$ . Therefore, $64 = ((\sqrt 2) 2) 6 = (\sqrt 2) 12$ . $(\sqrt 2) 12 = (\sqrt 2) 𝑛 - 1$ $𝑛 - 1 = 12$ $𝑛 = 13$ Assuming the sequence $2, 2 \sqrt 2, 4 . . .$ , 128 is the 13th term.

Question 7

Fig. 8.12 shows Stages 0 to 3 of the Sierpiński square carpet... 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 $𝑛 𝑡 ℎ$ 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 $𝑛 𝑡 ℎ$ stage. What happens to this area as $𝑛$ , the number of stages, goes on increasing?

Fig. 8.1: Triangular numbers as triangular arrays of dots (terms 1–7)

Fig. 8.12: Stages 0–3 of the Sierpiński square carpet

Solution

(i) Counting the red squares from the figure's logic:
Stage 0: 1 red square
Stage 1: 8 red squares (a $3 \times 3$ grid with the center removed)
Stage 2: 64 red squares (each of the 8 squares is replaced by 8 smaller ones)
Stage 3: 512 red squares ( $64 \times 8$ ) (ii) Every stage multiplies the number of squares by 8.
Stage 4: $512 \times 8 = 4096$ red squares
Stage 5: $4096 \times 8 = 32768$ red squares (iii) Let $𝑡 𝑛$ be the number of red squares at stage $𝑛$ .
Explicit formula: $𝑡 𝑛 = 8 𝑛$
Recursive formula: $𝑡 0 = 1$ , and $𝑡 𝑛 = 8 \times 𝑡 𝑛 - 1$ for $𝑛 \geq 1$ (iv) For the area, each stage removes $19$ of the area from the previous stage, meaning $89$ of the area remains.
Stage 1 area: $1 \times 89 = 89$
Stage 2 area: $89 \times 89 = (89) 2 = 6481$
Stage 3 area: $(89) 3 = 512729$
Stage 4 area: $(89) 4 = 40966561$
Stage 5 area: $(89) 5 = 3276859049$
Explicit formula for area ( $𝐴 𝑛$ ): $𝐴 𝑛 = (89) 𝑛$
Recursive formula for area: $𝐴 0 = 1$ , and $𝐴 𝑛 = 𝐴 𝑛 - 1 \times 89$ for $𝑛 \geq 1$
As $𝑛$ increases: Because we repeatedly multiply by a fraction less than 1 ( $89$ ), the area of the red region steadily decreases, getting closer and closer to 0.

END-OF-CHAPTER EXERCISES (Pages 194 - 195)

Question 1

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

Solution

We use the AP formula $𝑡 𝑛 = 𝑎 + (𝑛 - 1) 𝑑$ to set up two equations:
$𝑎 + 10 𝑑 = 38$
$𝑎 + 15 𝑑 = 73$ Subtract the first equation from the second: $(𝑎 + 15 𝑑) - (𝑎 + 10 𝑑) = 73 - 38$ $5 𝑑 = 35$ $𝑑 = 7$ Substitute $𝑑 = 7$ back into the first equation to find $𝑎$ : $𝑎 + 10 (7) = 38$ $𝑎 + 70 = 38$ $𝑎 = - 32$ Now find the 31st term ( $𝑛 = 31$ ): $𝑡 31 = 𝑎 + 30 𝑑$ $𝑡 31 = - 32 + 30 (7)$ $𝑡 31 = - 32 + 210 = 178$ The 31st term is 178.

Question 2

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

Solution

The 7th term is $𝑎 + 6 𝑑$ , and the 5th term is $𝑎 + 4 𝑑$ . The difference between them is 12: $(𝑎 + 6 𝑑) - (𝑎 + 4 𝑑) = 12$ $2 𝑑 = 12$ $𝑑 = 6$

We know the third term is 16: $𝑎 + 2 𝑑 = 16$ $𝑎 + 2 (6) = 16$ $𝑎 + 12 = 16$ $𝑎 = 4$

The first term is 4, and the common difference is 6. The AP is: 4, 10, 16, 22, 28, ...

Question 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.)

Solution

Smallest 3-digit number divisible by 7: $100 \div 7 \approx 14.28$ . The next whole number is 15. $15 \times 7 = 105$ . So, $𝑎 = 105$ .
Largest 3-digit number divisible by 7: $999 \div 7 \approx 142.71$ . Round down to 142. $142 \times 7 = 994$ . So, the last term $𝑡 𝑛 = 994$ .
Common difference ( $𝑑$ ): 7. Use the AP formula to find $𝑛$ : $𝑡 𝑛 = 𝑎 + (𝑛 - 1) 𝑑$ $994 = 105 + (𝑛 - 1) 7$ $889 = 7 (𝑛 - 1)$ $127 = 𝑛 - 1$ $𝑛 = 128$ There are 128 three-digit numbers divisible by 7.

Question 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.)

Solution

Smallest multiple of 4 strictly after 10: 12. So, $𝑎 = 12$ .
Largest multiple of 4 strictly before 250: 248. So, $𝑡 𝑛 = 248$ .
Common difference ( $𝑑$ ): 4. Use the AP formula to find $𝑛$ : $248 = 12 + (𝑛 - 1) 4$ $236 = 4 (𝑛 - 1)$ $59 = 𝑛 - 1$ $𝑛 = 60$ There are 60 multiples of 4 between 10 and 250.

Question 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.

Solution

Let the GP be $𝑎, 𝑎 𝑟, 𝑎 𝑟 2, 𝑎 𝑟 3, 𝑎 𝑟 4 . . .$
Sum of first two terms is 4: $𝑎 + 𝑎 𝑟 = 4 \Rightarrow 𝑎 (1 + 𝑟) = 4$
Fifth term is 4 times the third: $𝑎 𝑟 4 = 4 (𝑎 𝑟 2)$ Divide the second equation by $𝑎 𝑟 2$ (assuming $𝑎 \neq 0$ and $𝑟 \neq 0$ ): $𝑟 2 = 4$ $𝑟 = 2 o r 𝑟 = - 2$ Now find ' $𝑎$ ' for both cases using the first equation ( $𝑎 (1 + 𝑟) = 4$ ):
Case 1 ( $𝑟 = 2$ ): $𝑎 (1 + 2) = 4 \Rightarrow 3 𝑎 = 4 \Rightarrow 𝑎 = 4 / 3$ . The GP is: 4/3, 8/3, 16/3, 32/3...
Case 2 ( $𝑟 = - 2$ ): $𝑎 (1 - 2) = 4 \Rightarrow - 𝑎 = 4 \Rightarrow 𝑎 = - 4$ . The GP is: -4, 8, -16, 32...

Question 6 (*)

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

Solution

We need to find an AP where the sum is 100, the common difference $𝑑 = 1$ , and the first term $𝑎 \geq 1$ . The sum formula for an AP is: $𝑆 𝑛 = 𝑛 2 [2 𝑎 + (𝑛 - 1) 𝑑]$ $100 = 𝑛 2 [2 𝑎 + (𝑛 - 1) 1]$ $200 = 𝑛 (2 𝑎 + 𝑛 - 1)$ Here, $𝑛$ and $(2 𝑎 + 𝑛 - 1)$ must be factors of 200. Since their difference is $(2 𝑎 - 1)$ , which is always an odd number, one factor must be odd and the other even. The factors of 200 are: 1, 2, 4, 5, 8, 10, 20, 25, 40, 50, 100, 200. The odd factors are 1, 5, and 25. Because $𝑛$ is the number of terms, let's test the odd factors as $𝑛$ or the result of $(2 𝑎 + 𝑛 - 1)$ :
If the odd factor is 5: Let $𝑛 = 5$ . Then $2 𝑎 + 5 - 1 = 200 / 5 \Rightarrow 2 𝑎 + 4 = 40 \Rightarrow 2 𝑎 = 36 \Rightarrow 𝑎 = 18$ . This gives 5 consecutive numbers starting at 18: 18, 19, 20, 21, 22. (Sum = 100)
If the odd factor is 25: Because $𝑛$ cannot be 25 (if $𝑛 = 25$ , then $2 𝑎 + 24 = 8$ , which means $2 𝑎 = - 16$ , not possible for natural numbers), we let the other bracket be the odd factor. So, $2 𝑎 + 𝑛 - 1 = 25$ and $𝑛 = 200 / 25 = 8$ . $2 𝑎 + 8 - 1 = 25 \Rightarrow 2 𝑎 + 7 = 25 \Rightarrow 2 𝑎 = 18 \Rightarrow 𝑎 = 9$ . This gives 8 consecutive numbers starting at 9: 9, 10, 11, 12, 13, 14, 15, 16. (Sum = 100) (Note: Using 1 as the odd factor just gives the single number 100, which isn't a "sum of consecutive numbers".) There are two ways: $18 + 19 + 20 + 21 + 22$ and $9 + 10 + 11 + 12 + 13 + 14 + 15 + 16$ .

Question 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 $2 𝑛 𝑑$ hour, 4th hour and $𝑛 𝑡 ℎ$ hour?

Solution

This is a GP where the starting amount ( $𝑎$ ) is 30, and it doubles ( $𝑟 = 2$ ) every hour.
End of 1st hour: $30 \times 2 = 60$
End of 2nd hour: $60 \times 2 = 120$
End of 4th hour: $30 \times 24 = 30 \times 16 = 480$
End of $𝑛 𝑡 ℎ$ hour: We multiply 30 by 2 for exactly $𝑛$ hours. The formula is $30 \times 2 𝑛$ .

Question 8 (*)

The sum of the $4 𝑡 ℎ$ and $8 𝑡 ℎ$ terms of an AP is 24 and the sum of the $6 𝑡 ℎ$ and $10 𝑡 ℎ$ terms is 44. Find the first three terms of the AP.

Solution

Write the given information as equations using $𝑡 𝑛 = 𝑎 + (𝑛 - 1) 𝑑$ :
$𝑡 4 + 𝑡 8 = 24 \Rightarrow (𝑎 + 3 𝑑) + (𝑎 + 7 𝑑) = 24 \Rightarrow 2 𝑎 + 10 𝑑 = 24$ . Divide by 2: $𝑎 + 5 𝑑 = 12$
$𝑡 6 + 𝑡 10 = 44 \Rightarrow (𝑎 + 5 𝑑) + (𝑎 + 9 𝑑) = 44 \Rightarrow 2 𝑎 + 14 𝑑 = 44$ . Divide by 2: $𝑎 + 7 𝑑 = 22$ Subtract the first simplified equation from the second: $(𝑎 + 7 𝑑) - (𝑎 + 5 𝑑) = 22 - 12$ $2 𝑑 = 10$ $𝑑 = 5$ Substitute $𝑑 = 5$ into $𝑎 + 5 𝑑 = 12$ : $𝑎 + 5 (5) = 12$ $𝑎 + 25 = 12$ $𝑎 = - 13$ The first term is -13, and the common difference is 5. The first three terms are: -13, -8, -3.

Question 9 (*)

Find the smallest value of $𝑛$ such that the sum of the first $𝑛$ natural numbers is greater than 1,000.

Solution

The sum of the first $𝑛$ natural numbers is given by $𝑛 (𝑛 + 1) 2$ . We need this to be $> 1000$ : $𝑛 (𝑛 + 1) 2 > 1000$ $𝑛 (𝑛 + 1) > 2000$ We are looking for two consecutive numbers that multiply to just over 2000. Let's estimate using square roots. $\sqrt 2000 \approx 44.7$ . Let's test $𝑛 = 44$ : $44 \times 45 = 1980$ (This is less than 2000). Let's test $𝑛 = 45$ : $45 \times 46 = 2070$ (This is greater than 2000). The smallest value of $𝑛$ is 45.

Question 10 (*)

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

Solution

The first term ( $𝑎$ ) is 2. The common ratio ( $𝑟$ ) is $8 \div 2 = 4$ .
Explicit formula: $𝑡 𝑛 = 2 \times 4 𝑛 - 1$
Recursive formula: $𝑡 1 = 2$ , and $𝑡 𝑛 = 4 \times 𝑡 𝑛 - 1$ for $𝑛 \geq 2$
Which term is 131072? $2 \times 4 𝑛 - 1 = 131072$ Divide by 2: $4 𝑛 - 1 = 65536$ Keep multiplying by 4 to find the power: $41 = 4, 42 = 16, 43 = 64, 44 = 256, 45 = 1024, 46 = 4096, 47 = 16384, 48 = 65536$ . $𝑛 - 1 = 8$ $𝑛 = 9$ 131072 is the 9th term.

Question 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.

Solution

To make the math easier when a product of 3 GP terms is given, let the first three terms be $𝑎 𝑟, 𝑎, 𝑎 𝑟$ .
Product is -1: $(𝑎 𝑟) (𝑎) (𝑎 𝑟) = - 1 \Rightarrow 𝑎 3 = - 1 \Rightarrow 𝑎 = - 1$ Now we know the middle term is -1. The terms are $- 1 𝑟, - 1, - 𝑟$ .
Sum is 13/12: $- 1 𝑟 - 1 - 𝑟 = 1312$ Move the -1 over by adding it to both sides: $- 1 𝑟 - 𝑟 = 1312 + 1$ $- 1 - 𝑟 2 𝑟 = 2512$ Cross-multiply to solve for $𝑟$ : $- 12 - 12 𝑟 2 = 25 𝑟$ $12 𝑟 2 + 25 𝑟 + 12 = 0$ Factor the quadratic equation (find numbers that multiply to $12 \times 12 = 144$ and add to 25. The numbers are 16 and 9): $12 𝑟 2 + 16 𝑟 + 9 𝑟 + 12 = 0$ $4 𝑟 (3 𝑟 + 4) + 3 (3 𝑟 + 4) = 0$ $(4 𝑟 + 3) (3 𝑟 + 4) = 0$ This gives two possible common ratios: $𝑟 = - 34$ or $𝑟 = - 43$ .
The terms: If $𝑟 = - 34$ , the terms ( $- 1 𝑟, - 1, - 𝑟$ ) are: $43, - 1, 34$ . If $𝑟 = - 43$ , the terms are: $34, - 1, 43$ .

Question 12 (*)

If the $4 𝑡 ℎ, 10 𝑡 ℎ$ and $16 𝑡 ℎ$ terms of a GP are x, y and z respectively, prove that x, y, z are in GP.

Solution

Let the starting term of the GP be $𝐴$ and the common ratio be $𝑅$ .
$𝑥 = 𝐴 \times 𝑅 3$
$𝑦 = 𝐴 \times 𝑅 9$
$𝑧 = 𝐴 \times 𝑅 15$ For $𝑥, 𝑦, 𝑧$ to form their own GP, the ratio between consecutive terms must be identical. Let's check:
$𝑦 𝑥 = 𝐴 \times 𝑅 9 𝐴 \times 𝑅 3 = 𝑅 6$
$𝑧 𝑦 = 𝐴 \times 𝑅 15 𝐴 \times 𝑅 9 = 𝑅 6$ Because $𝑦 𝑥 = 𝑧 𝑦 = 𝑅 6$ , the terms $𝑥, 𝑦, 𝑧$ have a constant common ratio of $𝑅 6$ . Therefore, $𝑥, 𝑦, 𝑧$ are in a geometric progression.

Question 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.

Solution

Let the first three terms be $𝑎, 𝑎 𝑟, 𝑎 𝑟 2$ .
Sum: $𝑎 + 𝑎 𝑟 + 𝑎 𝑟 2 = 26 \Rightarrow 𝑎 (1 + 𝑟 + 𝑟 2) = 26$
Sum of squares: $𝑎 2 + 𝑎 2 𝑟 2 + 𝑎 2 𝑟 4 = 364 \Rightarrow 𝑎 2 (1 + 𝑟 2 + 𝑟 4) = 364$ There is a known algebraic identity: $1 + 𝑟 2 + 𝑟 4 = (1 + 𝑟 + 𝑟 2) (1 - 𝑟 + 𝑟 2)$ . Substitute this into the second equation: $𝑎 2 (1 + 𝑟 + 𝑟 2) (1 - 𝑟 + 𝑟 2) = 364$ We can split the $𝑎 2$ up to use our first equation: $[𝑎 (1 + 𝑟 + 𝑟 2)] \times [𝑎 (1 - 𝑟 + 𝑟 2)] = 364$ Since $𝑎 (1 + 𝑟 + 𝑟 2) = 26$ , we get: $26 \times [𝑎 (1 - 𝑟 + 𝑟 2)] = 364$ $𝑎 (1 - 𝑟 + 𝑟 2) = 14$ Now we have a system of two simplified equations: A) $𝑎 + 𝑎 𝑟 + 𝑎 𝑟 2 = 26$ B) $𝑎 - 𝑎 𝑟 + 𝑎 𝑟 2 = 14$ Subtract B from A: $2 𝑎 𝑟 = 12 \Rightarrow 𝑎 𝑟 = 6 \Rightarrow 𝑎 = 6 𝑟$ Add B to A: $2 𝑎 + 2 𝑎 𝑟 2 = 40 \Rightarrow 2 𝑎 (1 + 𝑟 2) = 40 \Rightarrow 𝑎 (1 + 𝑟 2) = 20$ Substitute $𝑎 = 6 𝑟$ into the second result: $(6 𝑟) (1 + 𝑟 2) = 20$ $6 + 6 𝑟 2 = 20 𝑟$ $6 𝑟 2 - 20 𝑟 + 6 = 0$ Divide by 2: $3 𝑟 2 - 10 𝑟 + 3 = 0$ Factor the quadratic: $(3 𝑟 - 1) (𝑟 - 3) = 0$ The common ratio $𝑟 = 3$ or $𝑟 = 13$ .
If $𝑟 = 3$ : $𝑎 = 63 = 2$ . The terms are 2, 6, 18.
If $𝑟 = 13$ : $𝑎 = 6 1 / 3 = 18$ . The terms are 18, 6, 2.

Question 14 (*)

Suppose $𝑃 1 = 1, 𝑃 2 = 2$ and for $𝑛 > 2$ , $𝑃 𝑛 = 𝑃 1 + 𝑃 2 + . . . + 𝑃 𝑛 - 1 + 1$ . Find the values of $𝑃 1, 𝑃 2, . . ., 𝑃 8$ . Can you find a simpler recursive formula for $𝑃 𝑛$ ? Can you give an explicit formula?

Solution

Calculate the terms using the rule (add all previous terms together, then add 1):
$𝑃 1 = 1$ (given)
$𝑃 2 = 2$ (given)
$𝑃 3 = 1 + 2 + 1 = 4$
$𝑃 4 = 1 + 2 + 4 + 1 = 8$
$𝑃 5 = 1 + 2 + 4 + 8 + 1 = 16$
$𝑃 6 = 1 + 2 + 4 + 8 + 16 + 1 = 32$
$𝑃 7 = 1 + 2 + 4 + 8 + 16 + 32 + 1 = 64$
$𝑃 8 = 1 + 2 + 4 + 8 + 16 + 32 + 64 + 1 = 128$ The values $𝑃 1$ to $𝑃 8$ are: 1, 2, 4, 8, 16, 32, 64, 128.
Simpler recursive formula: Notice that starting from $𝑛 = 2$ , every term is simply double the previous term. $𝑃 𝑛 = 2 \times 𝑃 𝑛 - 1$ for $𝑛 \geq 2$ .
Explicit formula: Because it doubles every time starting from $𝑃 1 = 1$ , the terms are powers of 2. $𝑃 𝑛 = 2 𝑛 - 1$ .

Question 15 (*)

Suppose $𝑊 1 = 1, 𝑊 2 = 2$ and for $𝑛 > 2, 𝑊 𝑛 = 𝑊 1 + 𝑊 2 + . . . + 𝑊 𝑛 - 2 + 2$ . Find the values of $𝑊 1, 𝑊 2, . . ., 𝑊 8$ . Do you recognise this sequence?

Solution

Calculate the terms using the rule (add all previous terms except the one immediately prior, then add 2):
$𝑊 1 = 1$ (given)
$𝑊 2 = 2$ (given)
$𝑊 3 = 𝑊 1 + 2 = 1 + 2 = 3$
$𝑊 4 = 𝑊 1 + 𝑊 2 + 2 = 1 + 2 + 2 = 5$
$𝑊 5 = 𝑊 1 + 𝑊 2 + 𝑊 3 + 2 = 1 + 2 + 3 + 2 = 8$
$𝑊 6 = 𝑊 1 + 𝑊 2 + 𝑊 3 + 𝑊 4 + 2 = 1 + 2 + 3 + 5 + 2 = 13$
$𝑊 7 = 𝑊 1 + 𝑊 2 + 𝑊 3 + 𝑊 4 + 𝑊 5 + 2 = 1 + 2 + 3 + 5 + 8 + 2 = 21$
$𝑊 8 = 𝑊 1 + 𝑊 2 + 𝑊 3 + 𝑊 4 + 𝑊 5 + 𝑊 6 + 2 = 1 + 2 + 3 + 5 + 8 + 13 + 2 = 34$ The values of $𝑊 1$ to $𝑊 8$ are: 1, 2, 3, 5, 8, 13, 21, 34. Yes, we recognise this sequence! This is the Virahānka-Fibonacci sequence introduced earlier in the chapter, where each term is the sum of the two preceding terms.

Section 8.1: Introduction to Sequences

Section 8.2: Explicit Rule for a Sequence

Section 8.3: Recursive Rule for a Sequence

EXERCISE SET 8.1 (Pages 179 - 180)

Section 8.4: Arithmetic Progressions

Section 8.5: Sum of the First 𝑛 Natural Numbers

EXERCISE SET 8.2 (Pages 185 - 186)

Section 8.6: Geometric Progressions

EXERCISE SET 8.3 (Pages 193 - 194)

END-OF-CHAPTER EXERCISES (Pages 194 - 195)

Section 8.5: Sum of the First $𝑛$ Natural Numbers