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

F R E E S T U D Y M AT E R I A L F O R E V E R Y S T U D E N T

C L A S S 9 · M AT H S

NCERT Solutions

Chapter 8: Predicting What
Comes Next: Exploring
Sequences and Progressions

NCERT Textbook — Ganita Manjari

BOOK PAGES SECTIONS QUESTIONS MEDIUM

174 – 196 20 62 English

Solutions, notes, sample papers & more at 64 pages

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

CLASS 9 · MATHS · GANITA MANJARI

NCERT Solutions — Chapter 8: Predicting What Comes
Next: Exploring Sequences and Progressions
Chapter 8 separates two things that look alike but are not: spotting a pattern and proving a rule. A sequence is
an ordered list of terms; a rule for it can be explicit (tn written in terms of the position n) or recursive (tn
written in terms of earlier terms). Two families of sequences are then singled out — the arithmetic
progression, built by repeated addition, and the geometric progression, built by repeated multiplication —
and the sum 1 + 2 + … + n is derived, not merely quoted.

TEXTBOOK BOOK PAGES

Ganita Manjari (Class 9) 174 – 196

SECTIONS QUESTIONS

20 62

MEDIUM

English

Think and Reflect — Page 174
Section 8.1 Introduction to Sequences

THINK AND REFLECT

Q1 The sequences printed just above this question are

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)

Can you describe the pattern in each of the above sequences? Can you predict the
next few numbers in these sequences?

Each of the four sequences is described by what happens from one term to the next, and that
alone fixes every later term.

Page 1 of 64

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

SEQUENCE PATTERN IN THE GAPS NEXT THREE RULE FOR THE NTH
TERMS TERM

1, 2, 3, 4, 5, 6, … add 1 each time 7, 8, 9 tn = n

1, 3, 5, 7, 9, 11, … add 2 each time 13, 15, 17 tn = 2n – 1

1, 3, 6, 10, 15, 21, gaps are 2, 3, 4, 5, 6 — they grow 28, 36, 45 tn = n(n + 1)/2
… by 1

1, 4, 9, 16, 25, 36, gaps are 3, 5, 7, 9, 11 — the odd 49, 64, 81 t n = n2
… numbers

Check the last two rows against the printed terms: 21 + 7 = 28, 28 + 8 = 36, 36 + 9 = 45; and 36 +
13 = 49, 49 + 15 = 64, 64 + 17 = 81. The gaps themselves form a sequence, and that is what
makes prediction possible.

Why it happens: the triangular numbers are the running totals of 1, 2, 3, 4, …, so
the gap between the (n – 1)th and the nth is exactly n. The square numbers are the
running totals of 1, 3, 5, 7, …, so the gap there is the nth odd number, 2n – 1. In both
cases you are not guessing — you are reading off a structure the sequence already
has.

Check it yourself: 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28, and 1 + 3 + 5 + 7 + 9 + 11 + 13 = 49.
The two lists of running totals really do give the triangular and square numbers.

In-text Questions — Pages 174–175
Section 8.1 Introduction to Sequences

Q1 Can you think of other finite sequences that you see in your daily life?

Yes — a finite sequence is any ordered list that stops, and everyday life is full of them.

The overs bowled in a T20 innings: 1, 2, 3, …, 20.
The Indian currency notes in circulation: 10, 20, 50, 100, 200, 500.
The marks of a student in six subjects, written in the order of the report card.
The number of days in the twelve months of 2026: 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31.
The stairs of a flight numbered from bottom to top, or the platform numbers at a railway
station.

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

Why it happens: two conditions make a list a sequence — the items must be in a
definite order, and each item must have a position number. A jumble of prices in a
shop is not a sequence; the same prices arranged from cheapest to costliest is. A
sequence is finite exactly when that list of positions runs out, as it does at over 20 or
at month 12.

Tip: the three dots … are what separate a finite sequence from an infinite one. 6, 12,
24, 48, 96 has five terms and ends; 6, 12, 24, 48, 96, … never does.

Q2 Can you draw the patterns for the next two terms of the sequence?

1 3 6 10 15

Fig. 8.1 — the first five triangular numbers, each shown as a triangular array of dots.

Fig. 8.1 shows the first five triangular numbers 1, 3, 6, 10, 15 as triangular arrays of dots. The
next two terms are 21 and 28, drawn by adding one more row each time.

Page 3 of 64

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

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

Q3 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 — a square array of dots divided into L-shaped bands of alternating colour.

Fig. 8.2 fills a square array of dots with L-shaped bands (gnomons), each one a different colour.
The first band is 1 dot, the second 3, the third 5, and so on — so building a square is the same
as adding up odd numbers.

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

A 6 × 6 array split into L-shaped bands of 1, 3, 5, 7, 9 and 11 dots. The grey lines mark where one
square ends and the next begins.

1 = 12

1 + 3 = 4 = 22

1 + 3 + 5 = 9 = 32

1 + 3 + 5 + 7 = 16 = 42

1 + 3 + 5 + 7 + 9 + 11 = 36 = 62

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

Why it happens: to enlarge an n × n square into an (n + 1) × (n + 1) square you must
add one new column of n dots, one new row of n dots, and one corner dot — that is
n + n + 1 = 2n + 1 dots, which is always an odd number. So each step up the square
numbers costs exactly one more odd number, and the sum of the first n odd
numbers must be n2. The picture is not an illustration of the fact; it is the proof.

Try This: add the first 10 odd numbers without adding them. The answer is 102 =
100.

In-text Questions — Page 176
Section 8.1 Introduction to Sequences

Q1 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? (Hint: The first term is 1. The second
term is 1 + 4 = 5, the third term is 1 + 4 + 7 = 12, and so on.)

The gap between consecutive terms is a constant 3, so the next four terms are 16, 19, 22, 25.

1, 4, 7, 10, 13, 16, 19, 22, 25, 28, …

nth term: tn = 1 + (n – 1) × 3 = 3n – 2

Now build the running totals. Take the first ten terms 1, 4, 7, 10, 13, 16, 19, 22, 25, 28 and add
them one at a time:

N 1 2 3 4 5 6 7 8 9 10

term of 1, 4, 7, … 1 4 7 10 13 16 19 22 25 28

running total Sn 1 5 12 22 35 51 70 92 117 145

So the required sequence is 1, 5, 12, 22, 35, 51, 70, 92, 117, 145.

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

Why it happens: the gaps of the new sequence are 4, 7, 10, 13, … — the original
sequence from its second term on. That is exactly what a running total does: the
jump from Sn–1 to Sn is the nth term you just added. Because those gaps themselves
increase by a constant 3, the totals grow faster and faster, and the new sequence is
not an AP even though the one it came from is.

Did you know? 1, 5, 12, 22, 35, 51, … are the pentagonal numbers. Their nth term
is n(3n – 1)/2 — check n = 4: 4 × 11 ÷ 2 = 22.

Q2 Exercise: Can you write t5, t6, t7 and t8 for the sequence of triangular numbers?

The triangular numbers are 1, 3, 6, 10, 15, 21, 28, 36, … , so

t5 = 1 + 2 + 3 + 4 + 5 = 15

t6 = 15 + 6 = 21

t7 = 21 + 7 = 28

t8 = 28 + 8 = 36

The same values come straight out of the explicit rule tn = n(n + 1)/2:

t5 = (5 × 6)/2 = 15 t6 = (6 × 7)/2 = 21

t7 = (7 × 8)/2 = 28 t8 = (8 × 9)/2 = 36

Why it happens: the subscript in tn is the position, not the value. Writing t7 = 28
states that the seventh triangular number is 28 — it does not say anything about the
number 7 itself. Keeping position and value apart is what lets you later solve
equations like tn = 300 to find a position.

Think and Reflect — Page 176

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

Section 8.1 Introduction to Sequences
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THINK AND REFLECT

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

Section 8.2 Explicit Rule for a Sequence

THINK AND REFLECT

Q1 Why is it useful to have an explicit formula for the nth term of a sequence?

Because it lets you reach any term directly, without first working out all the terms before it.

With un = 2n – 1:

u300 = 2 × 300 – 1 = 599 — in one step, not 299 steps

It is also the only convenient way to answer the reverse question. To ask "is 137 in this
sequence, and where?" you simply solve the equation:

2n – 1 = 137

2n = 138

n = 69 — so 137 is the 69th odd number

Why it happens: an explicit rule is a function of the position. Because n appears
on its own in the formula, substituting a value of n is a single calculation, and the
formula can equally be read backwards as an equation in n. A recursive rule cannot
do this: to reach the 300th term it must first know the 299th, which needs the 298th,
and so on all the way down to the start.

Tip: when you solve tn = k, always look at the n you get. Only a natural number
counts — a fraction or a negative value means k is not a term at all.

Q2 Can you find the rule describing the nth term of the sequence of square numbers?

The sequence is 1, 4, 9, 16, 25, 36, … , and the rule is

tn = n2

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

Test it against the printed terms: t1 = 12 = 1, t2 = 22 = 4, t3 = 9, t4 = 16, t5 = 25, t6 = 36. All six
agree.

Why it happens: the chapter has already shown that each square number is the
sum of the odd numbers up to that position — 1 + 3 + 5 + … + (2n – 1). Fig. 8.2 shows
why that sum is n2: the L-shaped bands assemble into an n × n array. So tn = n2 is
not just a formula that fits the first six terms; it has a reason behind it, and that
reason guarantees it for every n.

Check it yourself: the 20th square number should be 400, and 1 + 3 + 5 + … + 39
(twenty odd numbers) is indeed 400.

In-text Questions — Pages 177–178
Section 8.2 Explicit Rule for a Sequence

Q1 Exercise: Using the explicit rule un = 2n – 1, find the 53rd term, the 108th term, and
the 1170th term of the odd number sequence.

Substitute the position straight into the rule.

u53 = 2 × 53 – 1 = 106 – 1 = 105

u108 = 2 × 108 – 1 = 216 – 1 = 215

u1170 = 2 × 1170 – 1 = 2340 – 1 = 2339

Why it happens: the odd numbers are the even numbers shifted down by one. The
nth even number is 2n, so the nth odd number is 2n – 1. Nothing here depends on
knowing the earlier terms — this is exactly the advantage of an explicit rule, and it is
why the 1170th term costs no more work than the 53rd.

Check it yourself: 105, 215 and 2339 are all odd, and each is one less than an even
number that is twice its position.

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

Q2 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?

There are visible features, but no rule that predicts the next term. Compare the gaps:

2, 3, 5, 7, 11, 13, 17, 19, 23, 29

gaps: 1, 2, 2, 4, 2, 4, 2, 4, 6

The gaps do not settle to a constant (so this is not an AP), and the ratios 3/2, 5/3, 7/5, … are not
constant either (so it is not a GP). The next few primes — 31, 37, 41, 43, 47 — have to be found
by testing for divisibility, not produced by a formula.
What can be said with certainty:

2 is the only even prime; every later prime is odd.
Apart from 2 and 3, every prime is 1 more or 1 less than a multiple of 6.
The primes never run out — there are infinitely many of them.

Why it happens: a prime is defined by what it is not — it has no divisor other than 1
and itself. That is a condition about all the numbers below it, not a step forward from
the previous term. So the definition gives you no way to move from one prime to the
next, which is precisely why no explicit or recursive formula for the primes is known.
This is the chapter's own example of a perfectly well-defined sequence with no clear
regularity in its rule.

Did you know? the gaps between primes get arbitrarily large — somewhere beyond
100! there is a run of 99 consecutive numbers with no prime in it at all.

Q3 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) Substitute each position into tn = 3n – 7.

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

t1 = 3 – 7 = –4 t2 = 6 – 7 = –1 t3 = 9 – 7 = 2

t12 = 36 – 7 = 29 t18 = 54 – 7 = 47 t50 = 150 – 7 = 143

(ii) Solve tn = 332.

3n – 7 = 332

3n = 339

n = 113

So 332 is the 113th term.
(iii) Solve tn = 557 the same way.

3n – 7 = 557

3n = 564

n = 564 ÷ 3 = 188

188 is a natural number, so yes — 557 is a term of the sequence, the 188th term. Check: 3 ×
188 – 7 = 564 – 7 = 557.

Why it happens: every term of this sequence is 7 less than a multiple of 3, so a
number k is a term exactly when k + 7 is divisible by 3 — that is, when k leaves
remainder 2 on division by 3. Both 332 and 557 pass that test (332 = 3 × 110 + 2 and
557 = 3 × 185 + 2), so both belong. A number such as 558, which leaves remainder 0,
could never appear however far you continued.

Tip: a "why or why not" question is really asking for the test, not just the verdict.
Here the test is: does the equation 3n – 7 = k give a natural number n?

In-text Questions — Page 179

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

Section 8.3 Recursive Rule for a Sequence
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Q1
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[Virahānka–Fibonacci sequence 1, 2, 3, 5, 8, 13, 21, 34, …] Can you write the next two
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terms of this sequence?

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Use the recursive rule Vn = Vn–1 + Vn–2, adding the last two terms each time.
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V8 = 34, V7 = 21 a

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V9 = V8 + V7 = 34 + 21 = 55

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V10 = V9 + V8 = 55 + 34 = 89
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a l sequence continues 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, …
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Why it happens: this rule reaches back two steps, not one, so it needs two starting
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Exercise Set 8.1 — Pagesgl179–180 ase
Section 8.3 Recursive Rule for a Sequence
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Find the first five terms of the sequence in which the nth term
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Class 9 Maths Chapter 8 Predicting What Comes Next: Exploring Sequences and Progressions AglaSem · NCERT Solutions

t1 = 3 – 4 = –1 t2 = 6 – 4 = 2 t3 = 9 – 4 = 5 t4 = 12 – 4 = 8 t5 = 15 – 4 = 11

First five terms: –1, 2, 5, 8, 11

(ii) tn = 2 – 5n

t1 = 2 – 5 = –3 t2 = 2 – 10 = –8 t3 = 2 – 15 = –13 t4 = –18 t5 = –23

First five terms: –3, –8, –13, –18, –23

(iii) tn = n2 – 2n + 3

t1 = 1 – 2 + 3 = 2 t 2 = 4 – 4 + 3 = 3 t 3 = 9 – 6 + 3 = 6

t4 = 16 – 8 + 3 = 11 t5 = 25 – 10 + 3 = 18

First five terms: 2, 3, 6, 11, 18

Why it happens: the first two rules are linear in n, so their terms change by the
same amount every step — by +3 in (i) and by –5 in (ii). Both are arithmetic
progressions, one rising and one falling. The third rule is quadratic, so its gaps are 1,
3, 5, 7 — themselves growing — and it is not an AP. Reading the shape of the
formula tells you the shape of the sequence before you compute a single term.

Check it yourself: for (iii), n2 – 2n + 3 = (n – 1)2 + 2, so the terms are 2 more than the
square numbers 0, 1, 4, 9, 16 — giving 2, 3, 6, 11, 18 again.

Q2 Find the 10th and 15th terms of the sequence tn = 5n – 3 for n ≥ 1.

t10 = 5 × 10 – 3 = 50 – 3 = 47

t15 = 5 × 15 – 3 = 75 – 3 = 72

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

Why it happens: tn = 5n – 3 is an AP with first term t1 = 2 and common difference 5.
So the 15th term should sit five steps of 5 beyond the 10th: 47 + 5 × 5 = 72, which is
exactly what the direct substitution gives. Two independent routes to the same value
is a good check.

Q3 Determine whether 97 and 172 are terms of the sequence tn = 5n – 3 for n ≥ 1.

Solve tn = 97 and tn = 172 and see whether n comes out a natural number.

5n – 3 = 97 → 5n = 100 → n = 20

5n – 3 = 172 → 5n = 175 → n = 35

Both values of n are natural numbers, so 97 is the 20th term and 172 is the 35th term — both
are terms of the sequence.

Why it happens: every term of this sequence is 3 less than a multiple of 5, so it ends
in 2 or 7. Both 97 and 172 end in 7 and 2, and adding 3 gives 100 and 175, each
divisible by 5. Had we tested 98, we would have got 5n = 101 and n = 20.2 — not a
position, so not a term.

Tip: "is k a term?" is always the same question in disguise: does the equation tn = k
have a natural-number solution?

Q4 Which term of the sequence tn = 5n – 3 for n ≥ 1 is 607?

5n – 3 = 607

5n = 610

n = 122

So 607 is the 122nd term. Check: 5 × 122 – 3 = 610 – 3 = 607. ✓

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

Why it happens: solving for n reverses the rule. The formula 5n – 3 takes a position
and returns a value; the equation takes a value and returns its position. This two-way
use is what makes an explicit rule so much more powerful than a list of terms.

Q5 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?

Start at t1 = –5 and add 3 each time.

t2 = –5 + 3 = –2

t3 = –2 + 3 = 1

t4 = 1 + 3 = 4

t5 = 4 + 3 = 7

First five terms: –5, –2, 1, 4, 7

The terms rise by a constant 3, so this is an AP with a = –5 and d = 3. Convert to an explicit rule
and then solve:

tn = –5 + (n – 1) × 3 = 3n – 8

3n – 8 = 52

3n = 60

n = 20

Yes — 52 is a term, the 20th term. Check: 3 × 20 – 8 = 52. ✓

Why it happens: the recursive rule alone would force you to compute nineteen
terms before reaching 52. Turning it into the explicit rule tn = 3n – 8 first is the
efficient move — and the two forms describe the same sequence, since t1 = 3 – 8 = –
5 and each step in n adds 3.

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

Tip: the rule is written tn+1 = tn + 3 here, not tn = tn–1 + 3. Both say the same thing —
the term after any term is 3 more — only the label has shifted.

Q6 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.

Each term is the sum of the three terms before it.

T 4 = T3 + T2 + T1 = 4 + 2 + 1 = 7

T5 = T4 + T3 + T2 = 7 + 4 + 2 = 13

T6 = T5 + T4 + T3 = 13 + 7 + 4 = 24

T7 = T6 + T5 + T4 = 24 + 13 + 7 = 44

T8 = T7 + T6 + T5 = 44 + 24 + 13 = 81

The sequence is 1, 2, 4, 7, 13, 24, 44, 81, …

Why it happens: because the rule looks back three steps, it needs three starting
terms, and it can only begin at n = 4. Notice how the pattern of the Virahānka rule
extends: reaching back two terms gives 1, 2, 3, 5, 8, …; reaching back three gives 1, 2,
4, 7, 13, … The number of starting values a recursive rule needs is exactly the
number of steps it reaches back.

Did you know? a sequence built by adding the previous three terms is called a
tribonacci sequence, by analogy with the two-term rule.

Think and Reflect — Page 180

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

Section 8.4 Arithmetic Progressions
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THINK AND REFLECT

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Q1
gl Can
10, 11 and 12? In Stage 20? At any stage?

co m
e m . ag
g l as
a

co m
e m.
m l as
m .co a g
l a se
a g Stage 1 Stage 2 Stage 3 Stage 4

m
Fig. 8.3 — Stages 1 to 4 of a growing pattern of squares.
a s
m.co agl
l a se
ANSWER a g
Yes — and once the rule is justified, every one of these follows by substitution. Each stage adds
co m
4 squares, one to each arm of the X, so the counts are 1, 5, 9, 13, 17, 21, …
m .
m as e
.co a g l
se m
g l a
a
se m
com g l a
m . a
ase
agl

co m
m .
m ase
.co a g l
se m
g l a Stage 4 → 13 squares Stage 5 → 17 squares
a c
m .
m
Going from Stage 4 to Stage 5 lengthens each of the four diagonal arms by one square: 13 + 4 = 17.
a s e
e m . co agl
g l as
a

co m
m .
m as e
.co


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

Stage 1: 1

Stage n: 1 + (n – 1) × 4 = 4n – 3

STAGE 5 6 10 11 12 20 N

Number of squares 17 21 37 41 45 77 4n – 3

Stage 10: 4 × 10 – 3 = 37 Stage 11: 41 Stage 12: 45
Stage 20: 4 × 20 – 3 = 77

Why it happens: spotting "add 4" only lets you crawl forward one stage at a time.
The formula comes from seeing how many times the 4 has been added: at Stage n
the single central square has been joined by 4 squares on (n – 1) occasions, so the
count is 1 + (n – 1) × 4. Expanding gives 4n – 3, and it is this step — from a pattern in
the picture to an argument about the picture — that turns a guess into a rule you
can trust at Stage 200.

Check it yourself: Stage 1 must give 1. Putting n = 1 in 4n – 3 gives 4 – 3 = 1. A
formula that fails at n = 1 is the wrong formula.

Think and Reflect — Page 181
Section 8.4 Arithmetic Progressions

THINK AND REFLECT

Q1 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?

The test is single and mechanical: subtract each term from the one after it. If every one of those
differences is the same number, the sequence is an AP; one exception is enough to rule it out.

Page 20 of 64

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

SEQUENCE DIFFERENCES AP? JUSTIFICATION

1, 2, 3, 4, 5, … 1, 1, 1, … Yes a = 1, d = 1

1, 3, 5, 7, 9, … 2, 2, 2, … Yes a = 1, d = 2

1, 4, 7, 10, 13, … 3, 3, 3, … Yes a = 1, d = 3

1, 5, 9, 13, 17, … 4, 4, 4, … Yes a = 1, d = 4 (Fig. 8.3)

11, 7, 3, –1, –5, … –4, –4, –4, … Yes a = 11, d = –4; d may be negative

–7, –3, 1, 5, 9, … 4, 4, 4, … Yes a = –7, d = 4; terms may be negative

1, 3, 6, 10, 15, … (triangular) 2, 3, 4, 5 No the differences themselves change

1, 4, 9, 16, 25, … (square) 3, 5, 7, 9 No differences are the odd numbers, not a
constant

1, 1/2, 1/3, 1/4, … –1/2, –1/6, – No the gaps shrink towards 0
1/12

2, 3, 5, 7, 11, 13, … (primes) 1, 2, 2, 4, 2 No no regularity at all

1, 2, 3, 5, 8, 13, … 1, 1, 2, 3, 5 No the differences are the sequence itself again
(Virahānka)

1, 5, 13, 29, … (un = 2un–1 + 4, 8, 16 No differences double each time
3)

Why it happens: "increasing" and "arithmetic" are not the same thing. The
triangular and square numbers rise steadily, yet neither is an AP, because being an
AP is a statement about the gaps, not about the terms. Equivalently, an explicit rule
of the form tn = an + b (linear in n) always gives an AP, and any rule with n2, or with n
in an exponent, never does.

Tip: checking one pair of terms proves nothing. 1, 3, 6 begins with a difference of 2
but is not an AP — you must check that every difference matches.

In-text Questions — Pages 182–183

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

Section 8.4.1 Visualising an AP

Q1 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, … — differences 5 – 2 = 3, 8 – 5 = 3, 11 – 8 = 3. Constant, so it is an AP with a = 2, d
= 3.

tn = 2 + (n – 1) × 3 = 2 + 3n – 3 = 3n – 1

Check: t1 = 2, t2 = 5, t3 = 8, t4 = 11 ✓

(ii) –5, –1, 3, 7, … — differences –1 – (–5) = 4, 3 – (–1) = 4, 7 – 3 = 4. Constant, so it is an AP with a
= –5, d = 4.

tn = –5 + (n – 1) × 4 = –5 + 4n – 4 = 4n – 9

Check: t1 = –5, t2 = –1, t3 = 3, t4 = 7 ✓

What the plot shows: plotting (1, 2), (2, 5), (3, 8), (4, 11) — and likewise (1, –5), (2, –1), (3, 3), (4, 7)
— the points lie exactly on a straight line, just as in Fig. 8.4.

AP 2, 5, 8, 11, 14 GP 3, 6, 12, 24, 48

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

Left: an AP — equal steps up for equal steps right, so the points are collinear. Right: a GP for
comparison — the same points bend upwards.

Why it happens: moving one step to the right raises y by d every single time, and a
graph whose rise is the same for every unit of run is a straight line. In the language
of Chapter 2, tn = a + (n – 1)d is a linear expression in n, and its graph is a line of
slope d. So "AP" and "collinear points" are two descriptions of one fact.

Q2 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) Here a = 1/2 and d = 5/2 – 1/2 = 4/2 = 2 (also 9/2 – 5/2 = 2 and 13/2 – 9/2 = 2).

tn = 1/2 + (n – 1) × 2

= 1/2 + 2n – 2

= 2n – 3/2, or equivalently tn = (4n – 3)/2

Check: t1 = 1/2, t2 = 5/2, t3 = 9/2, t4 = 13/2 ✓

(ii) Here a = 1.5 and d = 3.5 – 1.5 = 2.

tn = 1.5 + (n – 1) × 2

= 1.5 + 2n – 2

= 2n – 0.5

Check: t1 = 1.5, t2 = 3.5, t3 = 5.5, t4 = 7.5 ✓

Why it happens: both progressions have the same common difference, d = 2, and
differ only in their first term — 0.5 against 1.5 — so the second is the first shifted up
by 1. That is the point of the formula tn = a + (n – 1)d: fractions, decimals and
negative numbers all go into it unchanged, because nothing in the derivation
assumed the terms were whole numbers. Only the size of the step matters, never
what kind of number the terms are.

Page 23 of 64

as e
a g l
Class 9 Maths Chapter 8 Predicting What Comes Next: Exploring Sequences and Progressions AglaSem · NCERT Solutions

co m
m.
Tip: always simplify to a single expression in n at the end. (4n – 3)/2 is easier to

m as e
l
substitute into than 1/2 + (n – 1) × 2.

m .co a g
l a se
a g
Q3 Exercise: Find recursive rules for the APs in the previous exercises.

co m
m . ag
l a se
agthings: the first term, and the instruction "add d".
A recursive rule for an AP needs two

co m
m.
AP A D RECURSIVE RULE

m as e
.co
2, 5, 8, 11, … 2 3
a g l
t1 = 2, tn = tn–1 + 3 for n ≥ 2

se m
g l a –5 4
a
–5, –1, 3, 7, … t1 = –5, tn = tn–1 + 4 for n ≥ 2

s
1/2, 5/2, 9/2, 13/2, … 1/2 2 t1 = 1/2, tn = tn–1 + 2 for n ≥ 2

m a
m .co2 agl
ase
1.5, 3.5, 5.5, 7.5, … 1.5 t1 = 1.5, tn = tn–1 + 2 for n ≥ 2

agl
Why it happens: the general AP has explicit rule tn = a + (n – 1)d and recursive rule

co m
t1 = a, tn = tn–1 + d. They agree because a + (n – 1)d minus a + (n – 2)d is exactly d.
m .
o m l a se
Notice the last two rows: identical recursive steps, different starting terms — which
.c the starting term can never be left out. Without it the
a g rule describes infinitely
m
semany different sequences.
is why
a
agl
se m
com
Tip: the condition n ≥ 2 matters. The rule tn = tn–1 + d is meaningless at n = 1,
g l a
m . a
ase
because there is no t0.

agl

co m
Think and Reflect — Page 184
m .
m ase
.co a g l
se m
g l a
a c
m .
m a s e
e m . co agl
g l as
a

co m
m .
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.co


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

Section 8.5 Sum of the First n Natural Numbers

THINK AND REFLECT

Q1 Can the same approach be used to find the sum of 1 + 2 + 3 + … + 100?

Yes — the method does not care how long the list is. Write the sum forwards, write it backwards
underneath, and add column by column.

S = 1 + 2 + 3 + … + 98 + 99 + 100
S = 100 + 99 + 98 + … + 3 + 2 + 1

2S = 101 + 101 + 101 + … + 101 (101 added 100 times)

2S = 100 × 101 = 10100

S = 5050

The same answer follows from the formula Sn = n(n + 1)/2 with n = 100: (100 × 101)/2 = 5050.

Why it happens: pairing the two lines works because as you move right the first line
grows by 1 while the second shrinks by 1, so every column carries the same total.
That total is always (first term + last term) = 1 + 100 = 101, and there are as many
columns as there are terms. This is why the trick generalises to any n at all — and it
is the argument that produces the formula, rather than merely checking it.

Did you know? the second half of Āryabhaṭa's rule in the Āryabhaṭīya is exactly this:
take the average of the first and last terms, then multiply by the number of terms.
Here that is (1 + 100)/2 × 100 = 50.5 × 100 = 5050.

Think and Reflect — Page 185

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

Section 8.5 Sum of the First n Natural Numbers

THINK AND REFLECT

Q1 Can you use this formula to find S20, S50 or S1000?

Yes. Substitute into Sn = n(n + 1)/2.

S20 = (20 × 21)/2 = 420/2 = 210

S50 = (50 × 51)/2 = 2550/2 = 1275

S1000 = (1000 × 1001)/2 = 1001000/2 = 500500

2 × (1+2+3+4+5+6) = 7 × 6 = 42
The picture behind the formula (Fig. 8.5). Red circles count 1 + 2 + 3 + 4 + 5 + 6 = 21; green circles
count the same again; together they fill a 7 × 6 rectangle.

Page 26 of 64

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

Why it happens: two copies of the staircase 1, 2, 3, …, n interlock into a rectangle n
+ 1 wide and n tall. That rectangle holds n(n + 1) circles, and it is exactly twice the
sum, so Sn = n(n + 1)/2. Because the argument is about how the two staircases fit, it
holds for n = 20 and n = 1000 just as it does for the n = 6 you can draw.

Check it yourself: one of n and n + 1 is always even, so n(n + 1)/2 is always a whole
number — as a count of circles must be.

Q2 Let us revisit the sequence tn of triangular numbers 1, 3, 6, 10, 15, … shown in Fig.
8.1. Note that the nth term of this sequence is the sum of the first n natural
numbers. Thus tn = n(n + 1)/2. Can you use this to find the 10th, 17th and 80th
triangular numbers?

1 3 6 10 15

Fig. 8.1 — the first five triangular numbers, each shown as a triangular array of dots.

t10 = (10 × 11)/2 = 110/2 = 55

t17 = (17 × 18)/2 = 306/2 = 153

t80 = (80 × 81)/2 = 6480/2 = 3240

Check the first of these directly: 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 = 55. ✓

Page 27 of 64

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

Why it happens: the two ideas in this section are one idea. A triangular array with n
rows has 1 dot in the top row, 2 in the next, and so on down to n — so counting its
dots is adding the first n natural numbers. Once you know that sum is n(n + 1)/2, you
know every triangular number without drawing a single dot. Drawing the 80th array
would need 3240 dots; the formula needs one multiplication and one division.

Try This: add two consecutive triangular numbers, say 55 + 66. You get 121 = 112.
Two triangles always fit together into a square.

Exercise Set 8.2 — Pages 185–186
Section 8.5 Sum of the First n Natural Numbers

Q1 Find the 10th and 26th terms of the AP: 3, 8, 13, 18, ….

First term a = 3; common difference d = 8 – 3 = 5 (and 13 – 8 = 5, 18 – 13 = 5).

tn = a + (n – 1)d = 3 + (n – 1) × 5 = 5n – 2

t10 = 5 × 10 – 2 = 48

t26 = 5 × 26 – 2 = 128

Why it happens: the 26th term is 16 steps beyond the 10th, so it must exceed it by
16 × 5 = 80. And 48 + 80 = 128 — the two answers are consistent with each other,
which is a stronger check than re-substituting into the same formula twice.

Q2 Which term of the AP : 21, 18, 15, … is – 81? Also, is 0 a term of this AP? Give reasons
for your answer.

Here a = 21 and d = 18 – 21 = –3, so the AP falls.

Page 28 of 64

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

co m
em.
tn = 21 + (n – 1)(–3) = 21 – 3n + 3 = 24 – 3n
m l as
m
24 – 3n = –81
.co a g
l a se
g
3n = 105
an = 35

. c om ag
a s
So –81 is the 35th term. Check: 24 – 3 × 35em
= 24 – 105 = –81. ✓
Is 0 a term? Solve 24 – 3n = 0.
agl

co m
m.
3n = 24

m as e
n=8
.co a g l
a s em
a l — 0 is a term, the 8th term, because n = 8 is a natural number. Listing confirms it: 21, 18,
gYes
15, 12, 9, 6, 3, 0, –3, …

om a s
. c
Why it happens: every term of thismAP is a multiple of 3 (21 is, and subtracting 3 agl
preserves that). 0 is a multiplela
se
ag
of 3 and lies below the first term, so the falling
sequence must land on it — and the equation says exactly where. A number like 10
is not a multiple of 3, so 24 – 3n = 10 would give n = 14/3, not a position, and 10
co m
m .
e
could never appear.
m l as
.co
m a negative common difference is perfectly normal. Do not force d to be positive a g
a s eTip:
agl by subtracting the wrong way round.

se m
com g l a
m . a
e
s 11, 8, 5, 2 … Write the recursive rule for this AP.
aAP:
Q3
a g l
Find the nth term of the

co m
.

e m
as
a = 11 and d = 8 – 11 = –3 (also 5 – 8 = –3, 2 – 5 = –3).
m l
.co
m t = 11 + (n – 1)(–3) a g
l a se
ag n

.c
= 11 – 3n + 3
s e m
m a
. co agl
= 14 – 3n
e m
g l as
Check: t1 = 11, t2 = 8, t3 = 5, t4 = 2 ✓
a

co m
m .
m as e
.co


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

Recursive rule:

t1 = 11, tn = tn–1 – 3 for n ≥ 2

Why it happens: the two rules answer different questions. "What is the 40th term?"
is a job for the explicit rule — 14 – 120 = –106, in one line. "What comes next?" is a
job for the recursive rule — subtract 3. They must agree, and they do: 14 – 3n minus
14 – 3(n – 1) equals –3, which is precisely the step the recursive rule takes.

Q4 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’.)

Write both facts using tn = a + (n – 1)d. The last term is the 50th, so n = 50 gives a + 49d.

a + 2d = 12 (1)

a + 49d = 106 (2)

Subtract (1) from (2) — this eliminates a:

47d = 94

d=2

From (1): a + 4 = 12, so a = 8

Now the 29th term:

t29 = a + 28d = 8 + 28 × 2 = 8 + 56 = 64

Why it happens: an AP is completely pinned down by two numbers, a and d. So any
two facts about it give two linear equations in a and d — and a pair of linear
equations in two unknowns is what Chapter 2 taught you to solve. Subtracting is the
efficient move here because a appears with coefficient 1 in both equations.

Page 30 of 64

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

Check it yourself: with a = 8 and d = 2 the AP is 8, 10, 12, 14, … Its 3rd term is 12 ✓
and its 50th is 8 + 98 = 106 ✓.

Q5 How many 2-digit numbers are divisible by 3? What is the sum of all these 2-digit
numbers?

The 2-digit multiples of 3 run from 12 to 99, and they form an AP with d = 3.

12, 15, 18, …, 99

tn = 12 + (n – 1) × 3 = 3n + 9

3n + 9 = 99

3n = 90

n = 30

So there are 30 such numbers. For the sum, pair the ends as in Section 8.5:

S = 12 + 15 + … + 96 + 99

S = 99 + 96 + … + 15 + 12

2S = 111 + 111 + … + 111 (111 added 30 times)

2S = 30 × 111 = 3330

S = 1665

Why it happens: the reversal trick from Section 8.5 was never really about 1, 2, 3, …
— it works for any AP, because when one line rises by d the other falls by d, so every
column has the same total (first term + last term). Here that total is 12 + 99 = 111,
and there are 30 columns. In words: sum = (number of terms) × (average of first and
last term) = 30 × 55.5 = 1665.

Check it yourself: another route — the 2-digit multiples of 3 are 3 × 4, 3 × 5, …, 3 ×
33, so the sum is 3 × (4 + 5 + … + 33) = 3 × (S33 – S3) = 3 × (561 – 6) = 3 × 555 = 1665. ✓

Page 31 of 64

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

Q6 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?

His annual salaries form an AP with a = ₹5,00,000 and d = ₹20,000.

tn = 500000 + (n – 1) × 20000

500000 + (n – 1) × 20000 = 700000

(n – 1) × 20000 = 200000

n – 1 = 10

n = 11

His salary is ₹7,00,000 in his 11th year of work — that is, after 10 years of increments.

Why it happens: the first year carries no increment, so the number of increments is
always one less than the year number. Reading (n – 1) = 10 as "10 raises have
happened" is the whole content of the question; answering "11 years" to "after how
many years" would silently count the starting year as a year of increase.

Check it yourself: ₹5,00,000 + 10 × ₹20,000 = ₹5,00,000 + ₹2,00,000 = ₹7,00,000. ✓

Q7 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?

The total is 1 + 2 + 3 + … + 25 — the sum of the first 25 natural numbers.

S25 = n(n + 1)/2 with n = 25

= (25 × 26)/2

= 650/2

= 325 marbles

Page 32 of 64

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

Why it happens: the arrangement is literally the triangular array of Fig. 8.1 with 25
rows, so the count is the 25th triangular number. Pairing gives the same answer
without the formula: 1 + 25 = 26, 2 + 24 = 26, …, 12 + 14 = 26 — that is 12 pairs worth
26 each, plus the unpaired middle marble row of 13. So 12 × 26 + 13 = 312 + 13 =
325. ✓

Tip: the last row has 25 marbles, not 325. The question asks for the total, so it is a
sum, not a term.

Think and Reflect — Page 186

Page 33 of 64

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

Section 8.6 Geometric Progressions
co m
em.
m as
THINK AND REFLECT

.co a g l
a s emyou predict the number of squares in Stages 5 and 6 of the pattern? In Stages
a
Q1
gl Can
10, 11 and 12? In Stage 20? At any stage? How is this different from the growing

m
pattern in Fig. 8.3?

. co ag
e m
g l as
a

co m
m.
Stage 1

m as e
.co
Stage 2
a g l
se m
g l a
a Stage 3

m a s
m.co agl
l a se
a g
m
.co
Stage 4

se m
com g l a
. a
Fig. 8.6 — Stages 1 to 4 of a growing pattern of squares.

m
ase
agl
se m
com g l a
m . a
ase
agl

co m
m .
m l a se Stage 4
.co
Stage 1 Stage 2 Stage 3
m ag
l a se
ag Fig. 8.3 — Stages 1 to 4 of a growing pattern of squares.

.c
s e m
m a
m . co agl
a se

Each stage of Fig. 8.6 is a g l
a rectangle 3 squares wide whose height doubles: 1, 2, 4, 8 rows. So the
counts double rather than grow by a fixed amount.

com
m .
m as e
.co


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

3, 6, 12, 24, 48, 96, …

t1 = 3, t2 = 3 × 2, t3 = 3 × 22, t4 = 3 × 23

tn = 3 × 2n–1

STAGE 5 6 10 11 12 20 N

Number of squares 48 96 1536 3072 6144 15,72,864 3 × 2n–1

Stage 10: 3 × 29 = 3 × 512 = 1536

Stage 20: 3 × 219 = 3 × 524288 = 15,72,864

How it differs from Fig. 8.3:

FIG. 8.3 FIG. 8.6

Step from one stage to the next add 4 multiply by 2

Sequence 1, 5, 9, 13, 17, … 3, 6, 12, 24, 48, …

Constant that describes it common difference d = 4 common ratio r = 2

nth term 4n – 3 3 × 2n–1

Type AP GP

At Stage 20 77 15,72,864

Why it happens: repeated addition puts n in the formula as a plain multiplier, so the
terms grow in proportion to n. Repeated multiplication puts n in the exponent, and
an exponent compounds — each doubling acts on everything already accumulated.
That is why two patterns which look similar at Stage 4 (13 against 24) are worlds
apart at Stage 20 (77 against more than fifteen lakh).

Check it yourself: 3 × 2n–1 must give 3 at n = 1. And 20 = 1, so it does.

In-text Questions — Page 187

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

Section 8.6 Geometric Progressions

Q1 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,
…

For a GP the test is on the ratios of consecutive terms, not the differences. Every consecutive
pair must give the same r.
(i) 2, 10, 50, 250, …

10 ÷ 2 = 5, 50 ÷ 10 = 5, 250 ÷ 50 = 5

Constant, so it is a GP with a = 2, r = 5

tn = 2 × 5n–1 (t1 = 2, t2 = 10, t3 = 50, t4 = 250 ✓)

(ii) 4, 8/3, 16/9, 32/27, …

(8/3) ÷ 4 = 8/12 = 2/3

(16/9) ÷ (8/3) = (16/9) × (3/8) = 2/3

(32/27) ÷ (16/9) = (32/27) × (9/16) = 2/3

Constant, so it is a GP with a = 4, r = 2/3

tn = 4 × (2/3)n–1

(iii) 3, –3/2, 3/4, –3/8, …

(–3/2) ÷ 3 = –1/2

(3/4) ÷ (–3/2) = (3/4) × (–2/3) = –1/2

(–3/8) ÷ (3/4) = (–3/8) × (4/3) = –1/2

Constant, so it is a GP with a = 3, r = –1/2

tn = 3 × (–1/2)n–1

Page 36 of 64

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

Why it happens: the common ratio may be greater than 1, between 0 and 1, or
negative, and each choice shows on the sequence. In (i), r = 5 makes the terms race
away. In (ii), 0 < r < 1 makes them shrink towards 0 while staying positive. In (iii), a
negative r makes the sign flip at every step, because (–1/2)n–1 is positive for odd n
and negative for even n — the terms shrink in size while alternating in sign.

Check it yourself: in (iii), put n = 4: 3 × (–1/2)3 = 3 × (–1/8) = –3/8. ✓

Q2 Exercise: Can you find a recursive rule for the formula tn = 3 × 10ⁿ⁻¹ that generates
the geometric progression 3, 30, 300, 3000, … ?

Each term is ten times the one before, so:

t1 = 3, tn = 10 × tn–1 for n ≥ 2

Running it: t2 = 10 × 3 = 30, t3 = 10 × 30 = 300, t4 = 10 × 300 = 3000. ✓

Why it happens: the two rules can be shown to agree, not merely observed to.
Divide the explicit rule at n by its value at n – 1:
tn ÷ tn–1 = (3 × 10n–1) ÷ (3 × 10n–2) = 10n–1–(n–2) = 10.

So tn = 10 tn–1 for every n ≥ 2, and since t1 = 3 × 100 = 3 the two rules start at the
same place as well. The same computation on tn = arn–1 gives the general fact tn = r ·
tn–1.

Tip: in a GP the common ratio is a multiplier, so the recursive step is "×r", never "+r".

Think and Reflect — Page 189

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

Section 8.6.1 Fun with Fractals

THINK AND REFLECT

Q1 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?

Stage 0 Stage 1 Stage 2 Stage 3

Fig. 8.7 — Stages 0 to 3 of the Sierpiński triangle.

(a) Counting the black triangles in Fig. 8.7:

Stage 0: 1 Stage 1: 3 Stage 2: 9 Stage 3: 27

(b) Each black triangle is replaced by 3 smaller black triangles at the next stage, so the count is
multiplied by 3.

Stage 4: 27 × 3 = 81 Stage 5: 81 × 3 = 243

(c) The counts 1, 3, 9, 27, 81, 243, … are a GP with first term 1 and common ratio 3, and they are
the powers of 3 with the exponent equal to the stage number:

Page 38 of 64

as e
a g l
Class 9 Maths Chapter 8 Predicting What Comes Next: Exploring Sequences and Progressions AglaSem · NCERT Solutions

co m
1 = 30, 3 = 31, 9 = 32, 27 = 33, 81 = 34, 243 = 35
e m.
om triangles at Stage n = 3n
Number of.cblack g l as
em t = 1, t = 3 × t a
a s
agl
Recursively: 0 n n–1

co m
. ag
(d) At each stage the triangle is cut into 4 equal parts and the middle one is removed, so exactly

e m
as
3/4 of the black area survives.

a g l
Stage 1: 1 × 3/4 = 3/4 = 0.75

co m
m.
Stage 2: (3/4) × (3/4) = 9/16 = 0.5625

m as e
.co
Stage 3: (3/4)3 = 27/64 ≈ 0.4219
a g l
a s em4: (3/4)4 = 81/256 ≈ 0.3164
gl
Stage
a Stage 5: (3/4)5 = 243/1024 ≈ 0.2373

m a s
.co agl
Area at Stage n = (3/4)n square units
se m
Recursively: s0 = 1, sn = (3/4) × sn–1
g l a
a
As n increases the area keeps being multiplied by 3/4, a number less than 1, so the black area
co m
m .
e
shrinks steadily and creeps closer and closer to 0 without ever becoming 0.
m l as
.co a g
a s emWhy it happens: the same figure produces two GPs pulling in opposite directions.

ag l Counting triangles multiplies by 3 (r > 1, so the count explodes); measuring area
m
multiplies by 3/4 (0 < r < 1, so the area collapses). There is no contradiction: at each

a se
com l
stage there are three times as many pieces, but each piece has only a quarter of the
. a g
e m
area of the piece it came from, and 3 × 1/4 = 3/4 < 1. That single number, 3/4, is why
s area move opposite ways.
g
the count of pieces and thel atotal
a

c o m
.
Did you know? the numbering starts at Stage 0, so here the exponent equals the
m what the first
stage number exactly — 3n and (3/4)n, with no "n – 1". Alwaysecheck
s
m
o is called before writing a formula.
cstage gl a
. a
as em
agl c
m .
In-text Questions — Page 189 m a s e
e m . co agl
g l as
a

co m
m .
m ase
.co


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

Section 8.6.1 Fun with Fractals

Q1 The number of black triangles increases very quickly as the stage numbers
increase. Can you explain why?

Because the growth is by multiplication, not addition. Every black triangle turns into three at the
next stage, so the whole count triples each time.

tn = 3 × tn–1 gives tn = 3n

Stage 5: 243 Stage 10: 59,049 Stage 15: 1,43,48,907

Compare this with the AP of Fig. 8.3, where each stage adds a fixed 4:

STAGE N 5 10 15

AP: 4n – 3 17 37 57

GP: 3n 243 59,049 1,43,48,907

Why it happens: when a quantity grows by addition, each step contributes the same
fixed amount no matter how large the quantity already is. When it grows by
multiplication, each step contributes an amount proportional to what is already there
— the bigger it is, the bigger the next jump. In 3n the position n sits in the exponent,
and every increase of 1 in n multiplies the whole answer by 3. That compounding is
what makes the count run away.

Did you know? this is the same reason a fixed rate of compound interest overtakes
a fixed rupee-amount of simple interest once enough time has passed.

Q2 Can you explain why the area of the black region at Stage n will be (3/4)ⁿ?

Because the same operation is applied to every black triangle at every stage, and it always
keeps exactly three of four equal parts.

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

One stage of the construction:

divide each black triangle into 4 equal small triangles

remove the middle one, keep 3

black area after the step = (3/4) × black area before the step

Applying this repeatedly from Stage 0, where the area is 1:

s1 = (3/4) × 1 = 3/4

s2 = (3/4) × (3/4) = (3/4)2

s3 = (3/4) × (3/4)2 = (3/4)3

…

sn = (3/4) × sn–1 = (3/4)n

Why it happens: joining the midpoints of a triangle cuts it into four triangles of
equal area, so removing one removes exactly one quarter — whatever the size of the
triangle. That is what lets the step be applied simultaneously to all 3n black triangles
at a stage and still be described by the single factor 3/4. Multiplying by 3/4 exactly n
times, starting from 1, is the meaning of (3/4)n. This is the recursive rule and the
explicit rule for the same GP, and the argument shows why they agree.

Check it yourself: at Stage n there are 3n black triangles, each of area (1/4)n. Total
area = 3n × (1/4)n = (3/4)n — the same answer by a different route.

Exercise Set 8.3 — Pages 193–194
Section 8.6 Geometric Progressions (questions marked * are the harder set)

Q1 Find the 12th term of a GP with common ratio 2, whose 8th term is 192.

The 12th term is four steps beyond the 8th, and each step multiplies by r = 2.

Page 41 of 64

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

t12 = t8 × 24 = 192 × 16 = 3072

The longer route gives the same answer. From t8 = ar7:

a × 27 = 192
128a = 192

a = 1.5

t12 = 1.5 × 211 = 1.5 × 2048 = 3072 ✓

Why it happens: in a GP, tm ÷ tn = rm–n — the first term cancels out. So a question
that links two terms of the same GP never needs a at all, which is why the short
route works and is safer arithmetic.

Q2 Find the 10th and nth terms of the GP: 5, 25, 125, … .

a = 5 and r = 25 ÷ 5 = 5 (also 125 ÷ 25 = 5).

tn = arn–1 = 5 × 5n–1 = 5n

t10 = 510 = 97,65,625

Check the rule on the printed terms: t1 = 51 = 5, t2 = 52 = 25, t3 = 53 = 125. ✓

Why it happens: here the first term happens to equal the common ratio, so 5 × 5n–1
collapses into the single power 5n by the law am × ak = am+k. This is a special feature
of this GP, not a general rule — for 3, 6, 12, … the nth term stays 3 × 2n–1 and cannot
be shortened.

Check it yourself: 510 = (55)2 = 31252 = 97,65,625.

Page 42 of 64

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

*Q3 A sequence is given by the recursive rule t1 = 2, tn+1 = 3tn – 2 for n ≥ 1. Which term
of the sequence is 730?

Generate the terms until 730 appears.

t1 = 2

t2 = 3 × 2 – 2 = 4

t3 = 3 × 4 – 2 = 10

t4 = 3 × 10 – 2 = 28

t5 = 3 × 28 – 2 = 82

t6 = 3 × 82 – 2 = 244

t7 = 3 × 244 – 2 = 730

So 730 is the 7th term.
There is also a way to see it without listing. Subtract 1 from each term: 1, 3, 9, 27, 81, 243, 729 —
a GP of powers of 3.

If un = tn – 1, then

un+1 = tn+1 – 1 = (3tn – 2) – 1 = 3(tn – 1) = 3un

with u1 = 1, so un = 3n–1

Hence tn = 3n–1 + 1

3n–1 + 1 = 730 → 3n–1 = 729 = 36 → n – 1 = 6 → n = 7

Why it happens: the rule tn+1 = 3tn – 2 is not a GP as it stands, because of the "– 2".
But shifting every term down by 1 removes the constant and leaves pure tripling —
the fixed point of the rule is x = 3x – 2, that is x = 1, and measuring from that fixed
point turns the sequence into a GP. Recognising this converts a listing problem into a
one-line equation, which is the only practical method if the question had asked
about the 30th term.

Page 43 of 64

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

co m
m.
Which term of the GP: 2, 6, 18, … is 4374? Write the explicit formula as well as the
e
Q4

m l as
.co
recursive formula for the nth term.

a g
se m
g l a
a

a = 2 and r = 6 ÷ 2 = 3 (also 18 ÷ 6 = 3).

co m
m . ag
Explicit: tn = 2 × 3n–1
l a se
Recursive: t1 = 2, tn = 3 × tn–1 agn ≥ 2
for

co m
Now solve tn = 4374:
em.
m l as
.co a g
2s×e3m
gl a
n–1 = 4374

a 3n–1 = 2187

m a s
.co agl
37 = 2187, so n – 1 = 7

se m
n=8
g l a
a
So 4374 is the 8th term. Check: 2 × 37 = 2 × 2187 = 4374. ✓
co m
m .
m as e
.co
Why it happens: in a GP the unknown position sits in the exponent, so the last step
a g l
a s eismnot a division but a matching of powers: you must recognise 2187 as 3 . Build the 7

ag l powers of 3 up from the bottom — 3, 9, 27, 81, 243, 729, 2187 — and count them. If
4374 had turned out not to be 2 times a power of 3, it would simply not be a term of
se m
this GP.
com g l a
m . a
ase
agl
A ball is dropped from a height of 80 metres. After hitting the ground, it bounces
om
Q5

. c
back to 60% of the height from which it fell. It continues bouncing in this way —

m a s
each time rising to 60% of the previous height. (i) What heightem does the ball reach
. gl the ball has travelled by
co after the 5th bounce? (ii) What is the total vertical adistance
as em the time it hits the ground for the 6th time?
agl c
m .
s e

m a
. co agl
The peak heights form a GP with r = 60% = 0.6, starting from the 80 m drop.
e m
g l as
a

co m
m .
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.co


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

After 1st bounce: 80 × 0.6 = 48 m

After 2nd bounce: 48 × 0.6 = 28.8 m

After 3rd bounce: 28.8 × 0.6 = 17.28 m

After 4th bounce: 17.28 × 0.6 = 10.368 m

After 5th bounce: 10.368 × 0.6 = 6.2208 m

(i) The height after the 5th bounce is 6.2208 m (about 6.22 m). Directly: 80 × (0.6)5 = 80 ×
0.07776 = 6.2208 m.
(ii) Count the journeys carefully. The ball falls 80 m, then each bounce is a rise followed by an
equal fall, and the 6th hit on the ground happens after the fall that follows the 5th bounce.

STAGE DISTANCE

First fall 80 m

Up and down after bounce 1 2 × 48 = 96 m

After bounce 2 2 × 28.8 = 57.6 m

After bounce 3 2 × 17.28 = 34.56 m

After bounce 4 2 × 10.368 = 20.736 m

After bounce 5 (ends with the 6th hit) 2 × 6.2208 = 12.4416 m

Total = 80 + 2 × (48 + 28.8 + 17.28 + 10.368 + 6.2208)

= 80 + 2 × 110.6688

= 80 + 221.3376

= 301.3376 m (about 301.34 m)

Why it happens: the heights make a GP, but the distance travelled is not a single
term of it — it is a total, and every rise after the first fall is matched by a fall of the
same length. That is why the five peak heights are doubled while the original 80 m is
counted once. Miscounting here is the usual error: the 6th hit comes after only 5
bounces.

Page 45 of 64

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

Tip: 60% means r = 0.6, so the ball loses 40% of its height each time. After 5 bounces
only (0.6)5 ≈ 7.8% of the original height is left.

Q6 Which term of the sequence 2, 2√2, 4, … is 128?

First check that this is a GP by taking ratios.

2√2 ÷ 2 = √2

4 ÷ 2√2 = 2/√2 = √2

Constant, so a = 2 and r = √2

Write every term as a power of 2, using √2 = 21/2:

tn = 2 × (√2)n–1 = 21 × 2(n–1)/2 = 2(n+1)/2

128 = 27
So (n + 1)/2 = 7

n + 1 = 14

n = 13

128 is the 13th term. Check: t13 = 2 × (√2)12 = 2 × 26 = 2 × 64 = 128. ✓

Why it happens: the sequence is 2, 2√2, 4, 4√2, 8, 8√2, 16, … — the whole numbers
double only at every second step, because two multiplications by √2 make one
multiplication by 2. Rewriting the surd as the power 21/2 turns an awkward-looking
sequence into plain powers of 2, and then matching exponents finishes the job in
one line.

Check it yourself: the terms with whole-number values sit at the odd positions — t1
= 2, t3 = 4, t5 = 8, t7 = 16, t9 = 32, t11 = 64, t13 = 128.

Page 46 of 64

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

Q7 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?

Stage 0 Stage 1 Stage 2 Stage 3

Fig. 8.12 — Stages 0, 1, 2 and 3 of the Sierpiński square carpet.

Every red square becomes 9 smaller squares of which 8 are kept, so both counts and areas are
governed by the same step.

Page 47 of 64

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

One step of the construction: 9 equal squares, the centre one removed, 8 kept — so the count is
multiplied by 8 and the area by 8/9.

(i) Counting the red squares in Fig. 8.12:

Stage 0: 1 Stage 1: 8 Stage 2: 8 × 8 = 64 Stage 3: 64 × 8 = 512

(ii)

Stage 4: 512 × 8 = 4096 Stage 5: 4096 × 8 = 32768

(iii) The counts 1, 8, 64, 512, … are a GP with common ratio 8, and they are the powers of 8 with
exponent equal to the stage number.

Explicit: tn = 8n

Recursive: t0 = 1, tn = 8 × tn–1 for n ≥ 1

Page 48 of 64

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

(iv) Each of the nine small squares has 1/9 of the area of the square it came from, and 8 of them
co m
are kept, so the red area is multiplied by 8/9 at every stage.
e m.
m l as
m .co a g
l a se
Stage 1: 8/9 ≈ 0.8889
g
aStage 2: (8/9)2 = 64/81 ≈ 0.7901

co m
ag
Stage 3: (8/9)3 = 512/729 ≈ 0.7023
m .
as e
Stage 4: (8/9)4 = 4096/6561 ≈ 0.6243
a g l
Stage 5: (8/9)5 = 32768/59049 ≈ 0.5549

co m
m.
Explicit: sn = (8/9)n square units

m as e
.co
Recursive: s0 = 1, sn = (8/9) × sn–1 for n ≥ 1
a g l
a s em
a l 8/9 is less than 1, the area falls at every stage and approaches 0 as n grows — though it
gSince
never actually reaches 0.

om a s
e
. c
m triangle's argument with different numbers. agl
s
Why it happens: this is the Sierpiński
Trisecting the sides makes 3g×la
a 3 = 9 congruent squares, so each has area 1/9 of the
original; keeping 8 of them leaves 8 × (1/9) = 8/9 of the area but multiplies the
number of pieces by 8. Because 8/9 is closer to 1 than the triangle's 3/4, the carpet's
co m
m .
area shrinks more slowly — but it shrinks all the same, while 8n outruns 3n. A fractal
m as e
.co
is exactly this: more and more pieces, less and less area.
a g l
sem
g l a
a Check it yourself: at Stage n there are 8n red squares, each of area (1/9)n. Their
se m
com
total is 8n × (1/9)n = (8/9)n ✓ — the two formulas fit together.
g l a
m . a
ase
agl
End-of-Chapter Exercises — Pages 194–195
co m
Chapter 8 review (questions marked * are the harder set)
m .
o m l a se
g 16th term is 73.
.c Find the 31st term of an AP whose 11th term is 38 and
a
se m
a
Q1

agl c
m .
e

m a s
. co agl
Two terms of an AP give two equations in a and d.

e m
g l as
a

co m
m .
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.co


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

a + 10d = 38 (1)

a + 15d = 73 (2)

(2) – (1): 5d = 35, so d = 7
From (1): a = 38 – 70 = –32

t31 = a + 30d = –32 + 210 = 178

Shorter: t31 is 15 steps past t16, so t31 = 73 + 15 × 7 = 73 + 105 = 178. ✓

Why it happens: in an AP the difference between the mth and nth terms is (m – n)d,
with a cancelling out. Here t16 – t11 = 5d = 35 straight away. The full solve is worth
doing once to see that a = –32 — the AP starts below zero and only becomes positive
later.

Q2 Determine the AP whose third term is 16 and whose 7th term exceeds the 5th term
by 12.

"The 7th term exceeds the 5th by 12" means t7 – t5 = 12, and those two terms are two steps
apart.

t7 – t5 = (a + 6d) – (a + 4d) = 2d

2d = 12

d=6

t3 = a + 2d = 16 → a + 12 = 16 → a = 4

The AP is 4, 10, 16, 22, 28, 34, 40, …
Check: the 3rd term is 16 ✓; t7 = 40 and t5 = 28, and 40 – 28 = 12 ✓.

Why it happens: the phrase "exceeds by" is a difference of terms, and in an AP any
such difference is a whole number of d's — never involving a. That is why the second
condition fixes d on its own, leaving the first condition free to fix a. "Determine the
AP" means state a and d, or list enough terms that the pattern is unambiguous.

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

*Q3 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.)

Find the two ends first.

Smallest: 100 ÷ 7 = 14.28…, so the first multiple of 7 at or above 100 is 7 × 15 = 105

Largest: 999 ÷ 7 = 142.71…, so the last one at or below 999 is 7 × 142 = 994

These form an AP with a = 105, d = 7, last term 994.

105 + (n – 1) × 7 = 994

(n – 1) × 7 = 889

n – 1 = 127

n = 128

There are 128 three-digit numbers divisible by 7.

Why it happens: the multiples of 7 are 7 × 15, 7 × 16, …, 7 × 142, so counting them is
counting the whole numbers from 15 to 142 — that is 142 – 15 + 1 = 128. The "+1" is
the fencepost correction, and it is the same correction that appears as the "– 1" in tn
= a + (n – 1)d. Both routes must agree, and they do.

Tip: counting the numbers from 15 to 142 as 142 – 15 = 127 forgets one end. Always
test the idea on a tiny case: from 3 to 5 there are 3 numbers, not 2.

*Q4 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.)

The multiples of 4 strictly between 10 and 250 run from 12 to 248.

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

Smallest: 4 × 3 = 12 (since 4 × 2 = 8 is below 10)

Largest: 4 × 62 = 248 (since 4 × 63 = 252 is above 250)

12 + (n – 1) × 4 = 248
(n – 1) × 4 = 236

n – 1 = 59

n = 60

There are 60 multiples of 4 between 10 and 250.

Why it happens: as in the previous question, the multiples of 4 here are 4 × 3 up to
4 × 62, so the count is 62 – 3 + 1 = 60. Note that 250 is not itself a multiple of 4, so the
word "between" causes no trouble at that end; had the limit been 248, you would
have to decide whether to include it.

*Q5 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 GP be a, ar, ar2, … Translate both conditions.

a + ar = –4 i.e. a(1 + r) = –4 (1)

t5 = 4 t3 → ar4 = 4ar2

r2 = 4 (dividing by ar2, which is not 0)

r = 2 or r = –2

Case r = 2: a(1 + 2) = –4 gives 3a = –4, a = –4/3.

GP: –4/3, –8/3, –16/3, –32/3, …

Check: –4/3 – 8/3 = –12/3 = –4 ✓; t5 = –64/3 and 4 t3 = 4 × (–16/3) = –64/3 ✓

Case r = –2: a(1 – 2) = –4 gives –a = –4, a = 4.

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

GP: 4, –8, 16, –32, 64, …

Check: 4 + (–8) = –4 ✓; t5 = 64 and 4 t3 = 4 × 16 = 64 ✓

Both are valid answers: –4/3, –8/3, –16/3, … and 4, –8, 16, –32, …

Why it happens: the condition t5 = 4t3 reduces to r2 = 4, and a squared quantity has
two square roots. Nothing in the problem rules out a negative ratio, so both must be
reported. Dividing by ar2 is legitimate here because a = 0 would make every term 0
and the sum could not be –4.

*Q6 Find all possible ways of expressing 100 as the sum of consecutive natural
numbers.

Suppose the sum has k terms starting at a. Using the AP sum idea, the total is k terms whose
average is (first + last)/2:

a + (a + 1) + … + (a + k – 1) = k × [2a + (k – 1)]/2 = 100

so k(2a + k – 1) = 200

Now k and (2a + k – 1) are two factors of 200. They always have opposite parity: if k is odd, k – 1
is even so 2a + k – 1 is even, and vice versa. So one factor must be odd. The odd factors of 200 =
23 × 52 are 1, 5 and 25.

K 2A + K – 1 A VALID?

1 200 100 the single number 100 — not a sum

5 40 (40 – 4)/2 = 18 yes

8 25 (25 – 7)/2 = 9 yes

25 8 negative rejected

40 5 negative rejected

200 1 negative rejected

Page 53 of 64

as e
a g l
Class 9 Maths Chapter 8 Predicting What Comes Next: Exploring Sequences and Progressions AglaSem · NCERT Solutions

co m
em.
100 = 18 + 19 + 20 + 21 + 22
m l as
.co
100 = 9 + 10 + 11 + 12 + 13 + 14 + 15 + 16
m a g
l a se
a g
Both check out: 18 + 22 = 40, and 5 terms averaging 20 give 100; 9 + 16 = 25, and 8 terms
averaging 12.5 give 100. So there are exactly two ways (three, if the single number 100 is

co m
. ag
allowed to count).

e m
g l as
a
Why it happens: the parity argument is what makes the search finite. Because k and
2a + k – 1 can never both be even, every valid k pairs with an odd factor of 200 —

c o m
and 200 has only three odd factors. This is also why a power of 2 such as 64 cannot
.
mfactor is 1.
s e
be written as a sum of consecutive natural numbers at all: its only odd

. com a gla
a s em it yourself: 18 + 19 + 20 + 21 + 22 = 100 and 9 + 10 + 11 + 12 + 13 + 14 + 15 +
gl
Check
a 16 = 100. Add them and see.

m a s
em
.co agl
a s
*Q7
a gl a certain culture doubles every hour. If there were 30
The number of bacteria in
bacteria present in the culture originally, how many bacteria will be present at the
end of the 2nd hour, 4th hour and nth hour?
co m
m .
m as e
.co a g l
se m
a
Doubling every hour is a GP with a = 30 and r = 2, counting from hour 0.

ag l
Start (hour 0): 30
se m
com g l a
m . a
ase
End of 1st hour: 30 × 2 = 60

End of 2nd hour: 30 × 22 = 120
agl
End of 3rd hour: 30 × 23 = 240
co m
m .
e
End of 4th hour: 30 × 24 = 480

comof nth hour: 30 × 2n g l as
m .End a
l a se
ag
.c
s e m
m a
e m . co agl
g l as
a

co m
m .
m as e
.co


a g l Page 54 of 64

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

Why it happens: the exponent counts how many doublings have taken place, and
after n hours there have been exactly n of them — so the exponent is n, not n – 1.
The n – 1 appears in tn = arn–1 only because there the first term is labelled n = 1; here
the starting count is labelled hour 0. Match the exponent to the number of steps
actually taken, and this never goes wrong.

Did you know? at this rate a single day of growth means 224 doublings — the 30
bacteria would become 30 × 1,67,77,216, over fifty crore.

*Q8 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.

Write each term as a + (n – 1)d and add.

t4 + t8 = (a + 3d) + (a + 7d) = 2a + 10d = 24 → a + 5d = 12 (1)

t6 + t10 = (a + 5d) + (a + 9d) = 2a + 14d = 44 → a + 7d = 22 (2)

(2) – (1): 2d = 10, so d = 5

From (1): a + 25 = 12, so a = –13

The first three terms are –13, –8, –3.
Check: t4 = –13 + 15 = 2 and t8 = –13 + 35 = 22, sum 24 ✓. t6 = –13 + 25 = 12 and t10 = –13 + 45 =
32, sum 44 ✓.

Why it happens: dividing each equation by 2 is worth doing early — it turns 2a +
10d = 24 into a + 5d = 12, which is just t6 = 12. In fact the sum of two terms equally
spaced about a middle term is always twice that middle term, so the two conditions
say t6 = 12 and t8 = 22 directly, and d = (22 – 12)/2 = 5 in one line.

*Q9 Find the smallest value of n such that the sum of the first n natural numbers is
greater than 1,000.

Require Sn = n(n + 1)/2 > 1000, that is n(n + 1) > 2000.

Page 55 of 64

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

n = 44: 44 × 45 = 1980, which is not more than 2000 → S44 = 990

n = 45: 45 × 46 = 2070, which is more than 2000 → S45 = 1035

So the smallest such value is n = 45, giving S45 = 1035 > 1000, while S44 = 990 < 1000.

Why it happens: Sn is roughly n2/2, so n2 must be near 2000 and n near √2000 ≈
44.7. That estimate tells you where to look; the two exact checks at n = 44 and n = 45
are what settle it. For a "smallest n" question you must always show that the value
below fails — otherwise you have not proved it is the smallest.

*Q10 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 = 2 and r = 8 ÷ 2 = 4 (also 32 ÷ 8 = 4).

Explicit: tn = 2 × 4n–1

Recursive: t1 = 2, tn = 4 × tn–1 for n ≥ 2

Solve tn = 131072 by turning everything into powers of 2:

2 × 4n–1 = 131072

4n–1 = 65536

4n–1 = 22(n–1) and 65536 = 216
2(n – 1) = 16

n–1=8

n=9

So 131072 is the 9th term. Check: 2 × 48 = 2 × 65536 = 131072. ✓

Page 56 of 64

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

Why it happens: both 2 and 4 are powers of the same base, so the equation can be
reduced to matching exponents of 2 — a step that removes all guesswork. Had the
numbers not shared a base, you would have to build the terms up one at a time: 2,
8, 32, 128, 512, 2048, 8192, 32768, 131072 — the ninth. ✓

*Q11 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.

Choosing the three terms as a/r, a, ar makes the product collapse.

Product: (a/r) × a × (ar) = a3 = –1
a = –1

Now use the sum:

–1/r – 1 – r = 13/12
–(1/r + r) = 13/12 + 1 = 25/12

1/r + r = –25/12

Multiply by 12r: 12 + 12r2 = –25r

12r2 + 25r + 12 = 0

Factorise: 12r2 + 16r + 9r + 12 = 4r(3r + 4) + 3(3r + 4) = (3r + 4)(4r + 3).

r = –4/3 or r = –3/4

With a = –1 and r = –3/4 the three terms are a/r = 4/3, a = –1, ar = 3/4:

4/3, –1, 3/4

Sum = 16/12 – 12/12 + 9/12 = 13/12 ✓

Product = (4/3) × (–1) × (3/4) = –1 ✓

The other root r = –4/3 gives the same three numbers in reverse order: 3/4, –1, 4/3.

Page 57 of 64

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

Why it happens: writing the terms as a/r, a, ar puts the middle term at the centre, so
r cancels in the product and one unknown disappears immediately. The two roots
are reciprocals of each other because reversing a GP replaces r by 1/r — they are not
two different progressions, only two ways of reading the same one.

Tip: the product being negative forces an odd number of negative terms. With r
negative, the middle term is the negative one — which is exactly what a = –1 says.

*Q12 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 GP have first term a and common ratio r.

x = t4 = ar3

y = t10 = ar9

z = t16 = ar15

Compare the two ratios of consecutive members of x, y, z:

y/x = ar9 ÷ ar3 = r6

z/y = ar15 ÷ ar9 = r6

So y/x = z/y = r6

The ratio of consecutive terms is the same constant, so x, y, z form a GP with common ratio r6.
Equivalently y2 = a2r18 = (ar3)(ar15) = xz, which is the same statement.

Why it happens: the positions 4, 10, 16 are themselves equally spaced — they form
an AP with common difference 6. Picking equally spaced terms out of a GP always
leaves a GP, because moving 6 places along multiplies by r6 every time. This is the
multiplicative twin of a familiar fact about APs: equally spaced terms of an AP form
an AP.

Page 58 of 64

as e
a g l
Class 9 Maths Chapter 8 Predicting What Comes Next: Exploring Sequences and Progressions AglaSem · NCERT Solutions

co m
m.
Tip: in a proof, do not stop at "the ratios look equal". Compute both ratios from the

m as e
l
definition and show they are the same expression — that is what makes it a proof.

m .co a g
l a se
a g
*Q13 The sum of the first three terms of a geometric progression is 26, and the sum of

co m
ag
their squares is 364. Find the terms of the GP.

m .
as e
a g l
Let the terms be a, ar, ar2.

co m
e m.
m
a(1 + r + r2) = 26 (1)
l as
m .c2o 4 a g
l a se
a2(1 + r + r ) = 364 (2)

a g
s
The key is the identity 1 + r2 + r4 = (1 + r + r2)(1 – r + r2). Substituting into (2):
m a
m.co agl
l a se
g
a(1 + r + r2) × a(1 – r + r2) = 364
a
26 × a(1 – r + r2) = 364

co m
a(1 – r + r2) = 14 (3)
m .
m as e
. c o a g l
s e m (3) from (1), then add them:
Subtract

agla
(1) – (3): 2ar = 12 → ar = 6
se m
com g l a
(1) + (3): 2a(1 + r2) = 40 → a + ar2 = 20
m . a
ase
Put a = 6/r into a + ar2 = 20: agl

co m
m .
6/r + 6r = 20
m ase
.6co+ 6r2 = 20r a g l
se m
g l a
a 3r2 – 10r + 3 = 0
c
m .
(3r – 1)(r – 3) = 0
m a s e
r = 3 or r = 1/3
e m . co agl
g l as
a

co m
m .
m as e
.co


a g l Page 59 of 64