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

NCERT Class 9 Mathematics — Ganita Manjari Part I, Predicting What Comes Next: Exploring Sequences and Progressions. All 11 Think and Reflect prompts answered.

Also here. Every exercise in this chapter is solved, one page per exercise.

The eleven boxes in this chapter trace the chapter’s own argument: first spot a pattern, then describe it in words, then capture it in a formula, and finally use the formula to answer questions the pattern alone could not.

Key insight. The question to ask of any new sequence is: what stays constant? If the difference between consecutive terms is constant, it is an AP and $t_n = a + (n-1)d$. If the ratio is constant, it is a GP and $t_n = ar^{\,n-1}$. If neither is, as with the triangular and square numbers, look at the differences of the differences — that is where the next layer of regularity hides.

Prompt 1 comes from the box on page 174, beside the four sequences the chapter opens with.

Question 1

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

Solution. The four sequences are the natural numbers, the odd numbers, the triangular numbers and the square numbers.

Sequence	Pattern	$n$th term	Next three
$1, 2, 3, 4, 5, 6, \dots$	add $1$ each time	$n$	$7, 8, 9$
$1, 3, 5, 7, 9, 11, \dots$	add $2$ each time	$2n – 1$	$13, 15, 17$
$1, 3, 6, 10, 15, 21, \dots$	add $2$, then $3$, then $4$, …	$\frac{n(n+1)}{2}$	$28, 36, 45$
$1, 4, 9, 16, 25, 36, \dots$	the squares	$n^2$	$49, 64, 81$

The first two are arithmetic progressions — a fixed amount is added each time, so their rules are linear in $n$.

The last two are not. Their differences grow. For the triangular numbers the differences are $2, 3, 4, 5, 6, \dots$ (each term adds the next natural number), and for the squares they are $3, 5, 7, 9, 11, \dots$ — the odd numbers. That second fact is a small delight in its own right: adding consecutive odd numbers builds the squares, $1 = 1$, $1 + 3 = 4$, $1 + 3 + 5 = 9$, and so on.

Prompt 2 comes from the box on page 176, after the chapter has shown sequences with fractions and with negative terms.

Question 2

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

Solution. The point of this box is that “sequence” is a much wider idea than the four opening examples suggest. Five genuinely different families, each with a property worth naming:

1. Geometric sequences — $2, 6, 18, 54, \dots$ Each term is a fixed multiple of the last ($t_n = 2 \times 3^{\,n-1}$). They grow far faster than any AP, which is the whole subject of Section 8.6.

2. Harmonic sequences — $1, \frac12, \frac13, \frac14, \dots$ The terms are the reciprocals of an AP. They decrease, and they get arbitrarily close to $0$ without ever reaching it.

3. Fibonacci-type sequences — $1, 2, 3, 5, 8, 13, \dots$ Each term is built from the previous two, so the rule is recursive and has no simple explicit form. Virahānka’s sequence is the famous one.

4. Alternating sequences — $1, -1, 1, -1, \dots$ or $2, -4, 8, -16, \dots$ The sign flips each time. The second of these is actually a GP with $r = -2$, which shows that a negative common ratio is perfectly allowed.

5. Sequences with no formula at all — the primes $2, 3, 5, 7, 11, 13, 17, \dots$ They are perfectly well defined, you can always find the next one, and yet no explicit rule of the kind this chapter writes is known. The chapter says as much on page 180.

Others worth mentioning in discussion: constant sequences ($5, 5, 5, \dots$, which are APs with $d = 0$ and GPs with $r = 1$); factorial sequences $1, 2, 6, 24, 120, \dots$; and finite sequences such as the twelve months or a cricket team’s batting order, where the list simply stops.

Prompt 3 comes from the box on page 177, right after the first explicit rule is written down.

Question 3

Why is it useful to have an explicit formula for the $n^{th}$ term of a sequence?

Solution. Because it lets you jump straight to any term without generating the ones before it, and it lets you run the question backwards.

Going forwards. With $u_n = 2n – 1$ you can write down the $1170$th odd number — $2(1170) – 1 = 2339$ — in one line. Without the formula you would have to list $1169$ terms first. A recursive rule can only ever climb one step at a time.

Going backwards. This is the part students undervalue. The formula turns “is $137$ in this sequence, and where?” into an equation:

$$2n – 1 = 137 \ \Longrightarrow \ n = 69$$

so $137$ is the $69$th odd number. And if the equation gives a non-whole answer, that proves the number is not a term at all — as with $5n – 2 = 471$, which gives $n = 94.6$.

Why $n$ must be a natural number. Positions are counted, not measured: there is no $94.6$th term of a sequence. That is why the check on $n$ is not a formality but the actual content of the answer.

Prompt 4 comes from the second box on page 177.

Question 4

Can you find the rule describing the $n^{th}$ term of the sequence of square numbers?

Solution. The sequence is $1, 4, 9, 16, 25, 36, \dots$, and the rule is simply

$$t_n = n^2$$

since $1 = 1^2$, $4 = 2^2$, $9 = 3^2$, and so on.

Check it is not an AP. The differences are $3, 5, 7, 9, 11$ — not constant, so no formula of the form $a + (n-1)d$ can produce this list. But the differences are themselves an AP (the odd numbers, common difference $2$), and that second-level regularity is the signature of a quadratic rule, which $n^2$ is.

It also gives a second way to write the same sequence:

$$n^2 = 1 + 3 + 5 + \dots + (2n – 1)$$

the sum of the first $n$ odd numbers — worth remembering, because it turns a question about squares into a question about an AP.

Prompt 5 comes from the box on page 180, beside Fig. 8.3 — the growing pattern of squares whose counts are $1, 5, 9, 13$.

Question 5

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. At each stage $4$ squares are added at the corners, so the counts form an AP with $a = 1$ and $d = 4$:

$$1, \ 5, \ 9, \ 13, \ \dots$$

At any stage — do this one first, because it answers all the others:

$$t_n = 1 + (n-1)4 = 4n – 3$$

Now everything is a substitution:

Stage $n$	$5$	$6$	$10$	$11$	$12$	$20$
Squares $4n – 3$	$17$	$21$	$37$	$41$	$45$	$77$

Stages $5$ and $6$ could have been got by adding $4$ twice, and Stages $10$ to $12$ by adding $4$ six more times — but Stage $20$ would be tedious that way and Stage $200$ impossible. That is the prompt’s real lesson: find the rule first, then the individual answers are free.

Prompt 6 comes from the box on page 181, immediately after arithmetic progressions are defined.

Question 6

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. The test is one line long: compute consecutive differences and see whether they are all the same.

These are APs:

Sequence	Justification
$1, 2, 3, 4, \dots$ (naturals)	differences all $1$
$1, 3, 5, 7, \dots$ (odds)	differences all $2$
$1, 4, 7, 10, 13, \dots$	differences all $3$
$-7, -3, 1, 5, 9, \dots$	differences all $4$
$1, 5, 9, 13, 17, 21$ (Fig. 8.3)	differences all $4$
$11, 7, 3, -1, -5, \dots$	differences all $-4$ (an AP may decrease)
$3, 8, 13, \dots$ (from $s_n = 5n – 2$)	differences all $5$

These are not:

Sequence	Justification
$1, 3, 6, 10, 15, \dots$ (triangular)	differences $2, 3, 4, 5$ — growing
$1, 4, 9, 16, 25, \dots$ (squares)	differences $3, 5, 7, 9$ — growing
$1, \frac12, \frac13, \frac14, \dots$	differences $-\frac12, -\frac16, -\frac{1}{12}$ — shrinking
$6, 12, 24, 48, 96$	differences $6, 12, 24, 48$ — doubling, not constant
$1, 2, 3, 5, 8, 13, \dots$ (Virahānka)	differences $1, 1, 2, 3, 5$ — the sequence itself
$1, 2, 4, 7, 13, 24, \dots$ (Tribonacci)	differences $1, 2, 3, 6, 11$
$2, 3, 5, 7, 11, 13, \dots$ (primes)	differences $1, 2, 2, 4, 2$ — irregular

A useful shortcut. A sequence is an AP exactly when its $n$th term is a linear expression in $n$ — of the form $pn + q$. So $4n – 3$ and $5n – 2$ are APs at a glance, while $n^2$ and $\frac{n(n+1)}{2}$ cannot be, because they are quadratic.

One caution. Checking a single difference is not enough. $1, 2, 3, 5$ starts with two gaps of $1$ and is not an AP. At least three terms — two differences — are needed before any claim, and for a sequence given by a rule it is better to check the rule than the terms.

Prompt 7 comes from the box on page 184, just after the reversing-and-adding trick is used on a shorter sum.

Question 7

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

Solution. Yes — and this is exactly what the method is for. Write the sum forwards, write it again backwards underneath, and add column by column:

$$S = 1 + 2 + 3 + \dots + 98 + 99 + 100$$ $$S = 100 + 99 + 98 + \dots + 3 + 2 + 1$$

Adding the two lines term by term, every column gives the same total:

$$1 + 100 = 101, \quad 2 + 99 = 101, \quad 3 + 98 = 101, \quad \dots, \quad 100 + 1 = 101$$

There are $100$ such columns, so

$$2S = 100 \times 101 = 10100 \ \Longrightarrow \ S = \frac{10100}{2} = 5050$$

Why it works, and when. Nothing about $100$ was special. Pairing the first term with the last works because as one end goes up by $1$ the other comes down by $1$, leaving the sum unchanged — and that is a property of any AP, not just the natural numbers. Doing it in general gives

$$S_n = \frac{n(n+1)}{2}$$

for the naturals, and $S_n = \frac{n}{2}(a + \ell)$ for an AP with first term $a$ and last term $\ell$.

This is the calculation famously done in seconds by a young Gauss, and it is the reason the $1 + 2 + \dots + n$ formula needs no memorising: the pairing regenerates it in two lines.

Prompt 8 comes from the first box on page 185.

Question 8

Can you use this formula to find $S_{20}$, $S_{50}$ or $S_{1000}$?

Solution. The formula is $S_n = \dfrac{n(n+1)}{2}$, and each of these is one substitution:

$$S_{20} = \frac{20 \times 21}{2} = \frac{420}{2} = 210$$

$$S_{50} = \frac{50 \times 51}{2} = \frac{2550}{2} = 1275$$

$$S_{1000} = \frac{1000 \times 1001}{2} = \frac{1001000}{2} = 500500$$

Notice how little work $S_{1000}$ takes compared with the alternative: adding a thousand numbers by hand would take an hour and would almost certainly contain an error. This is the same argument as prompt 3, now applied to sums rather than terms.

A quick check on the last one: the average of $1$ and $1000$ is $500.5$, and $500.5 \times 1000 = 500500$ ✓ — the “average times count” reading of the same formula.

Prompt 9 comes from the second box on page 185, which notes that the $n$th triangular number is the sum of the first $n$ natural numbers.

Question 9

Let us revisit the sequence $t_n$ of triangular numbers $1, 3, 6, 10, 15, \dots$ The $n$th term of this sequence is the sum of the first $n$ natural numbers, so $t_n = \dfrac{n(n+1)}{2}$. Can you use this to find the $10$th, $17$th and $80$th triangular numbers?

Solution.

$$t_{10} = \frac{10 \times 11}{2} = 55$$

$$t_{17} = \frac{17 \times 18}{2} = \frac{306}{2} = 153$$

$$t_{80} = \frac{80 \times 81}{2} = \frac{6480}{2} = 3240$$

What has just happened is worth noticing. The triangular numbers are not an AP — prompt 6 settled that — and yet they now have a perfectly good explicit formula. It came not from the AP machinery but from recognising each triangular number as the sum of an AP.

So summing an AP produces a quadratic sequence, and that is the general pattern: sums climb one level of degree. It also gives an instant check — $t_{10} = 55$ should be $1 + 2 + \dots + 10$, and it is.

Prompt 10 comes from the box on page 186, beside Fig. 8.6 — a different growing pattern, whose counts are $3, 6, 12, 24$.

Question 10

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?

Solution. Here each stage doubles the previous one, so the counts form a GP with $a = 3$ and $r = 2$:

$$3, \ 6, \ 12, \ 24, \ \dots$$

At any stage:

$$t_n = 3 \times 2^{\,n-1}$$

Stage $n$	$5$	$6$	$10$	$11$	$12$	$20$
Squares $3 \times 2^{\,n-1}$	$48$	$96$	$1536$	$3072$	$6144$	$1\,572\,864$

How it differs from Fig. 8.3

This is the heart of the prompt, and the contrast is worth setting out plainly.

	Fig. 8.3	Fig. 8.6
Counts	$1, 5, 9, 13, \dots$	$3, 6, 12, 24, \dots$
What is constant	the difference, $+4$	the ratio, $\times 2$
Type	arithmetic progression	geometric progression
Rule	$4n – 3$ (linear in $n$)	$3 \times 2^{\,n-1}$ ($n$ in the exponent)
Stage $20$	$77$	$1\,572\,864$

Addition versus multiplication is the whole difference, and the last row shows what it costs. By Stage $20$ the AP has reached $77$ while the GP has passed a million and a half. An AP grows in a straight line; a GP’s growth accelerates, because each step is multiplied by everything that came before. This is why compound interest, bacterial cultures and the Sierpiński fractals all behave so unlike a rising salary.

Prompt 11 comes from the box on page 189, beside Fig. 8.7 — the Sierpiński triangle.

Question 11

Observe the Sierpiński triangle and try to answer the following questions.

Solution. At each stage every black triangle is divided into four equal parts and the middle one is removed, leaving three black triangles where there was one.

(a) How many black triangles are there in Stages $0$ to $3$ of Fig. 8.7?

$$1, \ 3, \ 9, \ 27$$

Stage $0$ is the whole triangle; each stage triples the count.

(b) Can you predict the number of black triangles at Stages $4$ and $5$?

$$27 \times 3 = 81, \qquad 81 \times 3 = 243$$

(c) Can you find a rule for the number of black triangles at the $n^{th}$ stage?

The counts $1, 3, 9, 27, 81, 243, \dots$ are a GP of common ratio $3$, all of them powers of $3$ whose exponent matches the stage number:

$$N_n = 3^{\,n} \qquad \text{(recursively: } N_0 = 1, \ N_{n+1} = 3N_n\text{)}$$

(d) Suppose the area of the black region in Stage $0$ is $1$ square unit. What is the area at Stages $1$, $2$ and $3$? At Stages $4$ and $5$? Find a rule for the $n$th stage. What happens as $n$ increases?

At each stage the triangle is cut into four equal parts and one is thrown away, so three quarters of the black area survives:

Stage	Black triangles	Area of each	Total black area
$0$	$1$	$1$	$1$
$1$	$3$	$\frac14$	$\frac34 = 0.75$
$2$	$9$	$\frac{1}{16}$	$\frac{9}{16} = 0.5625$
$3$	$27$	$\frac{1}{64}$	$\frac{27}{64} \approx 0.422$
$4$	$81$	$\frac{1}{256}$	$\frac{81}{256} \approx 0.316$
$5$	$243$	$\frac{1}{1024}$	$\frac{243}{1024} \approx 0.237$

$$A_n = \left(\frac34\right)^{\!n} \qquad \text{(recursively: } A_0 = 1, \ A_{n+1} = \tfrac34 A_n\text{)}$$

What happens as $n$ increases. Since $\frac34 < 1$, the areas shrink towards zero: by Stage $10$ the black region is about $0.056$ of the original, by Stage $30$ about $0.0002$.

So the shape ends up with infinitely many triangles and no area at all — the number of pieces races to infinity while the paper they are made of vanishes. That is the defining oddity of a fractal, and it is exactly the behaviour Exercise Set 8.3 Question 7 finds in the Sierpiński square carpet, where the factor is $\frac89$ instead of $\frac34$.

Common mistakes

• Prompt 1, calling the triangular or square numbers APs. Their differences grow. Check two differences, not one.
• Prompt 5 and 10, listing terms instead of finding the rule. Stage $20$ of the doubling pattern is $1\,572\,864$; nobody reaches that by adding a column of numbers. Get $t_n$ first.
• Prompts 5 and 10, using $2^{\,n}$ instead of $2^{\,n-1}$. With the first stage numbered $1$, the exponent is $n – 1$; with Stage $0$, as in the fractals, it is $n$. State which convention you are using.
• Prompt 7, halving twice. $2S = 10100$, so $S = 5050$ — not $2525$.
• Prompt 9, using $n^2$ for the triangular numbers. $t_{10}$ is $55$, not $100$. The triangular numbers are $\frac{n(n+1)}{2}$; the squares are $n^2$.
• Prompt 11(d), multiplying the area by $3$ rather than $\frac34$. Three triangles survive but each has a quarter of the area, so the region must shrink. An area that grew would be impossible, since every stage only removes paper.

Practise next

• Exercise Set 8.1 — explicit and recursive rules put to work, which is prompts 3 and 4 as exercises.
• Exercise Set 8.3, Question 7 — the Sierpiński square carpet, the direct sequel to prompt 11.
• The End-of-Chapter Exercises, Questions 6 and 9 — two questions that lean on the $1 + 2 + \dots + n$ formula from prompts 7 to 9.