Predicting What Comes Next: Exploring Sequences and Progressions: Class 9 Ganita Manjari · Chapter 8

Ganita Manjari Part 1 · Companion notes, simulations and practice · Official NCERT chapter PDF ↗

Published 28 Sep 2026 · Updated 28 Sep 2026

95 min
1 simulation
41 practice questions
30 MCQs

What you will learn in Predicting What Comes Next: Exploring Sequences and Progressions

Describe a sequence in words and with the notation t_n, and tell finite sequences from infinite ones.
Use an explicit rule to find any term, and decide whether a given number is a term by solving for its position.
Generate terms from a recursive rule, including rules that use the two previous terms, such as the Virahānka–Fibonacci sequence.
Recognise an arithmetic progression, find its nth term a + (n − 1)d, and explain why its points lie on a straight line.
Find 1 + 2 + … + n = n(n + 1)/2 by pairing terms, and use it for sums of consecutive numbers and of simple APs.
Recognise a geometric progression, find its nth term ar^(n − 1), and compare how APs and GPs grow and look on a graph.
Use GPs to count the pieces and the area of fractals such as the Sierpiński triangle.

Original companion text aligned to NCERT Mathematics Class 9, Chapter 8, pp. 174–196. Not affiliated with or endorsed by NCERT.

8.1 Introduction to Sequences

Say the first Sunday of a month falls on the 3rd. Without opening a calendar you can list every Sunday of that month: 3, 10, 17, 24 and 31. You noticed a rule, add 7, and used it to look ahead. This chapter is about lists of numbers like these, the rules hidden inside them, and how a good rule lets us jump far ahead in a list.

You already know several such lists from earlier grades. Try the box below before reading on.

Each of these lists is a sequence (अनुक्रम): numbers written in a definite order, first, second, third and so on. The numbers that make up a sequence are its terms. Order matters. The list 4, 1, 9 holds the same numbers as 1, 4, 9, but it is a different sequence. Three dots at the end mean that the list carries on for ever.

Definition

Sequence

A list of numbers in a definite order. The numbers are its terms: the first term, the second term, and so on.

Example2, 4, 6, 8, … is the sequence of even numbers; its third term is 6.

Some sequences stop. The seat numbers in one row of a cinema hall, 1 to 24, form a finite sequence of 24 terms. A countdown 10, 9, 8, …, 1, 0 is finite too, with 11 terms. The multiples of 5, however, never run out, so 5, 10, 15, 20, … is an infinite sequence.

Table 8.1Four familiar sequences and the gaps between their terms

Sequence	First five terms	Gaps between terms	Another way to see it
natural numbers	1, 2, 3, 4, 5	1, 1, 1, 1	count on by 1
odd numbers	1, 3, 5, 7, 9	2, 2, 2, 2	count on by 2
triangular numbers	1, 3, 6, 10, 15	2, 3, 4, 5	$1 + 2 + \dots + n$
square numbers	1, 4, 9, 16, 25	3, 5, 7, 9	$n \times n$

Where does the name “triangular” come from? Stack laddoos on a plate in rows of 1, 2, 3, …, as in Fig. 8.1, and they pile up into a triangle. Each new bottom row is one laddoo longer than the row above it, which is why the gaps in Table 8.1 grow by one each time: 2, 3, 4, 5, … Counting row by row also shows that each triangular number is a running total of counting numbers, such as $10 = 1 + 2 + 3 + 4$ .

6/6Count row by row: $15 = 1 + 2 + 3 + 4 + 5$ .

Fig. 8.1Each new bottom row is one laddoo longer than the row above it.NCERT Fig. 8.1 · p. 175

A plate holds a pyramid of round laddoos, with a panel on the right giving the total, the sum that makes it, and the sequence so far. Step 1 shows a single laddoo: 1. Step 2 slides the stack up and adds a bottom row of 2, making 3 = 1 + 2. Steps 3, 4 and 5 add bottom rows of 3, 4 and 5 laddoos, making 6, 10 and 15; the newest row is shown in blue each time. Along the bottom the sequence 1, 3, 6, 10, 15 builds up with the jumps +2, +3, +4, +5 between neighbours. The last step labels each row with its count, 1 to 5, and shows 15 = 1 + 2 + 3 + 4 + 5.

Squares can be built from tiles in a similar step-by-step way. In Fig. 8.2 one tile grows into a $2 \times 2$ square when an L-shape of 3 tiles wraps around it, then into a $3 \times 3$ square with an L of 5, and so on. Each L-shape holds the next odd number of tiles. So adding up the first $n$ odd numbers always gives $n^{2}$ , and the gaps between neighbouring square numbers are the odd numbers 3, 5, 7, 9, …

5/5An L of 9 tiles gives $5 \times 5 = 25$ .

Fig. 8.2Every L-shaped border holds the next odd number of tiles and turns one square into the next.NCERT Fig. 8.2 · p. 175

Square tiles sit in the bottom-left corner, with a panel on the right. Step 1 shows one tile: 1 = 1 × 1. Step 2 adds an L-shaped border of 3 tiles along the top and right, in a new colour, making 1 + 3 = 4 = 2 × 2. Steps 3, 4 and 5 add L-shaped borders of 5, 7 and 9 tiles, each in its own colour, making squares of 9, 16 and 25 tiles. The panel lists the odd numbers added so far, coloured like their borders, and the total as a square: 1 + 3 + 5 + 7 + 9 = 25 = 5 × 5.

Try thisNCERT Exercise · p. 176

A reading streak

Meera starts a reading streak. She reads 3 pages on day 1, 5 pages on day 2 and 7 pages on day 3: each day, 2 pages more than the day before.

How many pages does she read on days 5, 6, 7 and 8?
Make a second list: the total number of pages she has read by the end of each day. It starts 3, 8, 15, … Continue it up to day 10.
Add 1 to every number in your second list. What do you notice? Can you explain it with the help of Fig. 8.2?

Naming the terms

Think of a sequence as a row of lockers numbered 1, 2, 3, …, with one number stored in each. Instead of saying “the number in locker 4 of the square numbers”, we write $t_{4}$ . The small 4, written low, is called a subscript: it is the locker number, the position. The symbol $t_{4}$ as a whole stands for what is stored there, the value. So for the square numbers $t_{4} = 16$ , while $t_{1} = 1$ and $t_{2} = 4$ . In general, $t_{n}$ is the term in position $n$ (Fig. 8.3).

Fig. 8.3The small number under

t

is the position; the number in the box is the value:

t_{4} = 16

Six boxes in a row hold the first six square numbers: 1, 4, 9, 16, 25 and 36. Above each box a numbered blue circle gives its position, 1 to 6, and below each box its name is written: t sub 1 under 1, t sub 2 under 4, and so on up to t sub 6 under 36. The fourth box is shaded and ringed with a dashed line, and the text below says that position 4 holds the value 16, so t sub 4 equals 16.

The letter $t$ is only a habit. When one problem involves two sequences, give them different letters, say $s_{n}$ for one and $u_{n}$ for the other, so that they cannot be mixed up. Locker numbers are always counting numbers 1, 2, 3, …: there is a 1st term and a 2nd term, but no 2.5th term. What is stored in a locker, though, can be any number at all. The sequences below behave quite differently from those in Table 8.1.

$\frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5}, \dots$ : fractions that creep up towards 1 but never reach it.
$- 12, - 9, - 6, - 3, 0, 3, \dots$ : negative terms that climb by 3 at a time.
$1, - 2, 3, - 4, 5, \dots$ : terms that swing between positive and negative.
$7, 7, 7, 7, \dots$ : a constant sequence, in which every term is the same.
$1, 2, 1, 2, 1, 2, \dots$ : a pattern that repeats.

Quick check

1 of 4

The next term of 1, 3, 6, 10, 15, 21, … is .

Show answer

Answer 28

The gaps are 2, 3, 4, 5, 6, so the next gap is 7 and

21 + 7 = 28

8.2 Explicit Rule for a Sequence

Knowing the pattern of a sequence is useful. Knowing a formula for $t_{n}$ in terms of $n$ is even better. Such a formula is called an explicit rule: feed in the position, and out comes the term.

Definition

Explicit rule

A formula that gives the $n$ th term of a sequence directly from its position $n$ .

Example $t_{n} = n^{2}$ gives the square numbers, so $t_{12} = 144$ .

Example 1NCERT Example 1 · p. 177

Numbers that leave remainder 3

The rule $t_{n} = 4 n - 1$ describes a sequence. Write its first five terms and its 25th term. What do all its terms have in common?

Try it yourself first. Then reveal the solution one step at a time.

With the rule, the 250th term takes one line: $t_{250} = 4 \times 250 - 1 = 999$ . Adding 4 over and over would take 249 additions. A rule can also be run in reverse, from a value back to its position. Where does 199 sit in 3, 7, 11, …? If $199 = t_{n}$ , then $4 n - 1 = 199$ , so $4 n = 200$ and $n = 50$ : 199 is the 50th term.

Example 2NCERT Example 2 · p. 177

Rest stops on a highway

Rest stops along a highway are numbered 1, 2, 3, …, and rest stop $n$ stands at the $(7 n + 3)$ km mark, so the first few are at the 10, 17 and 24 km marks. One car breaks down at the 262 km mark and another at the 450 km mark. Is either car exactly at a rest stop? For a car that is not, which stop is nearest, and how far away is it?

Try it yourself first. Then reveal the solution one step at a time.

3/3 $7 n + 3 = 450$ gives $n \approx 63.9$ : 450 falls between the 63rd and 64th terms.

Fig. 8.4The terms of

7 n + 3

are 7 apart. A number is a term only if it lands exactly on one of them.

The top row is a number line with the first six terms of 7n + 3 marked as dots: 10, 17, 24, 31, 38 and 45, with a jump of +7 drawn between each pair of neighbours. In step 2 a zoomed number line near 262 appears, with unit tick marks and the terms 255, 262 and 269 labelled 36th, 37th and 38th; an arrow marks 262 and it lands exactly on the 37th term, so the answer reads yes, n = 37. In step 3 a second zoomed line near 450 shows the terms 444 (63rd) and 451 (64th) with the gap between them shaded; the arrow for 450 lands inside the gap, one short of 451, so the answer reads no, n is about 63.9.

The test always works the same way. Set the rule equal to the number and solve for $n$ . If $n$ comes out as a counting number, the number is a term and $n$ is its position. If $n$ is a fraction, zero or negative, the number does not appear in the sequence.

For the squares the rule is simply $t_{n} = n^{2}$ . The primes are a different story. People have studied them for more than two thousand years, and no simple formula gives the $n$ th prime. Not every sequence has a tidy explicit rule.

Quick check

1 of 4

Using

t_{n} = 4 n - 1

, find the 60th term.

Show answer

Answer 239

t_{60} = 4 \times 60 - 1 = 239

8.3 Recursive Rule for a Sequence

Often the easiest thing to say about a sequence is how it gets from one term to the next. A savings plan says “put in ₹5 more than last week”, and a staircase plan says “each step is 5 cm higher than the one below”. Instructions like these build a sequence one term at a time. Written with letters, “start at 6 and keep adding 5” becomes $t_{1} = 6$ and $t_{n} = t_{n - 1} + 5$ for $n \geq 2$ , where $t_{n - 1}$ means the term just before $t_{n}$ . It produces 6, 11, 16, 21, 26, …, the same list that the explicit rule $t_{n} = 5 n + 1$ gives.

A rule of this kind is a recursive rule (पुनरावर्ती नियम). On its own it cannot get going: it must be told the first term, because the first term has no term before it to be built from. Fig. 8.5 compares the two kinds of rule.

Fig. 8.5A recursive rule walks from term to term; an explicit rule jumps straight to the term you want.

The top row, titled walk, shows the terms 6, 11, 16, 21, 26 and 31 as rounded stepping stones, with a curved arrow marked +5 from each stone to the next and a note that it takes 5 steps to reach the 6th term: this is the recursive rule. The bottom row, titled jump, shows a box with n = 40 pointing to a box with 5 × 40 + 1, which points to the answer 201, the 40th term, found without walking through the 39 terms before it.

Definition

Recursive rule

A rule that gives each term of a sequence from one or more terms before it, together with the starting term or terms.

Example $t_{1} = 6$ and $t_{n} = t_{n - 1} + 5$ give 6, 11, 16, 21, …

Each kind of rule has its strength. A recursive rule is often the easier one to spot, since it says what happens from one step to the next. An explicit rule is the quicker one for far-off terms, since it needs no earlier terms at all.

Example 3NCERT Example 3 · p. 178

A tank that settles down

At 6 a.m. a school’s water tank holds 200 litres. Every hour, half of the water in the tank is used, and then a pump adds 40 litres. Let $u_{1} = 200$ be the amount at 6 a.m., $u_{2}$ the amount at 7 a.m., and so on, so that $u_{n} = \frac{1}{2} u_{n - 1} + 40$ for $n \geq 2$ . (a) Find the first six terms. (b) Is 100 a term? (c) Will the tank ever hold exactly 80 litres?

Try it yourself first. Then reveal the solution one step at a time.

Example 4NCERT Example 4 · p. 178

A sequence that races away

Start from $s_{1} = 2$ . Every later term is 1 more than the product of all the terms before it: $s_{n} = s_{1} \times s_{2} \times \dots \times s_{n - 1} + 1$ for $n \geq 2$ . List the first five terms.

Try it yourself first. Then reveal the solution one step at a time.

The sequence in Example 4 hides a surprise. Add the reciprocals of its terms: $\frac{1}{2} + \frac{1}{3} = \frac{5}{6}$ , then $\frac{5}{6} + \frac{1}{7} = \frac{41}{42}$ , then $\frac{41}{42} + \frac{1}{43} = \frac{1805}{1806}$ . Each new term brings the total closer to 1, yet the total never quite gets there.

The Virahānka–Fibonacci sequence

Poets in ancient India met a counting puzzle that needs a rule looking back two terms. In Sanskrit and Prakrit verse a syllable is either short, taking one beat, or long, taking two beats. How many different rhythms fill a line of $n$ beats? Fig. 8.6 lists them for 2, 3 and 4 beats.

Every rhythm ends with either a short or a long syllable. Take away that last syllable and what remains is a rhythm one beat shorter, or two beats shorter. So the number of rhythms for $n$ beats is the number for $n - 1$ beats plus the number for $n - 2$ beats. Starting from 1 rhythm for 1 beat and 2 rhythms for 2 beats, the counts run 1, 2, 3, 5, 8, 13, 21, … Written with the letter $V$ , the rule is $V_{1} = 1$ , $V_{2} = 2$ and $V_{n} = V_{n - 2} + V_{n - 1}$ for every $n$ from 3 on. This is the Virahānka–Fibonacci sequence.

4/4 $3 + 2 = 5$ rhythms of 4 beats. Next: $5 + 3 = 8$ .

Fig. 8.6Every rhythm ends in a short or a long syllable, so each count is the sum of the two counts before it.

A legend shows a short syllable as a small blue block worth 1 beat and a long syllable as a wide orange block worth 2 beats. Step 1 lists all patterns filling 2 beats (short-short, long: 2 ways) and 3 beats (short-short-short, short-long, long-short: 3 ways). Step 2 starts a column for 4 beats with the three patterns that end in a short syllable; each is a 3-beat pattern with a short added, the added block outlined in black. Step 3 adds the two patterns that end in a long syllable, each a 2-beat pattern with a long added. Step 4 counts 3 + 2 = 5 ways and lists the running counts 1, 2, 3, 5, 8, 13, … with the new count, 5, highlighted.

Timeline

Who counted the rhythms?

7th century CE
Virahānka
Gives the rule for counting rhythms in the Vṛttajātisamuccaya, a work on metre.
c. 1135 CE
Gopāla
Explains the same counting rule again in a commentary on metres.
c. 1150 CE
Hemachandra
Uses the rule in his own treatise on metre.
1202 CE
Leonardo of Pisa (Fibonacci)
Meets the same numbers in a puzzle about breeding rabbits, in his book Liber Abaci.

The same numbers keep turning up in mathematics and in nature, for example in the spirals of seeds on a sunflower head.

Quick check

1 of 4

In the Virahānka–Fibonacci sequence, the three terms after 8 are 13, 21 and 34. The two terms after those are and .

Show answer

Answer (1) 55, (2) 89

21 + 34 = 55

and then

34 + 55 = 89

Practice · 7 questions

Practice 8.1: questions with answers

1
Short answerEasyNCERT Exercise Set 8.1, Q1
Write the first five terms of each sequence.
1. $t_{n} = 4 n - 9$
  Show solution
  Answer $- 5, - 1, 3, 7, 11$
  $4 - 9 = - 5$ , $8 - 9 = - 1$ , $12 - 9 = 3$ , $16 - 9 = 7$ , $20 - 9 = 11$ .
2. $t_{n} = \frac{n + 3}{2}$
  Show solution
  Answer $2, \frac{5}{2}, 3, \frac{7}{2}, 4$
  $\frac{4}{2} = 2$ , $\frac{5}{2}$ , $\frac{6}{2} = 3$ , $\frac{7}{2}$ , $\frac{8}{2} = 4$ : the terms go up by $\frac{1}{2}$ each time.
3. $t_{n} = (- 1)^{n} (2 n + 1)$
  Show solution
  Answer $- 3, 5, - 7, 9, - 11$
  $(- 1)^{n}$ is $- 1$ when $n$ is odd and $+ 1$ when $n$ is even, so the signs take turns: $- 3$ , $5$ , $- 7$ , $9$ , $- 11$ .
2
Short answerEasyNCERT Exercise Set 8.1, Q2
For the sequence $t_{n} = 6 n + 5$ , find the terms below.
1. the 12th term
  Show answer
  Answer 77
  $6 \times 12 + 5 = 77$ .
2. the 20th term
  Show answer
  Answer 125
  $6 \times 20 + 5 = 125$ .
Short answerMediumNCERT Exercise Set 8.1, Q3
Priya says that 100 and 150 both appear in the sequence 9, 16, 23, 30, … Is she right? Find an explicit rule first.
Show solution
AnswerOnly half right: 100 is the 14th term, but 150 is not a term.
The terms go up by 7 and $t_{1} = 9 = 7 (1) + 2$ , so $t_{n} = 7 n + 2$ .
$7 n + 2 = 100$ gives $7 n = 98$ and $n = 14$ , a counting number. So $100 = t_{14}$ .
$7 n + 2 = 150$ gives $7 n = 148$ and $n = 21 \frac{1}{7}$ , not a counting number. So 150 is not a term: it lies between $t_{21} = 149$ and $t_{22} = 156$ .
Short answerEasyNCERT Exercise Set 8.1, Q4
Rahul counts on in eights, starting from 5: 5, 13, 21, 29, … When he says 461, how many numbers has he said?
Show answer
Answer 58
$t_{n} = 5 + 8 (n - 1) = 8 n - 3$ . Then $8 n - 3 = 461$ gives $8 n = 464$ , so $n = 58$ : 461 is the 58th number he says.
5
Short answerMediumNCERT Exercise Set 8.1, Q5
A test tube taken out of a freezer is at $- 10$ °C. For the next half hour its temperature rises by 3 °C every 5 minutes. You read the thermometer at once and then every 5 minutes, so the first reading is $t_{1} = - 10$ .
1. Write a recursive rule for the readings and list the first five.
  Show solution
  Answer $t_{1} = - 10$ , $t_{n} = t_{n - 1} + 3$ ; $- 10, - 7, - 4, - 1, 2$
  Each reading is 3 more than the one before, so $t_{1} = - 10$ and $t_{n} = t_{n - 1} + 3$ for $n \geq 2$ . The readings are $- 10$ , $- 7$ , $- 4$ , $- 1$ and 2 (in °C).
2. Which reading is 8 °C?
  Show answer
  Answer 7
  $t_{n} = - 10 + 3 (n - 1) = 3 n - 13$ , and $3 n - 13 = 8$ gives $n = 7$ : the 7th reading, taken 30 minutes after the first.
3. Does any reading show exactly 0 °C? Explain.
  Show solution
  AnswerNo.
  $3 n - 13 = 0$ gives $n = 4 \frac{1}{3}$ , not a counting number. The temperature passes 0 °C between the 4th reading ( $- 1$ °C) and the 5th (2 °C).
6
Short answerMediumNCERT Exercise Set 8.1, Q6
Start with $A_{1} = A_{2} = A_{3} = 1$ . After that, each term is the sum of the three terms just before it: $A_{n} = A_{n - 1} + A_{n - 2} + A_{n - 3}$ for $n \geq 4$ .
1. Find $A_{4}, A_{5}, A_{6}, A_{7}$ and $A_{8}$ .
  Show solution
  Answer3, 5, 9, 17, 31
  $A_{4} = 1 + 1 + 1 = 3$ , $A_{5} = 3 + 1 + 1 = 5$ , $A_{6} = 5 + 3 + 1 = 9$ , $A_{7} = 9 + 5 + 3 = 17$ , $A_{8} = 17 + 9 + 5 = 31$ .
2. Every term so far is odd. Explain why every later term will be odd too.
  Show solution
  AnswerThe sum of three odd numbers is odd.
  Odd + odd is even, and even + odd is odd, so the sum of three odd numbers is odd. The first three terms are odd, so $A_{4}$ is odd. Then $A_{5}$ is again a sum of three odd terms, and the same step repeats for ever.
Quick answerEasy
Which explicit rule gives the sequence 9, 14, 19, 24, …?
$t_{n} = 9 n + 5$ $t_{n} = 4 n + 5$ $t_{n} = 5 n + 4$ $t_{n} = 5 n + 9$
Show answer
Answer (C) $t_{n} = 5 n + 4$
$5 (1) + 4 = 9$ and $5 (2) + 4 = 14$ . The rule $5 n + 9$ would make 14 the first term.

8.4 Arithmetic Progressions

The most orderly sequences of all change by exactly the same amount at every step. They are everywhere once you start looking: seat numbers along a row, the marks on a ruler, the height of a pile of identical plates.

Steel tumblers in many Indian kitchens are made to stack. Each tumbler is 12 cm tall, and a tumbler placed inside another sticks out 2 cm above the rim of the one below. Fig. 8.7 shows a stack being built.

5/5Four glasses: the first 12 cm plus three extra 2 cm pieces.

Fig. 8.7Every extra tumbler adds the same 2 cm to the stack.NCERT Fig. 8.3 · p. 180

A centimetre ruler stands on the left and a steel tumbler stands beside it; a table on the right lists the number of glasses and the height of the stack. Step 1 shows one tumbler, 12 cm tall, with a dashed line from its rim to the ruler. Steps 2, 3 and 4 drop a second, third and fourth tumbler into the stack; each sits inside the one below and sticks out by 2 cm, so the heights read 14 cm, 16 cm and 18 cm and the table grows row by row. The last step writes the height of four glasses as 12 + 2 + 2 + 2 = 12 + 3 × 2.

The heights are 12, 14, 16, 18, … centimetres. Think about what a stack of $n$ tumblers is made of: one whole tumbler, 12 cm tall, and above it $n - 1$ rims of 2 cm each, one for every tumbler after the first. So its height is $t_{n} = 12 + (n - 1) \times 2 = 2 n + 10$ centimetres. Five tumblers stand 20 cm tall and eight stand 26 cm, and since $2 n + 10 = 100$ gives $n = 45$ , a stack of 45 tumblers is exactly 1 metre tall. Notice the $n - 1$ : the first tumbler does not add a 2 cm rim.

In 12, 14, 16, 18, … every term is exactly 2 more than the term before it. Sequences with this property are so common that they have their own name.

Definition

Arithmetic progression (AP)

A sequence that goes up or down by one fixed amount $d$ at every step, so that $t_{n} - t_{n - 1} = d$ for all $n \geq 2$ . We call this constant step $d$ the common difference.

Example12, 14, 16, 18, … is an AP with first term 12 and common difference 2.

We shorten arithmetic progression to AP (समांतर श्रेढ़ी). The fixed amount added is the common difference $d$ . To find it, subtract any term from the term after it: $d = t_{2} - t_{1} = t_{3} - t_{2} = \dots$ . The common difference can be negative. In 25, 18, 11, 4, −3, … each term is 7 less than the one before, so $d = - 7$ .

Every AP is fixed by two numbers: its first term $a$ and its common difference $d$ . Its terms are $a, a + d, a + 2 d, a + 3 d, \dots$ To get from position 1 to position $n$ we make $n - 1$ equal hops of size $d$ , as Fig. 8.8 shows.

4/4 $t_{n} = a + (n - 1) d$ : first term plus $n - 1$ hops.

Fig. 8.8To reach the

n

th term you make

n - 1

equal hops of size

d

from the first term.

The top number line marks the terms 12, 14, 16, 18 and 20 as blue dots. In step 2 a green arrow labelled +2 hops from each term to the next, and a note says the 5th term is 4 hops from the first: 12 + 4 × 2 = 20. In step 3 a second number line appears below with the terms 25, 18, 11, 4 and −3 as purple dots; dashed orange arrows labelled −7 hop leftwards from each term to the next, and a note gives the 5th term as 25 + 4 × (−7) = −3. The last step adds the general rule: the nth term is the first term plus n − 1 hops of size d.

PropertyThe

n

th term of an AP

If an AP has first term $a$ and common difference $d$ , then its $n$ th term is $t_{n} = a + (n - 1) d$ .

An AP also has a very simple recursive rule: $t_{1} = a$ and $t_{n} = t_{n - 1} + d$ for $n \geq 2$ . For the tumblers, $t_{1} = 12$ and $t_{n} = t_{n - 1} + 2$ . The recursive rule describes a single step; the explicit rule $t_{n} = a + (n - 1) d$ tells us where all the steps lead.

Example 5NCERT Example 5 · p. 183

Filling a piggy bank

On her birthday Kavya puts ₹120 into a new piggy bank. After that she adds ₹25 every Sunday. (a) Write the amount in the bank after 1, 2, 3 and 4 Sundays. (b) Find the amount after $n$ Sundays, and after 8 Sundays. (c) After how many Sundays does the amount first go above ₹500?

Try it yourself first. Then reveal the solution one step at a time.

The Sequence builder below lets you make your own APs and see each one in three ways at once: as a list, as a table of changes and ratios, and as dots on a graph. You can also ask it for far-off terms and test whether a number is a term. We will come back to it for a second family of sequences in Section 8.6.

InteractiveSequence builder

A sequence builder. Buttons choose the kind of rule: add d (an AP), multiply by r (a GP) or add the last two terms. Steppers set the first term and d, a menu sets r, and a stepper sets the second term for the add-the-last-two rule. A table lists the first eight terms with the change and the ratio from the term before, and says under each column whether it stays the same. A dot plot shows the terms against their positions, with an optional dashed line through the first two dots that rings every dot lying on it. A panel writes the explicit and recursive rules, a status line says whether the sequence is an AP or a GP, and two tools find the term at any position and test whether a number is a term. A Mystery mode shows the first four terms of a hidden AP or GP, with hints, for you to match.

Fig. 8.9Build an AP, a GP or a “add the last two” sequence, then compare its changes, ratios and graph.

Quick check

1 of 4

Which of these sequences are APs?

Select all that apply

Show answer

Answer

(A) the odd numbers 1, 3, 5, 7, …
(C) 10, 17, 24, 31, …
(E) 35, 28, 21, 14, …

The odd numbers (

d = 2

), 10, 17, 24, … (

d = 7

) and 35, 28, 21, … (

d = - 7

) have a constant difference. The triangular numbers have gaps 2, 3, 4, … and the Virahānka–Fibonacci numbers have gaps 1, 1, 2, 3, …, which keep changing.

8.4.1 Visualising an AP

Let us put the tumbler heights on a graph. Take the number of tumblers as $x$ and the height of the stack as $y$ , and plot the pairs in Table 8.2.

Table 8.2Height of a stack of tumblers

Number of tumblers	1	2	3	4	5	$n$
Height (cm)	12	14	16	18	20	$2 n + 10$

4/4Going back one more step gives $12 - 2 = 10$ at 0 glasses: $t_{n} = 2 n + 10$ .

Fig. 8.10Equal steps across give equal steps up, so the points of an AP line up.NCERT Fig. 8.4 · p. 182

A grid shows the number of glasses from 0 to 6 across and the height from 0 to 24 cm up. Step 1 plots the points (1, 12), (2, 14), (3, 16), (4, 18) and (5, 20) as blue dots. Step 2 draws one straight line through all of them. Step 3 adds a staircase under the line: from each point, a green step of 1 glass across and a purple step of 2 cm up reaches the next point. Step 4 extends the line backwards as a dashed segment to 0 glasses, where it meets the height axis at 10 cm, ringed and labelled.

The five points lie on one straight line, and that is no accident. One step to the right always means one more tumbler, and one more tumbler always adds the same 2 cm, so every step along the graph climbs by the same amount. Extend the line backwards and it meets the height axis at 10, the value the rule $2 n + 10$ gives for $n = 0$ .

PropertyThe points of an AP lie on a line

For any AP with first term $a$ and common difference $d$ , every point $(n, t_{n})$ lies on the straight line $y = d x + (a - d)$ .

The common difference of an AP is therefore the slope of its line. A large $d$ gives a steep line, and a negative $d$ gives a line that falls from left to right. In the simulation below you plot the points yourself.

SimulationPlot an AP, then a GP

A coordinate grid from x = −1 to 7 and y = −4 to 9 with two fixed points, A at (1, −3) and B at (2, −1), the first two terms of an AP. Tapping the grid, or moving the cursor with the arrow keys and pressing Enter, plots a new point that snaps to whole numbers; each point is labelled and listed with its coordinates. Undo and Clear remove your own points but keep A and B.

Fig. 8.11A and B are the points

(1, t_{1})

and

(2, t_{2})

of the AP −3, −1, 1, 3, 5, … Tap the grid to plot more terms.NCERT Exercise · p. 182

Try this

Plot the 3rd, 4th and 5th terms of the AP −3, −1, 1, 3, 5 at $x = 3, 4$ and $5$ . Do all five points lie on one straight line?
Count grid squares from each point to the next: how many across, and how many up? Compare the step up with the common difference of the AP.
Press Clear (A and B stay; leave them out this time), then plot $(1, 2)$ , $(2, 4)$ and $(3, 8)$ from the doubling sequence 2, 4, 8. Do these three points line up, or do they bend?

Quick check

1 of 3

The points

(n, t_{n})

of the AP 50, 46, 42, 38, … lie on a straight line that falls from left to right.

TrueFalse

Show answer

Answer True

Every AP gives points on a straight line. Here

d = - 4

, so each step to the right goes 4 down.

8.5 Sum of the First n Natural Numbers

Imagine a girl who has had a cake with the right number of candles on every birthday from her 1st to her 12th. How many candles has she blown out in all? We need $1 + 2 + 3 + \dots + 12$ . Adding one number at a time works, but there is a neater way that works just as well for much longer sums.

Write the numbers on a strip of paper and fold it at the middle, as in Fig. 8.12. The fold puts 12 on top of 1, 11 on top of 2, and so on down to 7 on top of 6. Every pair adds up to 13, and there are 6 pairs, so the total is $6 \times 13 = 78$ candles.

4/46 pairs of 13: $6 \times 13 = 78$ .

Fig. 8.12Folding pairs the smallest number with the largest: every pair has the same sum.

A strip of twelve cells numbered 1 to 12 lies across the top: 1 to 6 in blue on the left, 7 to 12 in orange on the right, with a dashed fold line between 6 and 7. In step 2 the right half folds under the left half, so 12 sits under 1, 11 under 2, and so on down to 7 under 6. In step 3 each column of two numbers is joined by a green bracket and marked 13. In step 4 the text says there are 6 pairs, each 13, so 6 × 13 = 78, and 1 + 2 + … + 12 = 78.

Folding pairs up the numbers neatly when there is an even count of them. A small change of view makes the idea work for every $n$ : instead of folding the list, write it out twice, once forwards and once backwards.

TheoremSum of the first

n

natural numbers

The natural numbers from 1 to $n$ add up to $\frac{n (n + 1)}{2}$ . That is, $1 + 2 + 3 + \dots + n = \frac{n (n + 1)}{2}$ .

Fig. 8.13 makes the same argument with blocks. A staircase with rows of 1, 2, 3, 4 and 5 blocks, together with a copy turned half a turn, exactly fills a rectangle 5 blocks tall and 6 blocks wide. So two staircases hold $5 \times 6 = 30$ blocks, and one staircase holds 15.

4/4Together: 5 rows of 6, so $2 S = 5 \times 6 = 30$ and $S = 15$ .

Fig. 8.13Two equal staircases make a rectangle, so one staircase is half of it.NCERT Fig. 8.5 · p. 184

Step 1 shows a staircase of blue square blocks with rows of 1, 2, 3, 4 and 5 blocks, each row labelled with its count. Step 2 shows an equal orange staircase to its right. Step 3 turns the orange staircase half a turn, so its long row is at the top. Step 4 slides it into the gaps beside the blue staircase: together they make a rectangle 6 blocks wide and 5 blocks tall, labelled 6 and 5, holding 5 × 6 = 30 blocks, so each staircase holds 15.

From now on $S_{n}$ is our short name for $1 + 2 + \dots + n$ . So $S_{n} = \frac{n (n + 1)}{2}$ ; for example, $S_{5} = 15$ and $S_{12} = 78$ .

For $1 + 2 + \dots + 15$ there are 15 columns, each adding up to 16, so $2 S = 15 \times 16 = 240$ and $S = 120$ . Reversing $1 + 2 + \dots + 100$ gives 100 columns, each adding up to 101. So $S_{100} = \frac{100 \times 101}{2} = 5050$ .

Example

A block of consecutive numbers

Find $31 + 32 + 33 + \dots + 70$ .

Try it yourself first. Then reveal the solution one step at a time.

The triangular numbers of Section 8.1 are exactly these sums: the $n$ th triangular number is $S_{n} = \frac{n (n + 1)}{2}$ . So the 12th triangular number is 78, the number of candles we counted.

Example

Adding the multiples of 7

Find the sum of all the multiples of 7 from 7 to 98.

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 4

Find

S_{40} = 1 + 2 + \dots + 40

Show answer

Answer 820

S_{40} = \frac{40 \times 41}{2} = 820

Practice · 8 questions

Practice 8.2: questions with answers

1
Short answerEasyNCERT Exercise Set 8.2, Q1
For the AP 7, 13, 19, 25, …, find the terms below.
1. the 12th term
  Show answer
  Answer 73
  $7 + 11 \times 6 = 73$ .
2. the 30th term
  Show answer
  Answer 181
  $7 + 29 \times 6 = 181$ .
2
Short answerMediumNCERT Exercise Set 8.2, Q2
Meena counts down in fours from 45: 45, 41, 37, 33, … She says, “If I keep going, I will land exactly on 0.”
1. Is Meena right? Give a reason.
  Show solution
  AnswerNo.
  $49 - 4 n = 0$ gives $n = 12.25$ , which is not a counting number. She says $t_{12} = 1$ and then $t_{13} = - 3$ , stepping over 0.
2. At which position does she first say a negative number?
  Show answer
  Answer 13
  $49 - 4 n < 0$ needs $n > 12.25$ , so the first negative term is $t_{13} = 49 - 52 = - 3$ .
Short answerEasyNCERT Exercise Set 8.2, Q3
Find the $n$ th term of the AP 14, 9, 4, −1, …, write a recursive rule for it, and find its 15th term.
Show solution
Answer $t_{n} = 19 - 5 n$ ; recursive rule $t_{1} = 14$ , $t_{n} = t_{n - 1} - 5$ ; $t_{15} = - 56$ .
Here $a = 14$ and $d = 9 - 14 = - 5$ , so $t_{n} = 14 + (n - 1) (- 5) = 19 - 5 n$ . Check $n = 4$ : $19 - 20 = - 1$ .
Each term is 5 less than the one before: $t_{1} = 14$ and $t_{n} = t_{n - 1} - 5$ for $n \geq 2$ .
Using the explicit rule, $t_{15} = 19 - 5 \times 15 = - 56$ .
4
Short answerHardNCERT Exercise Set 8.2, Q4
An auditorium has 30 rows of seats. Each row has the same number of extra seats compared with the row in front of it. Row 6 has 32 seats and row 14 has 48 seats.
1. How many seats are there in row 1?
  Show answer
  Answer 22
  Subtracting, $8 d = 16$ , so $d = 2$ . Then $a = 32 - 5 \times 2 = 22$ .
2. How many seats are there in the back row, row 30?
  Show answer
  Answer 80
  $t_{30} = 22 + 29 \times 2 = 80$ .
3. How many seats does the auditorium have in all?
  Show answer
  Answer 1530
  By Āryabhaṭa’s rule the total is $30 \times \frac{22 + 80}{2} = 30 \times 51 = 1530$ .
5
Short answerMediumNCERT Exercise Set 8.2, Q5
Consider the three-digit multiples of 8 that are less than 200.
1. How many are there?
  Show answer
  Answer 12
  They are $8 \times 13$ up to $8 \times 24$ , that is $24 - 13 + 1 = 12$ numbers.
2. What is their sum?
  Show answer
  Answer 1776
  $8 (13 + 14 + \dots + 24) = 8 (S_{24} - S_{12}) = 8 (300 - 78) = 1776$ . Āryabhaṭa's rule agrees: $12 \times \frac{104 + 192}{2} = 12 \times 148 = 1776$ .
Short answerMediumNCERT Exercise Set 8.2, Q6
A cycle factory makes 1200 bicycles in its first year and raises production by 150 bicycles every year. In which year does it first make more than 2800 bicycles?
Show answer
Answer 12
$(n - 1) \times 150 > 1600$ gives $n - 1 > 10.67$ , so the smallest counting number $n$ that works is 12. Check: year 11 gives 2700 bicycles, not enough, and year 12 gives 2850.
Short answerEasyNCERT Exercise Set 8.2, Q7
On the 1st day of a month Arjun puts ₹1 in a jar, on the 2nd day ₹2, on the 3rd day ₹3, and so on. How many rupees are in the jar once he has added the money for the 24th day?
rupees
Show answer
Answer 300 rupees
$S_{24} = \frac{24 \times 25}{2} = 300$ , so the jar holds ₹300.
Quick answerMedium
The sum $2 + 4 + 6 + \dots + 100$ equals
2550505025005100
Show answer
Answer (A) 2550
It is $2 (1 + 2 + \dots + 50) = 2 \times 1275 = 2550$ . The value 5050 is $1 + 2 + \dots + 100$ , which also adds the odd numbers.

8.6 Geometric Progressions

Now for a sequence that grows in a very different way. Fold a sheet of paper in half, then in half again, and keep going. After each fold, open the sheet and count the panels marked out by the creases. Fig. 8.14 shows what you find.

6/6Every fold multiplies the panels by 2 and the size of a panel by $\frac{1}{2}$ .

Fig. 8.14Each fold doubles the number of panels and halves the size of each panel.NCERT Fig. 8.6 · p. 186

A rectangular sheet is shown on the left and a table on the right with the columns folds, panels and 1 panel. Step 1 shows the whole sheet, shaded: 0 folds, 1 panel of size 1. Step 2 adds a dashed fold line down the middle: 1 fold, 2 panels, one panel shaded, size one half. Steps 3, 4 and 5 add fold lines across, then down, then across again, giving 4, 8 and 16 panels; the shaded panel shrinks to one quarter, one eighth and one sixteenth of the sheet, and the table grows row by row. The last step marks the panels column times 2 and the size column times one half.

The panel counts are 2, 4, 8, 16, … A fold does not add a fixed number of panels. It doubles them, because every panel is split in two. So after $n$ folds there are $2 \times 2 \times \dots \times 2 = 2^{n}$ panels. Five folds give 32 panels. Nine folds give only 512, but ten folds give $2^{10} = 1024$ , more than a thousand, and twenty folds more than ten lakh ( $2^{20} = 1 048 576$ ). In practice, an ordinary sheet folded by hand gets too thick to fold again after about seven folds.

Here the trick that worked for the tumblers fails: subtracting neighbours gives 2, 4, 8, …, which are not equal. Dividing works instead: $\frac{4}{2} = \frac{8}{4} = \frac{16}{8} = 2$ , so each count is exactly twice the one before. A list that grows or shrinks by the same multiplier at every step is a geometric progression.

Definition

Geometric progression (GP)

A sequence of non-zero numbers where dividing any term by the one before it always gives the same number $r$ , so that $t_{n} = r \times t_{n - 1}$ for all $n \geq 2$ . This constant multiplier $r$ is the common ratio.

Example2, 4, 8, 16, … is a GP with first term 2 and common ratio 2.

We shorten geometric progression to GP (गुणोत्तर श्रेढ़ी). The fixed multiplier is the common ratio $r$ . To find it, divide any term by the term before it: $r = \frac{t_{2}}{t_{1}} = \frac{t_{3}}{t_{2}} = \dots$ . A GP with first term $a$ runs $a, a r, a r^{2}, a r^{3}, \dots$

PropertyThe

n

th term of a GP

If a GP has first term $a$ and common ratio $r$ , then its $n$ th term is $t_{n} = a r^{n - 1}$ .

The recursive rule of a GP is just as short: $t_{1} = a$ and $t_{n} = r \times t_{n - 1}$ for $n \geq 2$ . Put the two families side by side. An AP adds the same number at every step, while a GP multiplies by the same number.

Example 6NCERT Example 6 · p. 187

Money that grows by 10%

A bank adds 10% to the money in a savings account at the end of every year. Rohan’s balance at the start of years 1, 2, 3 and 4 is ₹2000, ₹2200, ₹2420 and ₹2662. Show that these amounts form a GP, and find his balance at the start of year 6 if he takes nothing out.

Try it yourself first. Then reveal the solution one step at a time.

Example 7NCERT Example 7 · p. 187

Halving each time

A ribbon is cut in half, then one of the halves is cut in half, and so on. The lengths, in cm, are 64, 32, 16, 8, … Check that these lengths form a GP, and write the next three.

Try it yourself first. Then reveal the solution one step at a time.

Example 8NCERT Example 8 · p. 187

Signs that flip

Two lists start in the same way: 2, −6, 18, −54, … and 2, −6, 18, −30, … Which of them is a GP?

Try it yourself first. Then reveal the solution one step at a time.

Example 9NCERT Example 9 · p. 187

A ratio that is a fraction

Show that 8, 12, 18, 27, … is a GP and find its $n$ th term. What is its 5th term?

Try it yourself first. Then reveal the solution one step at a time.

8.6.1 Fun with Fractals

The Sierpiński triangle, one of the best-known fractals, comes from a single instruction repeated for ever: in every shaded triangle, join the midpoints of the three sides and cut out the upside-down triangle in the middle. Fig. 8.15 begins at stage 0 with a solid triangle whose three sides are equal. One round of cutting gives stage 1, a second round on every piece that is left gives stage 2, and the rounds never have to stop. The Sierpiński triangle is the shape that these stages get closer and closer to.

5/5Triangles: $\times 3$ each stage. Area: $\times \frac{3}{4}$ each stage.

Fig. 8.15Each stage replaces every shaded triangle by three smaller ones and removes a quarter of the shaded area.NCERT Fig. 8.7 · p. 188

A large shaded equilateral triangle sits on the left; a panel on the right gives the stage, the number of shaded triangles and the shaded area. Step 1 is stage 0: one triangle, area 1. Step 2 is stage 1: joining the midpoints of the sides cuts out an upside-down middle triangle, leaving 3 shaded triangles with area three quarters. Step 3 is stage 2: the middle of each shaded triangle is cut out, leaving 9 triangles with area nine sixteenths. Step 4 is stage 3: 27 triangles, area twenty-seven sixty-fourths. The last step adds times 3 each stage beside the count and times three quarters each stage beside the area.

Think and reflectNCERT Think and Reflect · p. 189

Count and measure

Before reading on, fill in the first row from Fig. 8.15: count the shaded triangles at stages 0 to 3, then predict stages 4, 5 and $n$ .

For the second row, give the solid triangle at stage 0 an area of 1 square unit and work out how much each round of cutting leaves.

Stage	0	1	2	3	4	5	$n$
Shaded triangles
Shaded area

Your entries are saved automatically.

Now look along the area row. Where is the shaded area heading as the stages go on and on?

Think about what one cut does to a single shaded triangle: it leaves three shaded triangles behind, each half as wide. So whatever the count was, the next stage has three times as many: $1 \to 3 \to 9 \to 27$ . After $n$ stages we have multiplied by 3 exactly $n$ times, which gives $3^{n}$ triangles; stage 5 already has 243. (We number the stages from 0 here, so the first term is the count at stage 0.)

The area goes the other way. The midpoint lines split every shaded triangle into four equal pieces, and only three of them stay shaded. So each stage keeps $\frac{3}{4}$ of the shaded area it began with, and after $n$ stages the area is ${(\frac{3}{4})}^{n}$ of the original: $1, \frac{3}{4}, \frac{9}{16}, \frac{27}{64}, \dots$

Table 8.3Two GPs from the Sierpiński triangle

Stage	0	1	2	3	4	5	$n$
Shaded triangles	1	3	9	27	81	243	$3^{n}$
Shaded area	1	$\frac{3}{4}$	$\frac{9}{16}$	$\frac{27}{64}$	$\frac{81}{256}$	$\frac{243}{1024}$	${(\frac{3}{4})}^{n}$

So one fractal gives us two GPs at once, with common ratios 3 and $\frac{3}{4}$ . In symbols, $t_{n} = 3^{n}$ with $t_{0} = 1$ and $t_{n} = 3 t_{n - 1}$ , while $s_{n} = {(\frac{3}{4})}^{n}$ with $s_{0} = 1$ and $s_{n} = \frac{3}{4} s_{n - 1}$ . The number of triangles grows without limit, yet their total area keeps shrinking. By stage 10 the area is about 0.06, and by stage 20 it is below 0.004: it gets as close to 0 as we like.

Definition

Fractal

A shape made of smaller and smaller copies of itself, usually built by repeating one simple step again and again.

ExampleThe Sierpiński triangle: each shaded piece is a smaller copy of the whole.

8.6.2 Visualising a GP

What do the terms of a GP look like on a graph? Plot the paper-folding counts from Table 8.4, with the number of folds across and the number of panels up.

Table 8.4Panels made by folding a sheet in half again and again

Folds	0	1	2	3	4	5	6	$n$
Panels	1	2	4	8	16	32	64	$2^{n}$

4/4With $r = \frac{1}{2}$ the points fall and flatten out towards 0.

Fig. 8.16Doubling makes the points bend away from any straight line; halving makes them sink towards 0.NCERT Fig. 8.9 · p. 191

The left graph has folds from 0 to 6 across and panels from 0 to 64 up. Step 1 plots (0, 1), (1, 2), (2, 4), (3, 8), (4, 16), (5, 32) and (6, 64) as blue dots, and a list on the right gives the change from each point to the next: +1, +2, +4, +8, +16, +32, doubling each time. Step 2 draws a dashed straight line through the first two points; at 6 folds it reaches only 7, far below the point at 64. Step 3 joins the dots with a smooth curve that bends upwards more and more steeply. Step 4 replaces that list with a second graph of the size of one panel, from 1 down to one sixty-fourth: its green dots halve each time and the curve flattens towards 0 without reaching it.

This time the points do not lie on a straight line. A line through the first two points climbs by 1 panel per fold, but the real counts climb by 1, then 2, then 4, then 8: the rise itself doubles at every step. So the points bend upwards along a curve that gets steeper and steeper. A GP whose ratio lies between 0 and 1, like the size of one panel, bends the other way. Its points drop quickly at first, then level off, getting closer and closer to 0.

This is the clearest picture of the difference between the two families. The points of an AP always lie on a straight line. The points of a GP never do, unless its ratio is 1 and every term is the same.

Try thisNCERT Fig. 8.10 · p. 191

Back to the Sequence builder

Open the Sequence builder again.

Choose Multiply by r (GP) with first term 2 and $r = 2$ , the paper-folding panels. Tick Line through the first two dots. How many dots lie on the line now? Look at the change column: what do the changes themselves form?
Set the first term to 1 and $r = 3$ , like the Sierpiński counts. Then set $r = \frac{3}{4}$ , like their areas. Describe the two dot patterns.
Now set $r = - 2$ . What happens to the signs of the terms, and to the dots?

Example 10NCERT Example 10 · p. 192

Filtering muddy water

A purifier passes muddy water through a row of filter stages. Each stage lets through only 60% of the dirt that reaches it, and the water going in carries 500 units of dirt. (a) Find the dirt left after each of the first five stages, and after $n$ stages. (b) Water from the purifier may carry at most 50 units of dirt. How many stages does it need? (c) A better kind of stage lets through only 40% of the dirt. How many of those would be needed?

Try it yourself first. Then reveal the solution one step at a time.

4/4With a limit of 50 units, 5 stages are needed: 4 stages still leave 64.8.

Fig. 8.17Every stage lets through 60% of the dirt that reaches it.NCERT Fig. 8.11 · p. 193

Six jars of water stand in a row, joined by arrows through small filters: one jar for the water before filtering and one after each of five filter stages. The dirt settled at the bottom of each jar is drawn to scale, and its amount is written above the jar. Step 1 shows only the first jar, with 500 units of dirt. Step 2 adds the jars after stages 1, 2 and 3, holding 300, 180 and 108 units, with the note that each stage multiplies the dirt by 0.6. Step 3 adds the jars after stages 4 and 5, with 64.8 and 38.9 units. Step 4 draws a dashed line across all the jars at the limit of 50 units: the dirt after stage 4 still reaches above the line and is marked too much, while the jar after stage 5, ringed in green and marked OK, is the first below it.

Many real quantities shrink like this: a medicine in the blood, the resale value of a scooter, the loudness of a fading echo. Each keeps the same fraction of itself over each step of time, and that makes a GP with a ratio between 0 and 1.

Quick check

1 of 3

Which description fits the points

(n, t_{n})

of the GP 1, 3, 9, 27, …?

they lie on a curve that gets steeper and steeperthey lie on a straight line going upthey lie on a curve that levels off towards 0they zigzag above and below the axis

Show answer

Answer (A) they lie on a curve that gets steeper and steeper

The ratio 3 is bigger than 1, so each rise is 3 times the one before and the curve keeps getting steeper. A straight line needs equal rises, a curve levelling off needs a ratio between 0 and 1, and a zigzag needs a negative ratio.

Question 7 below uses the fractal in Fig. 8.18. At every stage each shaded square is cut into 9 equal squares, and only the centre square and the four squares in the middles of the sides are kept.

4/4Repeat once more.

Fig. 8.18Stages 0 to 3 of the cross fractal used in question 7.NCERT Fig. 8.12 · p. 194

A large shaded square is shown with a small key on the right: a 3 by 3 grid in which the centre square and the four squares touching the middle of each side are kept, and the four corner squares are removed. Step 1 is stage 0, the whole square. Step 2 is stage 1: the four corner squares are cut away, leaving a plus-shaped cross. Step 3 is stage 2: the same cut is made on every kept square, so each arm of the cross becomes a small cross. Step 4 is stage 3, with the cut repeated once more on every kept square.

Practice · 8 questions

Practice 8.3: questions with answers

1
Short answerEasyNCERT Exercise Set 8.3, Q1
In a GP, $t_{4} = 54$ and every term is 3 times the term before it.
1. Find $t_{1}$ .
  Show answer
  Answer 2
  Going back from the 4th term to the 1st divides by 3 three times: $54 \div 27 = 2$ .
2. Find $t_{7}$ .
  Show answer
  Answer 1458
  Going on from the 4th term to the 7th multiplies by 3 three times: $54 \times 27 = 1458$ .
2
Short answerEasyNCERT Exercise Set 8.3, Q2
For the GP 4, 12, 36, …:
1. Find the 8th term.
  Show answer
  Answer 8748
  $t_{8} = 4 \times 3^{7} = 4 \times 2187 = 8748$ .
2. Find the $n$ th term.
  Show solution
  Answer $t_{n} = 4 \times 3^{n - 1}$
  $t_{n} = a r^{n - 1} = 4 \times 3^{n - 1}$ .
3
Long answerHardNCERT Exercise Set 8.3, Q3
Tara starts with 1 sheet of paper. In each round she cuts every piece she has into 3 pieces and then gives 1 piece away. Let $t_{1} = 1$ be the number of pieces before any cutting, $t_{2}$ the number after round 1, and so on.
1. Write a recursive rule for $t_{n}$ and find $t_{2}$ to $t_{6}$ .
  Show solution
  Answer $t_{n} = 3 t_{n - 1} - 1$ ; 2, 5, 14, 41, 122
  Cutting every piece into 3 triples the count, and giving one away takes 1 off, so $t_{1} = 1$ and $t_{n} = 3 t_{n - 1} - 1$ for $n \geq 2$ . Then $t_{2} = 2$ , $t_{3} = 5$ , $t_{4} = 14$ , $t_{5} = 41$ and $t_{6} = 122$ .
2. Show that the numbers $2 t_{n} - 1$ form a GP, and use this to write an explicit rule for $t_{n}$ .
  Show solution
  Answer $2 t_{n} - 1 = 3^{n - 1}$ , so $t_{n} = \frac{3^{n - 1} + 1}{2}$ .
  From the rule, $2 t_{n} - 1 = 6 t_{n - 1} - 3 = 3 (2 t_{n - 1} - 1)$ . So each value of $2 t_{n} - 1$ is 3 times the one before: 1, 3, 9, 27, 81, …, a GP with first term 1 and ratio 3.
  Hence $2 t_{n} - 1 = 3^{n - 1}$ and $t_{n} = \frac{3^{n - 1} + 1}{2}$ . Check: $t_{6} = \frac{243 + 1}{2} = 122$ .
3. Which term of the sequence is 3281? After how many rounds does Tara have that many pieces?
  Show answer
  Answer 9
  $\frac{3^{n - 1} + 1}{2} = 3281$ gives $3^{n - 1} = 6561 = 3^{8}$ , so $n = 9$ : the 9th term, reached after 8 rounds, since $t_{1}$ counts the pieces before any cutting.
Short answerMediumNCERT Exercise Set 8.3, Q4
Aditi plays a doubling game on a chessboard: 7 grains of rice on square 1, 14 on square 2, 28 on square 3, and so on, doubling each time. Which square holds 3584 grains? Write the explicit rule and the recursive rule for the number of grains on square $n$ .
Show solution
AnswerSquare 10; $t_{n} = 7 \times 2^{n - 1}$ ; $t_{1} = 7$ , $t_{n} = 2 t_{n - 1}$ .
The grains form a GP with $a = 7$ and $r = 2$ , so $t_{n} = 7 \times 2^{n - 1}$ .
$7 \times 2^{n - 1} = 3584$ gives $2^{n - 1} = 512 = 2^{9}$ , so $n - 1 = 9$ and $n = 10$ : square 10.
Recursive rule: $t_{1} = 7$ and $t_{n} = 2 t_{n - 1}$ for $n \geq 2$ .
5
Long answerMediumNCERT Exercise Set 8.3, Q5
A frog makes a series of hops along a straight path. Its first hop is 125 cm long, and every hop is 80% as long as the one before.
1. How long is its 5th hop?
  cm
  Show answer
  Answer 51.2 cm
  $125 \times {0.8}^{4} = 125 \times 0.4096 = 51.2$ cm.
2. How far has it travelled after 6 hops?
  cm
  Show answer
  Answer 461.16 cm
  The hops are 125, 100, 80, 64, 51.2 and 40.96 cm, which add up to 461.16 cm.
Short answerMediumNCERT Exercise Set 8.3, Q6
Draw a square of side 1 cm. Draw a second square whose side is the diagonal of the first, a third whose side is the diagonal of the second, and so on. A square’s diagonal is $\sqrt{2}$ times its side, so the sides are $1, \sqrt{2}, 2, 2 \sqrt{2}, \dots$ cm. Which square has side 32 cm?
Show answer
Answer 11
The sides form a GP with $a = 1$ and $r = \sqrt{2}$ , so the $n$ th side is $(\sqrt{2})^{n - 1}$ . Since $32 = 2^{5} = (\sqrt{2})^{10}$ , we need $n - 1 = 10$ : the 11th square. (Their areas 1, 2, 4, 8, … double each time.)
7
Long answerHardNCERT Exercise Set 8.3, Q7
Look at the cross fractal in Fig. 8.18, and take the area of the whole square at stage 0 as 1 square unit.
1. Count the shaded squares at stage 3. Then write an explicit rule and a recursive rule for the count at stage $n$ .
  Show solution
  Answer125; $t_{n} = 5^{n}$ ; $t_{0} = 1$ , $t_{n} = 5 t_{n - 1}$ .
  Every shaded square is replaced by 5 smaller ones, so the count is multiplied by 5 at each stage: 1, 5, 25, 125. The counts are powers of 5, so $t_{n} = 5^{n}$ , with $t_{0} = 1$ and $t_{n} = 5 t_{n - 1}$ .
2. At which stage does the number of shaded squares first go above 10 000?
  Show answer
  Answer 6
  $5^{5} = 3125$ is still below 10 000, but $5^{6} = 15 625$ is above it: stage 6.
3. What is the shaded area at stage 2? Write an explicit rule and a recursive rule for the area at stage $n$ .
  Show solution
  Answer $\frac{25}{81}$ ; $s_{n} = {(\frac{5}{9})}^{n}$ ; $s_{0} = 1$ , $s_{n} = \frac{5}{9} s_{n - 1}$ .
  Each stage keeps 5 of the 9 equal parts of every shaded square, so the area is multiplied by $\frac{5}{9}$ : $1, \frac{5}{9}, \frac{25}{81}, \dots$ Hence $s_{n} = {(\frac{5}{9})}^{n}$ , with $s_{0} = 1$ and $s_{n} = \frac{5}{9} s_{n - 1}$ .
4. At which stage is less than one tenth of the square first left shaded? What happens to the shaded area in the long run?
  Show solution
  AnswerStage 4; it shrinks towards 0.
  ${(\frac{5}{9})}^{3} = \frac{125}{729} \approx 0.171$ is more than 0.1, but ${(\frac{5}{9})}^{4} = \frac{625}{6561} \approx 0.095$ is less: stage 4. Every later stage multiplies the area by $\frac{5}{9}$ again, so the area gets as close to 0 as we like, even though the number of squares keeps growing.
Quick answerMedium
The AP 5, 10, 15, … and the GP 5, 10, 20, … begin the same way. By how much is the 10th term of the GP bigger than the 10th term of the AP?
Show answer
Answer 2510
GP: $5 \times 2^{9} = 2560$ . AP: $5 + 9 \times 5 = 50$ . The difference is $2560 - 50 = 2510$ .

Recap

Predicting What Comes Next: Exploring Sequences and Progressions: key points and summary

A sequence is a list of numbers in a definite order. Its members are terms, and $t_{n}$ is the term in position $n$ (Section 8.1).
Sequences can be finite or infinite, and their terms can be fractions, negative numbers or any numbers at all (Naming the terms).
Triangular numbers are running totals $1 + 2 + \dots + n$ ; square numbers are running totals of the odd numbers (Section 8.1).
An explicit rule gives $t_{n}$ straight from $n$ . To test whether a number is a term, solve for $n$ and check that it is a counting number (Section 8.2).
A recursive rule builds each term from earlier ones and needs starting terms, as in the Virahānka–Fibonacci rule $V_{n} = V_{n - 1} + V_{n - 2}$ (Section 8.3).
An AP adds the same common difference $d$ at every step: $t_{n} = a + (n - 1) d$ , and recursively $t_{n} = t_{n - 1} + d$ (Section 8.4).
The points $(n, t_{n})$ of an AP lie on a straight line whose slope is $d$ (Section 8.4.1).
$1 + 2 + \dots + n = \frac{n (n + 1)}{2}$ , found by pairing numbers from the two ends; this is also the $n$ th triangular number (Section 8.5).
Āryabhaṭa's rule: the sum of an AP is the average of its first and last terms times the number of terms (Āryabhaṭa's rule).
A GP multiplies by the same common ratio $r$ at every step: $t_{n} = a r^{n - 1}$ , and recursively $t_{n} = r t_{n - 1}$ (Section 8.6).
The points of a GP do not lie on a line: with $r > 1$ they curve upwards, and with $0 < r < 1$ they fall towards 0 (Section 8.6.2).
Fractals such as the Sierpiński triangle give GPs: at stage $n$ there are $3^{n}$ shaded triangles with total area ${(\frac{3}{4})}^{n}$ (Section 8.6.1).

Flash cards · 10

Key terms

1 / 10

Practice · 18 questions

Practice questions with answers (NCERT “End-of-chapter exercises”)

1
Short answerMediumNCERT End-of-chapter Exercises, Q1
A hill road climbs steadily. At the 8 km stone it is 1250 m above sea level, and at the 20 km stone it is 1850 m above sea level. The height goes up by the same amount for every kilometre.
1. How high is the road at the 35 km stone?
  m
  Show answer
  Answer 2600 m
  Over $20 - 8 = 12$ km the road rises $1850 - 1250 = 600$ m, so it rises 50 m per km. From the 8 km stone to the 35 km stone is 27 km: $1250 + 27 \times 50 = 2600$ m.
2. At which kilometre stone is the road exactly 2500 m high?
  Show answer
  Answer 33
  $1250 + (n - 8) \times 50 = 2500$ gives $n - 8 = 25$ , so it is the 33 km stone.
Short answerMediumNCERT End-of-chapter Exercises, Q2
In an AP, the 5th term is three times the 2nd term, and the 8th term is 45. Write out the first few terms.
Show solution
Answer3, 9, 15, 21, …
$a + 4 d = 3 a + 3 d$ gives $d = 2 a$ . Then $a + 7 d = a + 14 a = 15 a = 45$ , so $a = 3$ and $d = 6$ . The AP is 3, 9, 15, 21, 27, … Check: $t_{2} = 9$ and $t_{5} = 27 = 3 \times 9$ .
Short answerMediumNCERT End-of-chapter Exercises, Q3
How many three-digit numbers leave a remainder of 2 when divided by 9?
Show answer
Answer 100
The smallest is $101 = 9 \times 11 + 2$ and the largest is $992 = 9 \times 110 + 2$ . Solving $101 + (n - 1) \times 9 = 992$ gives $n - 1 = 99$ , so there are 100 of them.
Short answerMediumNCERT End-of-chapter Exercises, Q4
Lamps stand along a straight road at the 0 m mark, the 6 m mark, the 12 m mark, and so on, one every 6 m. How many lamps stand between the 20 m mark and the 290 m mark?
Show answer
Answer 45
The lamps in this stretch stand at 24, 30, …, 288 m. Solving $24 + (n - 1) \times 6 = 288$ gives $n - 1 = 44$ , so there are 45 lamps.
Long answerHardNCERT End-of-chapter Exercises, Q5
A GP has 3rd term 18 and 5th term 162. Find every GP that fits.
Show solution
Answer2, 6, 18, 54, 162, … (with $r = 3$ ) or 2, −6, 18, −54, 162, … (with $r = - 3$ ).
$\frac{a r^{4}}{a r^{2}} = r^{2} = \frac{162}{18} = 9$ , so $r = 3$ or $r = - 3$ .
Then $a r^{2} = 18$ gives $9 a = 18$ , so $a = 2$ in both cases.
So the GP is 2, 6, 18, 54, 162, … or 2, −6, 18, −54, 162, … Both fit, because the 3rd and 5th terms use even powers of $r$ .
6
Long answerHardNCERT End-of-chapter Exercises, Q6
Some numbers can be written as a sum of two or more consecutive natural numbers, for example $9 = 4 + 5 = 2 + 3 + 4$ .
1. Write 45 as such a sum in every possible way.
  Show solution
  Answer $22 + 23$ , $14 + 15 + 16$ , $7 + 8 + 9 + 10 + 11$ , $5 + 6 + \dots + 10$ and $1 + 2 + \dots + 9$ .
  Suppose $k$ consecutive numbers start at $m$ . By Āryabhaṭa’s rule their sum is $k$ times the average $m + \frac{k - 1}{2}$ . Setting this equal to 45 and doubling gives $k (2 m + k - 1) = 90$ .
  So $k$ and $2 m + k - 1$ are two factors of 90 that multiply to 90. Their difference, $2 m - 1$ , is odd, so one factor is even and the other odd. Also $2 m + k - 1 > k$ , because $m \geq 1$ .
  With $k \geq 2$ the possible pairs are $2 \times 45$ , $3 \times 30$ , $5 \times 18$ , $6 \times 15$ and $9 \times 10$ . They give $m = 22$ , 14, 7, 5 and 1: five ways in all.
2. Now try 32. What do you find, and why?
  Show solution
  AnswerIt cannot be done.
  Now $k (2 m + k - 1) = 64$ , and one of the two factors must be odd. The only odd factor of 64 is 1, but $k \geq 2$ and $2 m + k - 1 > k$ , so neither factor can be 1. Hence 32 is not a sum of two or more consecutive natural numbers. The same is true of every power of 2.
7
Short answerMediumNCERT End-of-chapter Exercises, Q7
A new video is shared online. At the start 50 people have watched it, and the number of people who have watched it triples every hour.
1. How many people have watched it after 4 hours, and after $n$ hours?
  Show solution
  Answer4050 and $50 \times 3^{n}$ .
  After 4 hours: $50 \times 3^{4} = 50 \times 81 = 4050$ . After $n$ hours: $50 \times 3^{n}$ .
2. After how many whole hours have more than 1 lakh people watched it?
  hours
  Show answer
  Answer 7 hours
  $50 \times 3^{n} > 100 000$ needs $3^{n} > 2000$ . Since $3^{6} = 729$ and $3^{7} = 2187$ , it happens after 7 hours, when $50 \times 2187 = 109 350$ people have watched.
Long answerHardNCERT End-of-chapter Exercises, Q8
The first three terms of an increasing AP add up to 27 and multiply to 585. Find them.
Show solution
Answer5, 9, 13
Their sum is $(m - d) + m + (m + d) = 3 m = 27$ , so the middle term is $m = 9$ .
Their product is $(9 - d) \times 9 \times (9 + d) = 9 (81 - d^{2}) = 585$ , so $81 - d^{2} = 65$ and $d^{2} = 16$ .
The AP is increasing, so $d = 4$ and the terms are 5, 9 and 13. Check: $5 + 9 + 13 = 27$ and $5 \times 9 \times 13 = 585$ .
Short answerMediumNCERT End-of-chapter Exercises, Q9
Find the smallest $n$ for which $1 + 2 + \dots + n$ is greater than 2026.
Show answer
Answer 64
$S_{63} = \frac{63 \times 64}{2} = 2016$ , which is not enough, but $S_{64} = \frac{64 \times 65}{2} = 2080 > 2026$ .
Short answerMediumNCERT End-of-chapter Exercises, Q10
In a knockout chess tournament, half of the players are knocked out in each round. Round 1 has 1024 players, round 2 has 512, and so on. Which round has 8 players? Write the explicit rule and the recursive rule for the number of players in round $n$ .
Show solution
AnswerRound 8; $t_{n} = 1024 \times {(\frac{1}{2})}^{n - 1}$ ; $t_{1} = 1024$ , $t_{n} = \frac{1}{2} t_{n - 1}$ .
Here $a = 1024$ and $r = \frac{1}{2}$ , so $t_{n} = 1024 \times {(\frac{1}{2})}^{n - 1} = \frac{1024}{2^{n - 1}}$ .
$\frac{1024}{2^{n - 1}} = 8$ gives $2^{n - 1} = 128 = 2^{7}$ , so $n = 8$ : round 8.
Recursive rule: $t_{1} = 1024$ and $t_{n} = \frac{1}{2} t_{n - 1}$ for $n \geq 2$ .
Long answerHardNCERT End-of-chapter Exercises, Q11
The length, breadth and height of a cuboid box, in that order, form a GP. The box has volume $1000 c m^{3}$ and total surface area $700 c m^{2}$ . Find its three edge lengths.
Show solution
Answer5 cm, 10 cm and 20 cm (in one order or the other).
Volume: $\frac{a}{r} \times a \times a r = a^{3} = 1000$ , so $a = 10$ .
Surface area: $2 (\frac{a}{r} \times a + a \times a r + a r \times \frac{a}{r}) = 2 a^{2} (\frac{1}{r} + r + 1) = 200 (\frac{1}{r} + r + 1) = 700$ .
So $\frac{1}{r} + r = 2.5$ . Multiplying by $2 r$ gives $2 r^{2} - 5 r + 2 = 0$ , that is $(2 r - 1) (r - 2) = 0$ , so $r = 2$ or $r = \frac{1}{2}$ .
Either way the edges are 5 cm, 10 cm and 20 cm. Check: $5 \times 10 \times 20 = 1000$ and $2 (50 + 200 + 100) = 700$ .
ProofHardNCERT End-of-chapter Exercises, Q12
Take any GP $t_{1}, t_{2}, t_{3}, \dots$ with common ratio $r$ . Prove that the products of neighbouring terms, $t_{1} t_{2}, t_{2} t_{3}, t_{3} t_{4}, \dots$ , also form a GP, and find its common ratio.
Show solution
AnswerThey form a GP with common ratio $r^{2}$ .
Let the GP have first term $a$ , so $t_{n} = a r^{n - 1}$ and $t_{n + 1} = a r^{n}$ . The $n$ th product is $p_{n} = t_{n} t_{n + 1} = a^{2} r^{2 n - 1}$ .
$\frac{p_{n + 1}}{p_{n}} = \frac{a^{2} r^{2 n + 1}}{a^{2} r^{2 n - 1}} = r^{2}$
The ratio is $r^{2}$ for every $n$ , so the products form a GP with common ratio $r^{2}$ . For example, the GP 2, 6, 18, 54 gives the products 12, 108 and 972, each 9 times the one before.
Long answerMediumNCERT End-of-chapter Exercises, Q13
In a GP, the 1st and 3rd terms add up to 50, and the 2nd and 4th terms add up to 150. Find the GP.
Show solution
Answer5, 15, 45, 135, …
$a + a r^{2} = a (1 + r^{2}) = 50$ and $a r + a r^{3} = a r (1 + r^{2}) = 150$ .
Dividing the second equation by the first gives $r = 3$ (the factor $1 + r^{2}$ is never 0).
Then $a (1 + 9) = 50$ , so $a = 5$ . The GP is 5, 15, 45, 135, … Check: $5 + 45 = 50$ and $15 + 135 = 150$ .
14
Long answerHardNCERT End-of-chapter Exercises, Q14
Let $R_{1} = 2$ and, for $n \geq 2$ , let $R_{n} = 2 (R_{1} + R_{2} + \dots + R_{n - 1})$ , that is, twice the total of all the terms so far.
1. Find $R_{1}, R_{2}, \dots, R_{8}$ .
  Show solution
  Answer2, 4, 12, 36, 108, 324, 972, 2916
  $R_{2} = 2 \times 2 = 4$ , $R_{3} = 2 (2 + 4) = 12$ , $R_{4} = 2 (2 + 4 + 12) = 36$ , $R_{5} = 2 (54) = 108$ , $R_{6} = 2 (162) = 324$ , $R_{7} = 2 (486) = 972$ and $R_{8} = 2 (1458) = 2916$ .
2. Find a simpler recursive rule, and an explicit rule. From which term onwards do they work?
  Show solution
  Answer $R_{n} = 3 R_{n - 1}$ for $n \geq 3$ ; $R_{n} = 4 \times 3^{n - 2}$ for $n \geq 2$ (with $R_{1} = 2$ on its own).
  For $n \geq 3$ , the total inside the rule for $R_{n}$ is the total for $R_{n - 1}$ plus one more term, $R_{n - 1}$ . The old total, doubled, is $R_{n - 1}$ itself, so $R_{n} = R_{n - 1} + 2 R_{n - 1} = 3 R_{n - 1}$ .
  This fails at the start: $R_{2} = 4$ , but $3 R_{1} = 6$ . So the terms from $R_{2}$ on form a GP with first term 4 and ratio 3, giving $R_{n} = 4 \times 3^{n - 2}$ for $n \geq 2$ , while $R_{1} = 2$ stands apart. Check: $R_{8} = 4 \times 3^{6} = 2916$ .
15
Long answerHardNCERT End-of-chapter Exercises, Q15
Let $V_{1}, V_{2}, V_{3}, \dots$ be the Virahānka–Fibonacci sequence 1, 2, 3, 5, 8, …, and let $Q_{n} = V_{1} + V_{2} + \dots + V_{n}$ be its running totals.
1. Find $Q_{1}$ to $Q_{6}$ and compare them with the terms of the Virahānka–Fibonacci sequence. What do you notice?
  Show solution
  Answer1, 3, 6, 11, 19, 32; each is 2 less than the term two places further on: $Q_{n} = V_{n + 2} - 2$ .
  $Q_{1} = 1$ , $Q_{2} = 3$ , $Q_{3} = 6$ , $Q_{4} = 11$ , $Q_{5} = 19$ and $Q_{6} = 32$ . From $V_{3}$ on, the Virahānka–Fibonacci terms are 3, 5, 8, 13, 21 and 34, so $Q_{1} = 3 - 2$ , $Q_{2} = 5 - 2$ , $Q_{3} = 8 - 2$ , $Q_{4} = 13 - 2$ , $Q_{5} = 21 - 2$ and $Q_{6} = 34 - 2$ .
2. Explain why $Q_{n} = V_{n + 2} - 2$ keeps working for every $n$ .
  Show solution
  AnswerEach new total adds one more term, and the Virahānka–Fibonacci rule does the rest.
  It works for $n = 1$ : $Q_{1} = 1 = V_{3} - 2$ . Now suppose it works for some $n$ . Then $Q_{n + 1} = Q_{n} + V_{n + 1} = V_{n + 2} - 2 + V_{n + 1} = V_{n + 3} - 2$ , using the rule $V_{n + 3} = V_{n + 2} + V_{n + 1}$ . So whenever it works for one $n$ , it also works for the next one; starting from $n = 1$ , it works for every $n$ .
Quick answerEasy
Which sequence is both an AP and a GP?
1, 2, 4, 8, …2, 4, 6, 8, …1, −1, 1, −1, …4, 4, 4, 4, …
Show answer
Answer (D) 4, 4, 4, 4, …
A constant sequence has common difference 0 and common ratio 1. The sequence 1, −1, 1, −1 is a GP with $r = - 1$ , but its differences $- 2, 2, - 2$ are not constant.
Quick answerEasy
An AP has $t_{1} = 7$ and $t_{2} = 11$ , so $t_{100} =$ .
Show answer
Answer 403
$d = 4$ , so $t_{100} = 7 + 99 \times 4 = 403$ .
18
Quick answerMedium
Put these numbers in increasing order.
1. $2^{10}$
2. $1 + 2 + \dots + 10$
3. the 10th term of the AP 5, 10, 15, …
4. the 10th term of the GP 1, 2, 4, …
Show answer
Answer
the 10th term of the AP 5, 10, 15, …
$1 + 2 + \dots + 10$
the 10th term of the GP 1, 2, 4, …
$2^{10}$
They are 50, 55, $2^{9} = 512$ and 1024.

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Our practice questions are written for ncert.ai and follow the NCERT exercises. Find the one that matches the question in your textbook; its answer is with it.

In your NCERT textbook	Practice question here
NCERT Exercise Set 8.1, Q1	Practice 8.1 · Q1
NCERT Exercise Set 8.1, Q2	Practice 8.1 · Q2
NCERT Exercise Set 8.1, Q3	Practice 8.1 · Q3
NCERT Exercise Set 8.1, Q4	Practice 8.1 · Q4
NCERT Exercise Set 8.1, Q5	Practice 8.1 · Q5
NCERT Exercise Set 8.1, Q6	Practice 8.1 · Q6
NCERT Exercise Set 8.2, Q1	Practice 8.2 · Q1
NCERT Exercise Set 8.2, Q2	Practice 8.2 · Q2
NCERT Exercise Set 8.2, Q3	Practice 8.2 · Q3
NCERT Exercise Set 8.2, Q4	Practice 8.2 · Q4
NCERT Exercise Set 8.2, Q5	Practice 8.2 · Q5
NCERT Exercise Set 8.2, Q6	Practice 8.2 · Q6
NCERT Exercise Set 8.2, Q7	Practice 8.2 · Q7
NCERT Exercise Set 8.3, Q1	Practice 8.3 · Q1
NCERT Exercise Set 8.3, Q2	Practice 8.3 · Q2
NCERT Exercise Set 8.3, Q3	Practice 8.3 · Q3
NCERT Exercise Set 8.3, Q4	Practice 8.3 · Q4
NCERT Exercise Set 8.3, Q5	Practice 8.3 · Q5
NCERT Exercise Set 8.3, Q6	Practice 8.3 · Q6
NCERT Exercise Set 8.3, Q7	Practice 8.3 · Q7
NCERT End-of-chapter Exercises, Q1	End-of-chapter exercises · Q1
NCERT End-of-chapter Exercises, Q2	End-of-chapter exercises · Q2
NCERT End-of-chapter Exercises, Q3	End-of-chapter exercises · Q3
NCERT End-of-chapter Exercises, Q4	End-of-chapter exercises · Q4
NCERT End-of-chapter Exercises, Q5	End-of-chapter exercises · Q5
NCERT End-of-chapter Exercises, Q6	End-of-chapter exercises · Q6
NCERT End-of-chapter Exercises, Q7	End-of-chapter exercises · Q7
NCERT End-of-chapter Exercises, Q8	End-of-chapter exercises · Q8
NCERT End-of-chapter Exercises, Q9	End-of-chapter exercises · Q9
NCERT End-of-chapter Exercises, Q10	End-of-chapter exercises · Q10
NCERT End-of-chapter Exercises, Q11	End-of-chapter exercises · Q11
NCERT End-of-chapter Exercises, Q12	End-of-chapter exercises · Q12
NCERT End-of-chapter Exercises, Q13	End-of-chapter exercises · Q13
NCERT End-of-chapter Exercises, Q14	End-of-chapter exercises · Q14
NCERT End-of-chapter Exercises, Q15	End-of-chapter exercises · Q15

Quick check: MCQs on Predicting What Comes Next: Exploring Sequences and Progressions

Naming the terms4 questions
8.2 Explicit Rule for a Sequence4 questions
The Virahānka–Fibonacci sequence4 questions
8.4 Arithmetic Progressions4 questions
8.4.1 Visualising an AP3 questions
8.5 Sum of the First n Natural Numbers4 questions
8.6 Geometric Progressions4 questions
8.6.2 Visualising a GP3 questions

Glossary · 12

Key terms (English–Hindi)

arithmetic progressionसमांतर श्रेढ़ीsamantar shredhi: A sequence that adds the same number $d$ at every step: $a, a + d, a + 2 d, \dots$

common differenceसार्व अंतरsarva antar: The fixed number $d$ added at each step of an AP; $d = t_{2} - t_{1}$ .First seen in the chapter →
common ratioसार्व अनुपातsarva anupat: The fixed multiplier $r$ of a GP; $r = t_{2} \div t_{1}$ .First seen in the chapter →

explicit rule: A formula giving the $n$ th term straight from $n$ , such as $t_{n} = 7 n + 3$ .First seen in the chapter →

finite sequenceपरिमित अनुक्रमparimit anukram: A sequence that stops after some number of terms. One that never stops is an infinite sequence.First seen in the chapter →
fractal: A shape made of smaller and smaller copies of itself, such as the Sierpiński triangle.First seen in the chapter →

geometric progressionगुणोत्तर श्रेढ़ीgunottar shredhi: A sequence that multiplies by the same number $r$ at every step: $a, a r, a r^{2}, \dots$ First seen in the chapter →

recursive ruleपुनरावर्ती नियमpunaravarti niyam: A rule giving each term from earlier terms, plus the starting term(s), such as $t_{1} = 6$ , $t_{n} = t_{n - 1} + 5$ .First seen in the chapter →

sequenceअनुक्रमanukram: A list of numbers in a definite order, such as 3, 10, 17, 24, 31.First seen in the chapter →

termपदpad: One number in a sequence. The term in position $n$ is written $t_{n}$ .First seen in the chapter →
triangular numberत्रिभुजीय संख्याtribhujiya sankhya: A number of the form $1 + 2 + \dots + n = \frac{n (n + 1)}{2}$ : 1, 3, 6, 10, 15, …First seen in the chapter →

Virahānka–Fibonacci sequence: Each term is the sum of the two before it, starting 1, 2: so 1, 2, 3, 5, 8, 13, …First seen in the chapter →

8.1 Introduction to Sequences

Naming the terms

8.2 Explicit Rule for a Sequence

8.3 Recursive Rule for a Sequence

The Virahānka–Fibonacci sequence

8.4 Arithmetic Progressions

8.4.1 Visualising an AP

8.5 Sum of the First n Natural Numbers

8.6 Geometric Progressions

8.6.1 Fun with Fractals

8.6.2 Visualising a GP

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