Sequences and Series: Class 11 Maths · Chapter 8

Mathematics · Companion notes, simulations and practice · Official NCERT chapter PDF ↗

Published 18 Oct 2026 · Updated 18 Oct 2026

80 min
1 simulation
61 practice questions
20 MCQs

What you will learn in Sequences and Series

Treat a sequence as a function on the natural numbers, find its terms from an nth-term formula or a recurrence relation, and tell finite from infinite sequences.
Write a series in sigma notation and tell the series apart from its sum.
Recognise a geometric progression, find its common ratio and use the general term a_n = ar^(n−1).
Derive the sum S_n of the first n terms of a G.P. and use it to find sums, numbers of terms and terms from given conditions.
Find the geometric mean of two positive numbers and insert any number of geometric means between them.
Prove that A.M. ≥ G.M. for two positive numbers and use it to find numbers from their means and to find largest products.
Model growth, decay, compound interest, depreciation and instalment problems with geometric and arithmetic progressions.

Original companion text aligned to NCERT Mathematics Class 11, Chapter 8, pp. 135–150. Not affiliated with or endorsed by NCERT.

8.1 Introduction

Note the runs a batter scores in each over, the temperature in Nagpur at every hour of a summer day, or the page you reach each night in a novel. Each time you get numbers in a definite order, with a first, a second, a third and so on. Ordered lists like these are sequences, the subject of this chapter.

Two rules matter most in this chapter. Adding the same number every time gives an arithmetic progression (A.P.), familiar from Class 10. Multiplying by the same number every time gives a geometric progression (G.P.), our main subject. Rule-based sequences like these are called progressions. For the G.P. we will find any term and the sum of many terms, meet the geometric mean, and see why the arithmetic mean of two positive numbers can never be smaller than their geometric mean.

8.2 Sequences

Fold a sheet of paper in half and it has 2 layers. Fold it again: 4 layers. Then 8, then 16. Seven folds give 128 layers, a stack about as thick as a notebook. The counts 2, 4, 8, 16, 32, 64, 128 come in a fixed order, one for each fold, and that order is what makes them a sequence.

Now list the square numbers: 1, 4, 9, 16, 25, …. This list never stops, because after every square there is a bigger one.

Each number in a sequence is a term. We name terms by their positions: $a_{1}$ is the first term, $a_{2}$ the second, and in general $a_{n}$ is the term in position $n$ . The subscript says where a term stands; the value of $a_{n}$ says what stands there. In the folding sequence $a_{1} = 2$ , $a_{3} = 8$ and $a_{7} = 128$ . A formula for $a_{n}$ in terms of $n$ , when one exists, is the general term (or $n$ th term) of the sequence.

A sequence with a fixed number of terms is finite, like the seven layer counts or the dates of the Mondays in March. A sequence that goes on without end is infinite, like the square numbers or the multiples of 7.

Often one formula produces every term. The multiples of 5 are 5, 10, 15, 20, …, and the term in position $n$ is $5 n$ , so $a_{n} = 5 n$ . The sequence 3, 7, 11, 15, … starts at 3 and adds 4 each time, so $a_{n} = 4 n - 1$ ; check that $a_{1} = 3$ and $a_{4} = 15$ . Fig. 8.1 pictures this rule as a matching of positions with values.

3/3The graph is a set of separate dots $(n, a_{n})$ : there is no term at $n = 2.5$ .

Fig. 8.1A sequence pairs each position

n

with one term

a_{n}

On the left is a column of five boxes holding the positions 1, 2, 3, 4 and 5. In the first step only these positions are shown. In the second step an arrow runs from each position to a second column of boxes holding the terms 3, 7, 11, 15 and 19, and the rule a sub n equals 4n minus 1 appears on the right. In the third step a graph appears on the right with n from 1 to 5 on the horizontal axis and the term value up to 20 on the vertical axis; the five pairs are drawn as separate dots climbing in equal steps, with nothing between them.

Sometimes it is easier to say how each term is built from the terms before it. Such a rule, together with the starting terms, is a recurrence relation. The best-known example is the Fibonacci sequence: start with $a_{1} = a_{2} = 1$ and, for $n \geq 3$ , use

Fibonacci rule

a_{n} = a_{n - 1} + a_{n - 2}

This gives 1, 1, 2, 3, 5, 8, 13, 21, …, each term the sum of the two before it. In Fig. 8.2 each new term becomes the side of a square that fits exactly along the long edge of the rectangle built so far.

4/4A square of side $3 + 5 = 8$ completes a $13 \times 8$ rectangle.

Fig. 8.2Each new side is the sum of the two before it, so each square fits the rectangle's long edge.

In the first step two small squares of side 1 sit side by side, and the line below reads 1, 1. In the second step a square of side 2 is added under them, matching their combined width, and the line reads 1, 1, 2. In the third step a square of side 3 is added beside the first three, then a square of side 5 above them all, and the line reads 1, 1, 2, 3, 5. In the fourth step a large square of side 8 fills the left side, completing a 13 by 8 rectangle, and the line reads 1, 1, 2, 3, 5, 8.

Not every sequence comes with a formula. The digits of $π$ , read in order, give 3, 1, 4, 1, 5, 9, 2, 6, …, an infinite sequence with no neat formula for the $n$ th digit; we can only state the rule in words. The primes 2, 3, 5, 7, 11, … are similar. What every sequence does need is a definite way to decide which number stands in each position.

Definition

Sequence

A function defined on all of $𝐍$ (an infinite sequence) or on an initial block ${1, 2, 3, \dots, k}$ of natural numbers (a finite sequence with $k$ terms). Its value at position $n$ is written $a_{n}$ , or $a (n)$ in function notation.

Example $a (n) = n^{2}$ gives the infinite sequence $1, 4, 9, 16, \dots$

Seeing a sequence as a function links this chapter to Chapter 2: the input is a position and the output is a term. That is why the graph in Fig. 8.1 is a set of separate dots, one above each natural number, rather than a continuous line.

8.3 Series

Save ₹10 in the first week of the year, ₹20 in the second, ₹30 in the third, and so on. The weekly amounts form a sequence, but the question 'how much after 8 weeks?' is about $10 + 20 + \dots + 80$ . Terms of a sequence joined by plus signs make a series; it has as many terms as the sequence it comes from, so it may be finite or infinite.

Writing out long sums is tiring, so we use sigma notation. The capital Greek letter $Σ$ (sigma, the Greek S, for 'sum') packs the whole sum into one symbol:

sigma notation

\begin{aligned} \sum_{k = 1}^{n} a_{k} \\ = a_{1} + a_{2} + \dots + a_{n} \end{aligned}

(8.1)

where

$k$: the index, which runs through $1, 2, \dots, n$
$a_{k}$: the term in position $k$

Read it as 'the sum of $a_{k}$ for $k$ from 1 to $n$ '. The letter $k$ is only a counter, so $\sum_{j = 1}^{n} a_{j}$ means the same sum. For example, $\sum_{k = 1}^{5} (2 k - 1)$ stands for $1 + 3 + 5 + 7 + 9$ , and the savings above are $\sum_{k = 1}^{8} 10 k$ . Fig. 8.3 shows why the sum of the first five odd numbers is $5^{2} = 25$ .

5/5A layer of 9 makes a $5 \times 5$ square: $\sum_{k = 1}^{5} (2 k - 1) = 25.$

Fig. 8.3Each odd number is one L-shaped layer, so the first

n

odd numbers fill an

n \times n

square.

On the left a single unit square sits in a corner and the line 1 = 1 is written on the right. In the second step an L-shaped layer of 3 squares wraps around it, making a 2 by 2 square, and 1 + 3 = 4 appears. In the third step a layer of 5 makes a 3 by 3 square and 1 + 3 + 5 = 9 appears. In the fourth step a layer of 7 makes a 4 by 4 square and 1 + 3 + 5 + 7 = 16 appears. In the fifth step a layer of 9 completes a 5 by 5 square, 1 + 3 + 5 + 7 + 9 = 25 appears, and the sigma form of the sum is written below.

Keep two ideas apart. A series is the expression: terms joined by plus signs, like $10 + 20 + 30 + 40$ . Its sum is the number you get when you actually add them, 100 here. 'Find the sum of the series' always asks for that number.

Example 8.1NCERT Example 1 · p. 137

Terms from a formula

Write the first three terms of each sequence: (i) $a_{n} = 5 - 3 n$ ; (ii) $a_{n} = \frac{n^{2}}{n + 1}$ .

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

Example 8.2NCERT Example 2 · p. 138

A term far down the list

Find the 25th term of the sequence $a_{n} = \frac{(- 1)^{n} (n^{2} - 1)}{4 n} .$

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

Example 8.3NCERT Example 3 · p. 138

Terms from a recurrence

A sequence has $a_{1} = 64$ and $a_{n} = - \frac{1}{2} a_{n - 1}$ for $n \geq 2$ . Write its first five terms and the corresponding series.

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

Quick check

1 of 3

Evaluate

\sum_{k = 1}^{4} k^{2}

Show answer

Answer 30

1 + 4 + 9 + 16 = 30

. Squaring the sum

1 + 2 + 3 + 4 = 10

to get 100 is the common slip: the square applies to each term.

Practice · 8 questions

Practice 8.1: questions with answers

1
Short answerEasyNCERT Exercise 8.1, Q1–Q3
Write the first five terms of each sequence.
1. $a_{n} = n^{2} - 2 n$
  Show solution
  $- 1, 0, 3, 8, 15$
2. $a_{n} = \frac{2 n}{n + 2}$
  Show solution
  $\frac{2}{3}, 1, \frac{6}{5}, \frac{4}{3}, \frac{10}{7}$
3. $a_{n} = 3^{n} - 1$
  Show solution
  $2, 8, 26, 80, 242$
2
Short answerEasyNCERT Exercise 8.1, Q4–Q6
Write the first five terms of each sequence.
1. $a_{n} = \frac{3 n - 2}{4}$
  Show solution
  $\frac{1}{4}, 1, \frac{7}{4}, \frac{5}{2}, \frac{13}{4}$
2. $a_{n} = (- 1)^{n} 2^{n + 1}$
  Show solution
  $- 4, 8, - 16, 32, - 64$
3. $a_{n} = \frac{n (n^{2} - 1)}{6}$
  Show solution
  $0, 1, 4, 10, 20$ . A term may be 0.
3
Quick answerEasyNCERT Exercise 8.1, Q7
For the sequence $a_{n} = 7 n - 5$ , find the terms below.
1. $a_{12}$
  Show answer
  Answer 79
  $7 (12) - 5 = 79$ .
2. $a_{30}$
  Show answer
  Answer 205
  $7 (30) - 5 = 205$ .
Quick answerEasyNCERT Exercise 8.1, Q8
For $a_{n} = \frac{2^{n}}{n^{2}}$ , find $a_{8}$ .
Show answer
Answer 4
$a_{8} = \frac{256}{64} = 4$ .
Quick answerEasyNCERT Exercise 8.1, Q9
For $a_{n} = (- 1)^{n} n^{2}$ , find $a_{11}$ .
Show answer
Answer -121
11 is odd, so $(- 1)^{11} = - 1$ and $a_{11} = - 121$ .
Quick answerEasyNCERT Exercise 8.1, Q10
For $a_{n} = \frac{n^{2} - 4}{n + 5},$ find $a_{15}$ as a decimal.
Show answer
Answer 11.05
$a_{15} = \frac{221}{20} = 11.05.$
7
Short answerMediumNCERT Exercise 8.1, Q11–Q13
Write the first five terms of each sequence and the corresponding series.
1. $a_{1} = 2$ , $a_{n} = 2 a_{n - 1} + 3$ for $n \geq 2$
  Show solution
  $a_{2} = 7$ , $a_{3} = 17$ , $a_{4} = 37$ , $a_{5} = 77$ . Series: $2 + 7 + 17 + 37 + 77 + \dots$
2. $a_{1} = 3$ , $a_{n} = \frac{(n - 1) a_{n - 1}}{n}$ for $n \geq 2$
  Show solution
  $a_{2} = \frac{3}{2}$ , $a_{3} = 1$ , $a_{4} = \frac{3}{4}$ , $a_{5} = \frac{3}{5}$ ; in fact $a_{n} = \frac{3}{n}$ . Series: $3 + \frac{3}{2} + 1 + \frac{3}{4} + \frac{3}{5} + \dots$
3. $a_{1} = 1$ , $a_{2} = 4$ , $a_{n} = a_{n - 1} - a_{n - 2}$ for $n \geq 3$
  Show solution
  $a_{3} = 3$ , $a_{4} = - 1$ , $a_{5} = - 4$ . Series: $1 + 4 + 3 - 1 - 4 + \dots .$ (Carry on: the terms repeat every 6 places.)
Short answerMediumNCERT Exercise 8.1, Q14
The sequence $b_{1} = 1$ , $b_{2} = 3$ , $b_{n} = b_{n - 1} + b_{n - 2}$ ( $n \geq 3$ ) uses the Fibonacci rule with different starting values. Find $\frac{b_{n + 1}}{b_{n}}$ for $n = 1, 2, 3, 4, 5$ .
Show solution
The terms are 1, 3, 4, 7, 11, 18.
The ratios are $3$ , $\frac{4}{3}$ , $\frac{7}{4}$ , $\frac{11}{7}$ and $\frac{18}{11} \approx 1.636.$ They swing above and below a limit and close in on $1.618$ , the same golden ratio as for the Fibonacci numbers.

8.4 Geometric Progression (G.P.)

Compare two ways of growing. Start at 2 and keep adding 3: 2, 5, 8, 11, 14, 17. Start at 1 and keep multiplying by 2: 1, 2, 4, 8, 16, 32. The first list starts ahead, but by the sixth term the second is nearly twice as large (Fig. 8.4). Repeated multiplication is how money grows at compound interest, how a forwarded message spreads, and how a medicine's level in the blood falls.

3/3By $n = 6$ the doubling list (32) is far ahead of the adding list (17).

Fig. 8.4Adding a fixed amount gives equal steps; multiplying by a fixed factor gives steps that keep growing.

Six pairs of bars stand over n = 1 to 6. In the first step only the solid blue bars of the list that adds 3 are shown, rising in equal steps from 2 to 17. In the second step hatched orange bars for the list that multiplies by 2 appear beside them: 1, 2, 4, 8, 16, 32, small at first and then shooting up. In the third step a dashed ring marks n = 6, where 32 is far above 17.

Two more examples. A ball dropped from 8 m bounces back to half its previous height each time, reaching 4 m, 2 m, 1 m, 0.5 m, …. A message forwarded by every person to three new people reaches 1, 3, 9, 27, … people at successive stages. In each list, dividing any term by the term before it always gives the same number: 2 for the doubling list, $\frac{1}{2}$ for the ball and 3 for the message.

Definition

Geometric progression (G.P.)

A sequence is a geometric progression when none of its terms is 0 and every term divided by the term before it gives one and the same number $r$ , so that $\frac{a_{k + 1}}{a_{k}} = r$ for all $k \geq 1$ . This number is the common ratio.

ExampleIn $5$ , $- 15$ , $45$ , $- 135$ , … each term is $- 3$ times the last, so $r = - 3$ .

If the first term is $a$ , the G.P. reads $a, a r, a r^{2}, a r^{3}, \dots$ . The terms must be non-zero because we divide by them; for the same reason $r = 0$ is ruled out, since it would make every later term 0. The ratio can be negative, which makes the signs alternate, or a fraction, which makes the terms shrink.

first term: $a$
common ratio: $r$
number of terms: $n$
last term: $l$ (or $a_{n}$ )
sum of the first $n$ terms: $S_{n}$

Try this

Fold and count

You’ll need

a sheet of newspaper
a ruler

Predict first

If a sheet could be folded in half 10 times, how many layers would it have?

201005121024

Fold the sheet in half and count the layers. Repeat for as many folds as you can.
Record each count in the table, then divide each count by the one before it.
Measure the thickness of the stack after your last fold.

Table 8.1Folds and layers

Folds	Layers
1
2
3
4
5

Your entries are saved automatically.

Plotting the pairs $(n, a_{n})$ shows the difference at a glance. The terms of an A.P. change by equal steps, so their dots lie on a straight line. The terms of a G.P. change by equal factors, so their dots bend away from any line. Try it below.

SimulationA.P. or G.P.?

A grid from 0 to 9 on both axes with four draggable points a1 to a4 at (1, 1), (2, 2), (3, 3) and (4, 4), each showing its coordinates and snapping to whole numbers. They start as the A.P. 1, 2, 3, 4, lying on a straight line. Dragging the points vertically changes the terms, so the learner can build a G.P. such as 1, 2, 4, 8 or 8, 4, 2, 1 and see the dots bend instead of lining up.

Fig. 8.5Each dot is a pair

(n, a_{n})

. Keep each dot above its own

n

and drag it up or down to change the term.

Try this

Drag $a_{3}$ up to 4 and $a_{4}$ up to 8 to make the G.P. $1, 2, 4, 8$ . Do the four dots still lie on one straight line?
Now build the G.P. $8, 4, 2, 1$ with ratio $\frac{1}{2}$ . Which way does the chain of dots bend?
Can three dots of a G.P. with positive terms ever lie on one line? Try heights 3, 3, 3. Which ratio is that?

8.4.1 General term of a G.P.

Let a G.P. have first term $a$ and common ratio $r$ . The second term is $a$ multiplied by $r$ once, the third is $a$ multiplied by $r$ twice, and the fourth three times (Fig. 8.6). To reach position $n$ we start at position 1 and make $n - 1$ jumps, each a multiplication by $r$ . This gives (8.2):

3/3Reaching position $n$ takes $n - 1$ jumps, so $a_{n} = a r^{n - 1}$ .

Fig. 8.6The power of

r

counts the jumps, which is one less than the position.

In the first step a single box labelled 1st holds the first term a. In the second step more boxes appear to its right, labelled 2nd, 3rd, 4th and nth, holding ar, ar squared, ar cubed and, after a gap with dots, ar to the power n minus 1; curved arrows under the first boxes are each labelled times r. In the third step a bracket under the whole row is labelled n minus 1 jumps, each times r, showing why the power of r is one less than the position.

nth term of a G.P.

a_{n} = a r^{n - 1}

(8.2)

Three remarks. A G.P. may stop or go on for ever; a finite one with $n$ terms ends at $a r^{n - 1}$ . Adding up the terms of a G.P. gives a geometric series, finite or infinite to match. And a quick check on any answer: the power of $r$ is always one less than the position.

8.4.2 Sum to n terms of a G.P.

Adding terms one by one is fine for six terms but hopeless for sixty. A neat trick works for any $n$ . Write $S_{n}$ for the sum of the first $n$ terms of a G.P. with first term $a$ and ratio $r$ .

DerivationSum of the first n terms

Case $r = 1$ . Every term equals $a$ , so $S_{n} = a + a + \dots + a = n a .$

Case $r \neq 1$ . Write the sum, then multiply every term by $r$ :

\begin{aligned} S_{n} = a + a r + \dots \\ + a r^{n - 1} \end{aligned}

\begin{aligned} r S_{n} = a r + \dots \\ + a r^{n - 1} + a r^{n} \end{aligned}

The two lines share every term from $a r$ to $a r^{n - 1}$ (Fig. 8.7). Subtracting the second line from the first cancels all of them; only the first term on top and the last term below survive:

\begin{aligned} (1 - r) S_{n} \\ = a - a r^{n} \end{aligned}

As $r \neq 1$ , we may divide by $1 - r$ :

sum of n terms, r ≠ 1

\begin{aligned} S_{n} & = \frac{a (1 - r^{n})}{1 - r} \\ = \frac{a (r^{n} - 1)}{r - 1} \end{aligned}

(8.3)

3/3Only $a$ and $a r^{n}$ survive: $(1 - r) S_{n} = a - a r^{n}$ .

Fig. 8.7Shift by one place, subtract, and almost everything cancels.

The top row reads S sub n equals a plus ar plus ar squared plus dots plus ar to the n minus 1. The second row reads r S sub n equals ar plus ar squared plus dots plus ar to the n minus 1 plus ar to the n, written one place to the right so that equal terms stand in shaded columns. In the second step every matching pair in the shaded columns is struck through. In the third step the two survivors, a on top and ar to the n below, are circled, and the lines one minus r times S sub n equals a minus ar to the n and the final formula for S sub n appear underneath.

The two forms in (8.3) are the same fraction with top and bottom multiplied by $- 1$ . The first is handy when $r < 1$ and the second when $r > 1$ , because then the numbers you handle stay positive. Remember the exception: for $r = 1$ , $S_{n} = n a$ . The lab below lets you change $a$ , $r$ and $n$ and compare the formula with plain adding.

InteractiveProgression lab

Two modes. In Terms and sums, sliders set the first term a from −5 to 5, the common ratio r from −3 to 3 in quarter steps and the number of terms n from 1 to 12. Bars show the terms (blue for positive, orange for negative), an optional set of green dots shows the partial sums, and a panel gives the nth term, the sum found by adding and the sum from the formula, with the r = 1 case and the a = 0 or r = 0 cases explained. In Insert means, sliders set two positive numbers and how many means to insert (1 to 5); two number lines show the arithmetic means in equal steps and the geometric means in equal ratios, and for one mean the panel compares A and G.

Fig. 8.8Set

a

r

and

n

to see the terms and their sum, or insert arithmetic and geometric means between two numbers.

Example 8.4NCERT Example 4 · p. 140

The 10th and the nth term

Find the 10th term and the $n$ th term of the G.P. $3, 12, 48, \dots$

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

Example 8.5NCERT Example 5 · p. 141

Which term is it?

Which term of the G.P. $5, 10, 20, \dots$ is 5120?

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

Example 8.6NCERT Example 6 · p. 141

Two terms fix the G.P.

The 2nd term of a G.P. is 6 and its 5th term is $- 162$ . Find its 9th term.

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

Example 8.7NCERT Example 7 · p. 141

A sum with a fractional ratio

Find the sum of the first $n$ terms of $4 + 3 + \frac{9}{4} + \dots$ , and the sum of its first 4 terms.

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

Ratios between 0 and 1 have a pleasing picture. With $a = \frac{1}{2}$ and $r = \frac{1}{2}$ , (8.3) gives $S_{n} = 1 - {(\frac{1}{2})}^{n} .$ In Fig. 8.9 each new term fills half of the space still empty in a unit square, and the gap left over is always exactly as big as the last piece added.

4/4 $S_{4} = \frac{15}{16}$ ; the gap $\frac{1}{16}$ equals the last piece.

Fig. 8.9Each piece fills half of what is left, so

S_{n} = 1 - {(\frac{1}{2})}^{n} .

A square of area 1 is outlined. In the first step its left half is shaded and labelled one half, and the right side reads S1 equals one half. In the second step the top of the right half is shaded as a quarter, and S2 equals three quarters. In the third step half of the remaining space is shaded as an eighth, and S3 equals seven eighths. In the fourth step a sixteenth is added, a dashed square marks the empty gap of one sixteenth, and the text reads S4 equals fifteen sixteenths and S sub n equals 1 minus one half to the power n.

Example 8.8NCERT Example 8 · p. 141

How many terms?

How many terms of the G.P. $2, 6, 18, \dots$ must be added to get 728?

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

Example 8.9NCERT Example 9 · p. 142

Three terms from their sum and product

Three consecutive terms of a G.P. add up to $\frac{19}{2}$ and multiply to 27. Find them.

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

Example 8.10NCERT Example 10 · p. 142

A sum that hides a G.P.

Find the sum of the first $n$ terms of $2 + 22 + 222 + \dots$

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

Example 8.11NCERT Example 11 · p. 143

Matches in a knockout

A badminton knockout tournament starts with 128 players. Every match knocks one player out, and winners play on until one champion is left. How many matches are played?

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

4/4 $8 + 4 + 2 + 1 = 15 = 16 - 1$ matches.

Fig. 8.10The rounds of a knockout form a G.P. with ratio

\frac{1}{2}

Sixteen dots on the left stand for the players. In the first step brackets pair them into the 8 matches of round 1, and the text reads Round 1: 8. In the second step dashed brackets join the winners into 4 matches, and Round 2: 4 is added. In the third step thicker brackets show the 2 matches of round 3 and then the final, which leads to a champion, and the lines Round 3: 2 and Final: 1 are added. In the fourth step a line under the list gives the total of 15 matches, which equals 16 minus 1.

Quick check

1 of 4

Which of these are geometric progressions?

5, - 10, 20, - 40, \dots

2, 4, 6, 8, \dots

81, 27, 9, 3, \dots

3, 0, 0, 0, \dots

Select all that apply

Show answer

Answer

(A) $5, - 10, 20, - 40, \dots$
(C) $81, 27, 9, 3, \dots$

The ratios are constant in A (

- 2

) and C (

\frac{1}{3}

). B has a constant difference, so it is an A.P. D is not a G.P. because its terms must be non-zero;

0 \div 0

has no meaning.

8.4.3 Geometric Mean (G.M.)

When three positive numbers $a, G, b$ are in G.P., the middle one satisfies $\frac{G}{a} = \frac{b}{G},$ that is, $G^{2} = a b$ . This middle number is the geometric mean of $a$ and $b$ :

G.M. of a, b > 0

G = \sqrt{a b}

(8.4)

For example, the G.M. of 4 and 25 is $\sqrt{100} = 10$ , and $4, 10, 25$ does have the common ratio $\frac{5}{2}$ . We keep $a$ and $b$ positive so that $a b > 0$ and the root is real, and we take the positive root.

To fit $n$ numbers $G_{1}, \dots, G_{n}$ between $a$ and $b$ so that the whole list is a G.P., count the terms: there are $n + 2$ , and $b$ comes last, so $b = a r^{n + 1}$ . Hence

inserting n G.M.s

\begin{aligned} r & = {(\frac{b}{a})}^{1 / (n + 1)} \\ G_{k} & = a r^{k} \end{aligned}

(8.5)

where

$k$: $1, 2, \dots, n$

These $G_{1}, \dots, G_{n}$ are the $n$ geometric means between $a$ and $b$ . Fig. 8.11 compares them with arithmetic means: between 2 and 32, three A.M.s go up in equal steps of 7.5, while three G.M.s go up in equal factors of 2. The Insert means mode of Fig. 8.8 lets you try other numbers.

3/3Equal factors: $r^{4} = \frac{32}{2} = 16$ , so $r = 2$ .

Fig. 8.11Arithmetic means are evenly spaced; geometric means are evenly multiplied.

Two horizontal number lines run from 2 to 32, the upper one titled Arithmetic means and the lower one Geometric means. In the first step only the end points 2 and 32 are marked on both. In the second step the upper line gains points at 9.5, 17 and 24.5, with four equal arcs underneath each labelled plus 7.5. In the third step the lower line gains points at 4, 8 and 16, with arcs labelled times 2 that grow longer from left to right, the last one reaching from 16 all the way to 32.

Example 8.12NCERT Example 12 · p. 144

Inserting three geometric means

Insert three numbers between 2 and 162 so that the five numbers form a G.P.

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

Quick check

1 of 3

Find the geometric mean of 3 and 27.

Show answer

Answer 9

\sqrt{81} = 9

; indeed

3, 9, 27

has ratio 3. The arithmetic mean, 15, is a different average.

8.5 Relationship Between A.M. and G.M.

For two positive numbers $a$ and $b$ we now have two averages: the arithmetic mean $A = \frac{a + b}{2}$ , the middle term of an A.P., and the geometric mean $G = \sqrt{a b}$ , the middle term of a G.P. Try a few pairs. For 2 and 8, $A = 5$ and $G = 4$ ; for 9 and 16, $A = 12.5$ and $G = 12$ ; for 7 and 7, both are 7. The A.M. never seems to fall below the G.M.

TheoremA.M.–G.M. inequality

If $a$ and $b$ are positive real numbers with arithmetic mean $A$ and geometric mean $G$ , then $A \geq G$ , and $A = G$ exactly when $a = b$ .

A.M. ≥ G.M. for a, b > 0

\frac{a + b}{2} \geq \sqrt{a b}

(8.6)

There is a picture proof too (Fig. 8.12). Draw a semicircle on a diameter of length $a + b$ , split into pieces $a$ and $b$ . Its radius is $A$ . The perpendicular from the split point up to the arc has length $G$ : the angle in a semicircle is a right angle, and the altitude to the hypotenuse of a right triangle is the G.M. of the two pieces it cuts. No half-chord is longer than the radius, so $G \leq A$ .

3/4In the right triangle, the altitude satisfies $G^{2} = a b$ , and $G < A$ .

Fig. 8.12The radius is the A.M.; the half-chord at the split point is the G.M.

A shaded semicircle stands on a horizontal diameter divided by a tick into a long piece a on the left and a shorter piece b on the right. In the second step a vertical radius rises from the centre to the top of the arc, labelled A equals a plus b over 2. In the third step a dashed vertical segment rises from the split point to the arc, labelled G equals root ab, and dashed lines join its top to both ends of the diameter, forming a right triangle; the segment is visibly shorter than the radius. In the fourth step the split point slides to the centre, so a equals b, and the dashed segment stretches until it covers the radius exactly, showing A equals G.

Example 8.13NCERT Example 13 · p. 144

Numbers from their means

The A.M. of two positive numbers is 15 and their G.M. is 12. Find the numbers.

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

The inequality also settles 'best value' questions. Suppose 40 m of fencing must enclose a rectangular plot with sides $x$ and $y$ , so $x + y = 20$ . By (8.6), $\sqrt{x y} \leq \frac{x + y}{2} = 10,$ so the area $x y$ is at most 100 m², and it reaches 100 m² only when $x = y = 10$ . For a fixed sum, the product is largest when the two numbers are equal.

Quick check

1 of 4

For two unequal positive numbers, which statement is always true?

A = G

A > G

A < G

It depends on the numbers

Show answer

Answer (B) $A > G$

A - G

is half a square, and the square is positive because

\sqrt{a} \neq \sqrt{b}

. Equality needs

a = b

, which is excluded here.

Practice · 29 questions

Practice 8.2: questions with answers

Short answerEasyNCERT Exercise 8.2, Q1
Find the 20th term and the $n$ th term of the G.P. $\frac{3}{4}, \frac{3}{2}, 3, \dots$
Show solution
$a = \frac{3}{4}$ and $r = 2$ , so $a_{n} = \frac{3}{4} \cdot 2^{n - 1} = 3 \cdot 2^{n - 3} .$
Then $a_{20} = 3 \cdot 2^{17} = 393 216.$
Quick answerEasyNCERT Exercise 8.2, Q2
In a G.P. with common ratio 2, the 10th term is 1536. Find the 13th term.
Show answer
Answer 12288
$a_{13} = a_{10} r^{3} = 1536 \times 8 = 12 288.$ You never need $a$ (it is 3).
ProofMediumNCERT Exercise 8.2, Q3
In a G.P., call the 3rd term $p$ , the 9th term $q$ and the 15th term $s$ . Show that $q^{2} = p s$ .
Show solution
$p = a r^{2}$ , $q = a r^{8}$ and $s = a r^{14}$ .
So $p s = a^{2} r^{16} = (a r^{8})^{2} = q^{2} .$
Short answerMediumNCERT Exercise 8.2, Q4
A G.P. begins with $- 2$ . Squaring its 3rd term gives its 6th term. Find the 8th term.
Show solution
$a r^{5} = a^{2} r^{4}$ . Terms are non-zero, so divide by $a r^{4}$ : $r = a = - 2$ .
Then $a_{8} = a r^{7} = (- 2) (- 2)^{7} = 256.$ Check: $a_{3} = - 8$ and $a_{6} = 64 = (- 8)^{2}$ .
5
Quick answerMediumNCERT Exercise 8.2, Q5
Which term of each G.P. is the given number?
1. $\sqrt{2}, 2, 2 \sqrt{2}, \dots$ is 64
  Show answer
  Answer 12
  Here $a = r = \sqrt{2}$ , so $a_{n} = (\sqrt{2})^{n} = 2^{n / 2} .$ Then $2^{n / 2} = 2^{6}$ gives $n = 12$ .
2. $\sqrt{5}, 5, 5 \sqrt{5}, \dots$ is 3125
  Show answer
  Answer 10
  $a_{n} = (\sqrt{5})^{n} = 5^{n / 2}$ and $3125 = 5^{5}$ , so $n = 10$ .
3. $\frac{2}{3}, \frac{2}{9}, \frac{2}{27}, \dots$ is $\frac{2}{6561}$
  Show answer
  Answer 8
  $a_{n} = \frac{2}{3^{n}}$ and $6561 = 3^{8}$ , so $n = 8$ .
Short answerEasyNCERT Exercise 8.2, Q6
For which values of $x$ are $- \frac{3}{5}, x, - \frac{20}{3}$ in G.P.?
Show solution
$x^{2} = (- \frac{3}{5}) (- \frac{20}{3}) = 4,$ so $x = 2$ or $x = - 2$ .
Both work: the ratios are $- \frac{10}{3}$ and $\frac{10}{3}$ respectively.
7
Short answerMediumNCERT Exercise 8.2, Q7–Q10
Find each sum.
1. $0.4 + 0.04 + 0.004 + \dots$ to 15 terms
  Show solution
  $a = 0.4$ , $r = 0.1$ : $S_{15} = \frac{0.4 (1 - {0.1}^{15})}{0.9} = \frac{4}{9} (1 - 10^{- 15}) .$
2. $\sqrt{2} + \sqrt{6} + 3 \sqrt{2} + \dots$ to $n$ terms
  Show solution
  $a = \sqrt{2}$ , $r = \sqrt{3}$ : $S_{n} = \frac{\sqrt{2} (3^{n / 2} - 1)}{\sqrt{3} - 1} .$
  Rationalising, $S_{n} = \frac{\sqrt{6} + \sqrt{2}}{2} (3^{n / 2} - 1) .$
3. $1 - 2 x + 4 x^{2} - 8 x^{3} + \dots$ to $n$ terms ( $x \neq - \frac{1}{2}$ )
  Show solution
  $a = 1$ , $r = - 2 x \neq 1$ : $S_{n} = \frac{1 - (- 2 x)^{n}}{1 + 2 x} .$
4. $y^{2} + y^{5} + y^{8} + \dots$ to $n$ terms ( $y \neq 1$ )
  Show solution
  $a = y^{2}$ , $r = y^{3} \neq 1$ : $S_{n} = \frac{y^{2} (y^{3 n} - 1)}{y^{3} - 1} .$
Quick answerMediumNCERT Exercise 8.2, Q11
Evaluate $\sum_{k = 1}^{8} (5 - 2^{k})$ .
Show answer
Answer -470
$\sum 5 = 40$ and $2 + 4 + \dots + 256 = 2 (2^{8} - 1) = 510,$ so the sum is $40 - 510 = - 470$ .
Short answerMediumNCERT Exercise 8.2, Q12
Find three numbers in G.P. whose product is 8 and whose sum is 7. What is the common ratio?
Show solution
The product is $a^{3} = 8$ , so $a = 2$ .
Then $\frac{2}{r} + 2 + 2 r = 7$ , i.e. $2 r^{2} - 5 r + 2 = 0$ , so $r = 2$ or $r = \frac{1}{2}$ .
The numbers are 1, 2, 4 (ratio 2) or 4, 2, 1 (ratio $\frac{1}{2}$ ).
Quick answerEasyNCERT Exercise 8.2, Q13
Starting from 4 and adding $4^{2}$ , $4^{3}$ , … in turn, how many powers of 4 must you add to reach exactly 1364?
Show answer
Answer 5
$\frac{4 (4^{n} - 1)}{3} = 1364$ gives $4^{n} - 1 = 1023$ , so $4^{n} = 1024 = 4^{5}$ and $n = 5$ .
Short answerMediumNCERT Exercise 8.2, Q14
The 1st, 2nd and 3rd terms of a G.P. total 7, while the 4th, 5th and 6th total 56. Find $a$ , $r$ and a formula for $S_{n}$ .
Show solution
$a (1 + r + r^{2}) = 7$ and $a r^{3} (1 + r + r^{2}) = 56$ . Dividing, $r^{3} = 8$ , so $r = 2$ and $a = 1$ .
Then $S_{n} = \frac{2^{n} - 1}{2 - 1} = 2^{n} - 1.$
Quick answerMediumNCERT Exercise 8.2, Q15
The 6th term of a G.P. is 243 and its first term is 1024. Add up its first six terms.
Show answer
Answer 3367
$r^{5} = \frac{3^{5}}{4^{5}}$ , so $r = \frac{3}{4}$ . $S_{6} = \frac{1024 [1 - (\frac{3}{4})^{6}]}{\frac{1}{4}} = 4096 - 729 = 3367.$
Short answerMediumNCERT Exercise 8.2, Q16
A G.P. has its 4th term equal to nine times its 2nd term, and its first two terms total 12. Find every such G.P.
Show solution
$a r^{3} = 9 a r$ , so $r^{2} = 9$ and $r = \pm 3$ .
If $r = 3$ : $a (1 + 3) = 12$ , so $a = 3$ and the G.P. is $3, 9, 27, \dots$
If $r = - 3$ : $a (1 - 3) = 12$ , so $a = - 6$ and the G.P. is $- 6, 18, - 54, \dots$
ProofMediumNCERT Exercise 8.2, Q17
Let $x$ , $y$ and $z$ be the 2nd, 7th and 12th terms of a G.P. Prove that $x, y, z$ also form a G.P.
Show solution
$x = a r$ , $y = a r^{6}$ , $z = a r^{11}$ , so $x z = a^{2} r^{12} = y^{2}$ , and all three are non-zero. Hence $x, y, z$ are in G.P. with ratio $r^{5}$ .
The same works whenever the positions form an A.P. (here 2, 7, 12): equal gaps in position give equal powers of $r$ .
Short answerMediumNCERT Exercise 8.2, Q18
Find the sum of the first $n$ terms of $3 + 33 + 333 + \dots$
Show solution
$S_{n} = \frac{3}{9} [(10 + 10^{2} + \dots + 10^{n}) - n] .$
$S_{n} = \frac{1}{3} [\frac{10 (10^{n} - 1)}{9} - n] = \frac{10}{27} (10^{n} - 1) - \frac{n}{3} .$
Quick answerEasyNCERT Exercise 8.2, Q19
Multiply the corresponding terms of $3, 6, 12, 24$ and $64, 16, 4, 1$ , and add the four products.
Show answer
Answer 360
The products are $192, 96, 48, 24$ , a G.P. with ratio $\frac{1}{2}$ ; their sum is 360.
ProofMediumNCERT Exercise 8.2, Q20
Show that dividing the terms of the G.P. $a, a r, a r^{2}, \dots$ by the corresponding terms of the G.P. $A, A R, A R^{2}, \dots$ gives a G.P., and find its common ratio.
Show solution
The $k$ th quotient is $\frac{a r^{k - 1}}{A R^{k - 1}} = \frac{a}{A} {(\frac{r}{R})}^{k - 1} .$
That is a G.P. with first term $\frac{a}{A}$ and common ratio $\frac{r}{R}$ (all terms non-zero).
Short answerMediumNCERT Exercise 8.2, Q21
Find four numbers in G.P. such that the 3rd exceeds the 1st by 40 and the 4th exceeds the 2nd by 120.
Show solution
$a (r^{2} - 1) = 40$ and $a r (r^{2} - 1) = 120$ . Dividing, $r = 3$ ; then $8 a = 40$ , so $a = 5$ .
The numbers are 5, 15, 45, 135.
ProofHardNCERT Exercise 8.2, Q22
A G.P. of positive terms has first term $A$ and ratio $R$ . Its terms in positions $p$ , $q$ and $r$ are $x$ , $y$ and $z$ . Prove that $y^{r - p} = x^{r - q} z^{q - p} .$
Show solution
$x = A R^{p - 1}$ , $y = A R^{q - 1}$ , $z = A R^{r - 1}$ .
Powers of $A$ : left $r - p$ ; right $(r - q) + (q - p) = r - p$ .
Powers of $R$ : left $(q - 1) (r - p)$ ; right $(p - 1) (r - q) + (r - 1) (q - p),$ which expands to $q r - p q - r + p = (q - 1) (r - p) .$ The two sides agree.
20
ProofHardNCERT Exercise 8.2, Q23
A G.P. has $n$ positive terms, first term $a$ and last term $l$ .
1. Show that the $k$ th term from the start times the $k$ th term from the end is always $a l$ .
  Show solution
  They are $a r^{k - 1}$ and $l r^{- (k - 1)}$ ; their product is $a l$ .
  So, writing the product $P$ of all the terms forwards and backwards and multiplying the two lines term by term, $P^{2} = (a l)^{n}$ .
2. Hence find the product of the six terms of $2, 6, 18, \dots, 486$ as a power of 972.
  Show solution
  $a = 2$ , $l = 486$ , $n = 6$ : $P^{2} = 972^{6}$ , so $P = 972^{3}$ (the terms are positive). Check: the product is $2^{6} \cdot 3^{15} = (2^{2} \cdot 3^{5})^{3} .$
ProofMediumNCERT Exercise 8.2, Q24
For a G.P., let $S_{1}$ be the sum of the first $n$ terms, $S_{2}$ the sum of the next $n$ terms and $S_{3}$ the sum of the $n$ terms after those. Prove that $S_{2}^{2} = S_{1} S_{3}$ .
Show solution
Every term of the second block is $r^{n}$ times the matching term of the first, so $S_{2} = r^{n} S_{1}$ ; likewise $S_{3} = r^{2 n} S_{1}$ .
Hence $S_{1} S_{3} = r^{2 n} S_{1}^{2} = S_{2}^{2} .$
ProofMediumNCERT Exercise 8.2, Q25
If $a, b, c$ are in G.P., prove that $(a + b + c) (a - b + c) = a^{2} + b^{2} + c^{2} .$
Show solution
The left side is $(a + c)^{2} - b^{2} = a^{2} + 2 a c + c^{2} - b^{2} .$
With $a c = b^{2}$ this is $a^{2} + 2 b^{2} + c^{2} - b^{2} = a^{2} + b^{2} + c^{2} .$
Short answerEasyNCERT Exercise 8.2, Q26
Insert two numbers between 5 and 320 so that the four numbers are in G.P.
Show solution
$r^{3} = 64$ , so $r = 4$ , and the numbers are 20 and 80.
Short answerHardNCERT Exercise 8.2, Q27
For positive numbers $a \neq b$ , find $n$ such that $\frac{a^{n} + b^{n}}{a^{n - 1} + b^{n - 1}}$ is the G.M. of $a$ and $b$ .
Show solution
We need $a^{m + 1} + b^{m + 1} = \sqrt{a b} (a^{m} + b^{m})$ with $m = n - 1$ .
Rearrange: $a^{m + \frac{1}{2}} (\sqrt{a} - \sqrt{b}) - b^{m + \frac{1}{2}} (\sqrt{a} - \sqrt{b}) = 0.$
As $a \neq b$ , $a^{m + \frac{1}{2}} = b^{m + \frac{1}{2}},$ which forces $m + \frac{1}{2} = 0$ . So $m = - \frac{1}{2}$ and $n = \frac{1}{2}$ .
ProofHardNCERT Exercise 8.2, Q28
Two positive numbers add up to four times their geometric mean. Prove that the larger is to the smaller as $(2 + \sqrt{3})$ is to $(2 - \sqrt{3})$ .
Show solution
$a + b = 4 \sqrt{a b}$ becomes $t^{2} - 4 t + 1 = 0$ , so $t = 2 \pm \sqrt{3}$ and $\frac{a}{b} = t^{2} = (2 \pm \sqrt{3})^{2} .$
Since $(2 + \sqrt{3}) (2 - \sqrt{3}) = 1,$ $(2 + \sqrt{3})^{2} = \frac{2 + \sqrt{3}}{2 - \sqrt{3}} .$ So $a : b = (2 + \sqrt{3}) : (2 - \sqrt{3}),$ or the reverse.
26
Short answerMediumNCERT Exercise 8.2, Q29
Two positive numbers have A.M. $A$ and G.M. $G$ .
1. Show that they are the roots of $x^{2} - 2 A x + G^{2} = 0$ .
  Show solution
  Their sum is $2 A$ and their product is $G^{2}$ , so they are the roots of $x^{2} - 2 A x + G^{2} = 0$ .
2. Find the numbers when $A = 17$ and $G = 8$ .
  Show solution
  $x^{2} - 34 x + 64 = 0$ gives $x = 17 \pm \sqrt{289 - 64} = 17 \pm 15,$ so the numbers are 32 and 2.
  In general the formula gives $A \pm \sqrt{(A + G) (A - G)}$ for the two numbers.
Short answerEasyNCERT Exercise 8.2, Q30
A cell culture triples every hour and starts with 50 cells. How many cells are there after 3 hours, after 5 hours and after $n$ hours?
Show solution
After $n$ hours: $50 \cdot 3^{n}$ cells.
After 3 hours: 1350; after 5 hours: 12 150.
NumericalEasyNCERT Exercise 8.2, Q31
₹20,000 is deposited at 8% a year, compounded annually. What is the amount after 5 years, in rupees?
Show answer
Answer 29386.56
The amounts form a G.P. with ratio 1.08, so after 5 years the amount is $20 000 \times {1.08}^{5} \approx$ ₹29,386.56.
Short answerMediumNCERT Exercise 8.2, Q32
A quadratic equation $x^{2} + p x + q = 0$ has positive roots whose arithmetic mean is 13 and geometric mean is 5. Find $p$ and $q$ .
Show solution
Sum of roots $= 26 = - p$ and product $= 25 = q$ , so $p = - 26$ , $q = 25$ : the equation is $x^{2} - 26 x + 25 = 0$ , with roots 1 and 25.

Miscellaneous Examples

Example 8.14NCERT Example 14 · p. 147

When a sum of squares is not positive

Let $a, b, c$ be non-zero real numbers, and suppose some real $x$ satisfies $(a^{2} + b^{2}) x^{2} - 2 b (a + c) x + (b^{2} + c^{2}) \leq 0.$ Show that $a, b, c$ are in G.P. with common ratio $x$ .

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

Two everyday kinds of problem in the exercises use progressions directly. In an instalment plan, the interest each year is charged on the amount still unpaid, and that amount falls by a fixed instalment each year, so the interest payments form an A.P.

Example

Paying in instalments

Meera buys a laptop priced ₹60,000. She pays ₹20,000 at once and clears the remaining ₹40,000 in yearly instalments of ₹8,000, each paid together with 10% interest on the amount still unpaid at the start of that year. How much does the laptop cost her in all?

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

Interest shows the two progressions side by side. At simple interest, ₹10,000 at 10% a year earns the same ₹1,000 every year, so the amounts at the end of each year form an A.P. At compound interest, each year's interest joins the principal, so the amount is multiplied by 1.1 every year: ₹11,000, ₹12,100, ₹13,310, …, a G.P. Depreciation works the same way with a ratio below 1: a machine that loses 10% of its value each year keeps 0.9 of it (Fig. 8.13).

3/3After 5 years: ₹15,000 against ₹16,105, and the gap keeps widening.

Fig. 8.13Simple interest adds a fixed amount (A.P.); compound interest multiplies by a fixed factor (G.P.).

A graph with years 0 to 5 across and amounts from 10,000 to 16,000 rupees up the side. In the first step square markers joined by a dashed straight line show simple interest: 10,000, 11,000, 12,000 and so on up to 15,000, rising by 1,000 a year. In the second step round markers joined by a solid line show compound interest: 10,000, 11,000, 12,100, 13,310, 14,641 and 16,105, curving upward. In the third step a short bar at year 5 marks the gap between 15,000 and 16,105.

Quick check

1 of 2

₹5,000 grows at 10% a year, compounded annually. What is it worth after 2 years, in rupees?

Show answer

Answer 6050

5000 \times {1.1}^{2} = 6050

. Simple interest would give only 6000, because it never earns interest on interest.

Recap

Sequences and Series: key points and summary

A sequence is a list in a definite order; formally a function on $𝐍$ or on ${1, \dots, k}$ . It is finite or infinite (Section 8.2).
Terms come from a formula for $a_{n}$ , from a recurrence relation with starting terms (Fibonacci), or from a rule in words (digits of $π$ ).
A series is the expression $a_{1} + a_{2} + \dots$ ; its sum is the number obtained. $\sum_{k = 1}^{n} a_{k}$ is sigma notation (Section 8.3).
A G.P. has non-zero terms with a constant ratio $r = \frac{a_{k + 1}}{a_{k}}$ (Section 8.4).
The $n$ th term is $a_{n} = a r^{n - 1}$ ((8.2)).
$S_{n} = \frac{a (r^{n} - 1)}{r - 1}$ for $r \neq 1$ , and $S_{n} = n a$ for $r = 1$ ((8.3)).
Useful set-ups: $\frac{a}{r}, a, a r$ for three terms; $(10^{k} - 1)$ for $7 + 77 + \dots$ (Section 8.4.2).
The G.M. of positive $a$ and $b$ is $\sqrt{a b}$ ; $n$ G.M.s use $r = (b / a)^{1 / (n + 1)}$ (Section 8.4.3).
For positive numbers $A \geq G$ , since $A - G = \frac{1}{2} (\sqrt{a} - \sqrt{b})^{2};$ equality only when $a = b$ (Section 8.5).
Compound interest and depreciation are G.P.s; simple interest and instalments are A.P.s (Miscellaneous Examples).

Timeline

How progressions were studied

c. 1800 BCE
Babylonian tablets
Scribes worked with lists that grow by adding and by doubling, including problems on growing loans.
c. 300 BCE
Euclid
Book IX of the Elements gives a rule for adding up a geometric progression.
499 CE
Aryabhata
The Aryabhatiya gives the sum of an A.P. and the sums of the squares and cubes of natural numbers.
c. 510
Boethius
His textbook of arithmetic carries Greek work on progressions and means into medieval Europe.
c. 850
Mahavira
The Ganitasarasangraha gives rules for sums of geometric progressions; Brahmagupta and later Bhaskara II also studied sums of squares and cubes.
c. 1150
Hemachandra
Counts poetic rhythms of short and long syllables; the counts follow the Fibonacci rule.
1202
Fibonacci
Liber Abaci poses a rabbit-breeding puzzle that produces 1, 1, 2, 3, 5, 8, …
c. 1400
Madhava
Kerala mathematicians find infinite series for π and for sine and cosine.
1671
James Gregory
Finds the infinite series for the arctangent; infinite series become a central tool of analysis.

Find your NCERT exercise question

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 8.1, Q1–Q3	Practice 8.1 · Q1
NCERT Exercise 8.1, Q4–Q6	Practice 8.1 · Q2
NCERT Exercise 8.1, Q7	Practice 8.1 · Q3
NCERT Exercise 8.1, Q8	Practice 8.1 · Q4
NCERT Exercise 8.1, Q9	Practice 8.1 · Q5
NCERT Exercise 8.1, Q10	Practice 8.1 · Q6
NCERT Exercise 8.1, Q11–Q13	Practice 8.1 · Q7
NCERT Exercise 8.1, Q14	Practice 8.1 · Q8
NCERT Exercise 8.2, Q1	Practice 8.2 · Q1
NCERT Exercise 8.2, Q2	Practice 8.2 · Q2
NCERT Exercise 8.2, Q3	Practice 8.2 · Q3
NCERT Exercise 8.2, Q4	Practice 8.2 · Q4
NCERT Exercise 8.2, Q5	Practice 8.2 · Q5
NCERT Exercise 8.2, Q6	Practice 8.2 · Q6
NCERT Exercise 8.2, Q7–Q10	Practice 8.2 · Q7
NCERT Exercise 8.2, Q11	Practice 8.2 · Q8
NCERT Exercise 8.2, Q12	Practice 8.2 · Q9
NCERT Exercise 8.2, Q13	Practice 8.2 · Q10
NCERT Exercise 8.2, Q14	Practice 8.2 · Q11
NCERT Exercise 8.2, Q15	Practice 8.2 · Q12
NCERT Exercise 8.2, Q16	Practice 8.2 · Q13
NCERT Exercise 8.2, Q17	Practice 8.2 · Q14
NCERT Exercise 8.2, Q18	Practice 8.2 · Q15
NCERT Exercise 8.2, Q19	Practice 8.2 · Q16
NCERT Exercise 8.2, Q20	Practice 8.2 · Q17
NCERT Exercise 8.2, Q21	Practice 8.2 · Q18
NCERT Exercise 8.2, Q22	Practice 8.2 · Q19
NCERT Exercise 8.2, Q23	Practice 8.2 · Q20
NCERT Exercise 8.2, Q24	Practice 8.2 · Q21
NCERT Exercise 8.2, Q25	Practice 8.2 · Q22
NCERT Exercise 8.2, Q26	Practice 8.2 · Q23
NCERT Exercise 8.2, Q27	Practice 8.2 · Q24
NCERT Exercise 8.2, Q28	Practice 8.2 · Q25
NCERT Exercise 8.2, Q29	Practice 8.2 · Q26
NCERT Exercise 8.2, Q30	Practice 8.2 · Q27
NCERT Exercise 8.2, Q31	Practice 8.2 · Q28
NCERT Exercise 8.2, Q32	Practice 8.2 · Q29
NCERT Miscellaneous Exercise, Q1	Miscellaneous exercises · Q1
NCERT Miscellaneous Exercise, Q2	Miscellaneous exercises · Q2
NCERT Miscellaneous Exercise, Q3	Miscellaneous exercises · Q3
NCERT Miscellaneous Exercise, Q4	Miscellaneous exercises · Q4
NCERT Miscellaneous Exercise, Q5	Miscellaneous exercises · Q5
NCERT Miscellaneous Exercise, Q6	Miscellaneous exercises · Q6
NCERT Miscellaneous Exercise, Q7	Miscellaneous exercises · Q7
NCERT Miscellaneous Exercise, Q8	Miscellaneous exercises · Q8
NCERT Miscellaneous Exercise, Q9	Miscellaneous exercises · Q9
NCERT Miscellaneous Exercise, Q10	Miscellaneous exercises · Q10
NCERT Miscellaneous Exercise, Q11	Miscellaneous exercises · Q11
NCERT Miscellaneous Exercise, Q12	Miscellaneous exercises · Q12
NCERT Miscellaneous Exercise, Q13	Miscellaneous exercises · Q13
NCERT Miscellaneous Exercise, Q14	Miscellaneous exercises · Q14
NCERT Miscellaneous Exercise, Q15	Miscellaneous exercises · Q15
NCERT Miscellaneous Exercise, Q16	Miscellaneous exercises · Q16
NCERT Miscellaneous Exercise, Q17	Miscellaneous exercises · Q17
NCERT Miscellaneous Exercise, Q18	Miscellaneous exercises · Q18

Quick check: MCQs on Sequences and Series

8.2 Sequences4 questions
8.3 Series3 questions
8.4.2 Sum to n terms of a G.P.4 questions
8.4.3 Geometric Mean (G.M.)3 questions
8.5 Relationship Between A.M. and G.M.4 questions
Miscellaneous Examples2 questions

Glossary · 13

Key terms (English–Hindi)

arithmetic meanसमांतर माध्यsamantar madhya: For numbers $a$ and $b$ , the number $A = \frac{a + b}{2}$ ; then $a, A, b$ is an A.P.First seen in the chapter →
arithmetic progressionसमांतर श्रेढ़ीsamantar shredhi: A sequence in which each term is the previous term plus a fixed common difference $d$ .First seen in the chapter →

common ratioसार्व अनुपातsarv anupat: The constant $r = \frac{a_{k + 1}}{a_{k}}$ of a G.P.; it is never 0.First seen in the chapter →

Fibonacci sequence: The sequence $1, 1, 2, 3, 5, 8, \dots$ in which each term after the second is the sum of the two before it.First seen in the chapter →

general termव्यापक पदvyapak pad: A formula for $a_{n}$ in terms of the position $n$ , such as $a_{n} = 4 n - 1$ .First seen in the chapter →
geometric meanगुणोत्तर माध्यgunottar madhya: For positive $a$ and $b$ , the number $G = \sqrt{a b}$ ; then $a, G, b$ is a G.P.First seen in the chapter →
geometric progressionगुणोत्तर श्रेढ़ीgunottar shredhi: A sequence of non-zero numbers in which each term is the previous term times a fixed common ratio $r$ .
geometric seriesगुणोत्तर श्रेणीgunottar shreni: The series $a + a r + a r^{2} + \dots$ formed from a G.P.First seen in the chapter →

recurrence relation: A rule that builds each term from earlier terms, given together with the first term or terms.First seen in the chapter →

sequenceअनुक्रमanukram: A list of numbers in a definite order; formally a function defined on $𝐍$ or on ${1, 2, \dots, k}$ .First seen in the chapter →
seriesश्रेणीshreni: The expression $a_{1} + a_{2} + a_{3} + \dots$ formed by adding the terms of a sequence; its sum is the number obtained.First seen in the chapter →
sigma notation: Short form for a sum: $\sum_{k = 1}^{n} a_{k}$ means $a_{1} + a_{2} + \dots + a_{n}$ .First seen in the chapter →

termपदpad: One number of a sequence; $a_{n}$ is the term in position $n$ .First seen in the chapter →

Sequences and Series: Class 11 Maths · Chapter 8

What you will learn in Sequences and Series

8.1 Introduction

8.2 Sequences

8.3 Series

Practice 8.1: questions with answers

8.4 Geometric Progression (G.P.)

8.4.1 General term of a G.P.

8.4.2 Sum to n terms of a G.P.

8.4.3 Geometric Mean (G.M.)

8.5 Relationship Between A.M. and G.M.

Practice 8.2: questions with answers

Miscellaneous Examples

Sequences and Series: key points and summary

Practice questions with answers (NCERT “Miscellaneous exercises”)

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Key terms (English–Hindi)

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