1. Sequences and Series
- Books Name
- AMARENDRA PATTANAYAK Mathmatics Book
- Publication
- KRISHNA PUBLICATIONS
- Course
- CBSE Class 11
- Subject
- Mathmatics
Chapter 9
Sequences and Series
Sequences and Series:
A sequence can be regarded as a function whose domain is the set of natural numbers or some subset of it. Sometimes, we use the functional notation a(n) for an.
The different numbers occurring in any particular sequence are known as terms. The terms of a sequence are denoted by
a1, a2, a3,….,an
If a sequence has a finite number of terms then it is known as a finite sequence. A sequence is termed as infinite if it is not having a definite number of terms.
Series: Let a1, a2, a3,…,an, be a given sequence. Then, the expression
a1 + a2 + a3 +,…+ an + ...
is called the series associated with the given sequence .The series is finite or infinite according as the given sequence is finite or infinite. Series are often represented in compact form, called sigma notation, using the Greek letter å (sigma) as means of indicating the summation involved. Thus, the series a1 + a2 + a3 + ... + an is abbreviated as
Sequence relates to the organization of terms in a particular order (i.e. related terms follow each other) and series is the summation of the elements of a sequence.
Examples
Example: Write the first three terms in each of the following sequences defined by
the following: an = 2n + 5,
Ans.: (i) Here an = 2n + 5
Substituting n = 1, 2, 3, we get
a1 = 2(1) + 5 = 7, a2 = 9, a3 = 11
Therefore, the required terms are 7, 9 and 11.
Example Let the sequence an be defined as follows:
a1 = 1, an = an – 1 + 2 for n 2.
Find first five terms and write corresponding series.
Solution We have
a1 = 1, a2 = a1 + 2 = 1 + 2 = 3, a3 = a2 + 2 = 3 + 2 = 5,
a4 = a3 + 2 = 5 + 2 = 7, a5 = a4 + 2 = 7 + 2 = 9.
Hence, the first five terms of the sequence are 1,3,5,7 and 9. The corresponding series
is 1 + 3 + 5 + 7 + 9 +...
Examples
Q.1: If an = 2n, then find the first five terms of the series.
Solution: Given: an = 2n
On substituting n = 1, 2, 3, 4, 5, we get
a1 = 21 = 2
a2 = 22 = 4
a3 = 23 = 8
a4 = 24 = 16
a5 = 25 = 32
Therefore, the required terms are 2, 4, 8, 16, and 32.
Q.2: Find the sum of odd integers from 1 to 2001.
Solution:
The odd integers from 1 to 2001 are 1, 3, 5, …1999, 2001.
It clearly forms a sequence in A.P.
Where, the first term, a = 1
Common difference, d = 2
Now,
a + (n -1)d = 2001
1 + (n-1)(2) = 2001
2n – 2 = 2000
2n = 2000 + 2 = 2002
n = 1001
We know,
Sn = n/2 [2a + (n-1)d]
Sn = 1001/2[2×1+(1001-1)2]
Sn = 1001/2[2+1000×2]
Sn = 1001/2[2002]
Sn = 1001 x 1001
Sn = 1002001
2. Arithmetic Progression (AP) and Arithmetic Mean (AM)
- Books Name
- AMARENDRA PATTANAYAK Mathmatics Book
- Publication
- KRISHNA PUBLICATIONS
- Course
- CBSE Class 11
- Subject
- Mathmatics
Arithmetic Progression (AP) and Arithmetic Mean (AM)
A sequence a1, a2, a3,…, an,… is called arithmetic sequence or arithmetic progression
if an + 1 = an + d, n Î N, where a1 is called the first term and the constant term d is called
the common difference of the A.P.
Let us consider an A.P. (in its standard form) with first term a and common
difference d, i.e., a, a + d, a + 2d, ...
The nth term of an AP is given by
nth term = an= a + (n-1) d.
a = the first term, l = the last term, d = common difference,
n = the number of terms.
Sum of nth term:
Sn= the sum to n terms of A.P.
Let a, a + d, a + 2d, …, a + (n – 1) d be an A.P.
Then
l = a + (n – 1) d
Arithmetic mean: Given two numbers a and b. We can insert a number A between them
so that a, A, b is an A.P. Such a number A is called the arithmetic mean (A.M.) of the numbers
a and b. Note that, in this case, we have
A – a = b – A
- A =(a+b)/2
Between any two numbers ‘a’ and ‘b’, n numbers can be inserted such that the resulting sequence is an Arithmetic Progression. A1, A2, A3,……,An be n numbers between a and b such that a, A1 , A2 , A3,……,An, b is in A.P.
Here, a is the 1st term and b is (n+2)th term. Therefore,
b = a + d[(n + 2) – 1] = a + d (n + 1).
Hence, common difference (d) = (b-a)/(n+1)
Now, A1= a+d= a+((b-a)/(n+1))
A2= a+2d = a + ((2(b-a)/(n+1))
An = a+nd= a + ((n(b-a)/(n+1))}
The nth term of a geometric progression is given by an = arn-1
Example: Insert 6 numbers between 3 and 24 such that the resulting sequence is
an A.P.
Solution: Let A1, A2, A3, A4, A5 and A6 be six numbers between 3 and 24 such that
3, A1, A2, A3, A4, A5, A6, 24 are in A.P. Here, a = 3, b = 24, n = 8.
Therefore, 24 = 3 + (8 –1) d, so that d = 3.
Thus A1 = a + d = 3 + 3 = 6; A2 = a + 2d = 3 + 2 × 3 = 9;
A3 = a + 3d = 3 + 3 × 3 = 12; A4 = a + 4d = 3 + 4 × 3 = 15;
A5 = a + 5d = 3 + 5 × 3 = 18; A6 = a + 6d = 3 + 6 × 3 = 21.
Hence, six numbers between 3 and 24 are 6, 9, 12, 15, 18 and 21.
simple properties of an A.P. :
(i) If a constant is added to each term of an A.P., the resulting sequence is
also an A.P.
(ii) If a constant is subtracted from each term of an A.P., the resulting
sequence is also an A.P.
(iii) If each term of an A.P. is multiplied by a constant, then the resulting
sequence is also an A.P.
(iv) If each term of an A.P. is divided by a non-zero constant then the
resulting sequence is also an A.P.
Example 1. Show that the sum of (m + n)th and (m – n)th terms of an A.P. is equal to twice the mth term.
Solution:
Let’s take a and d to be the first term and the common difference of the A.P. respectively.
We know that, the kth term of an A. P. is given by
ak = a + (k –1) d
So, am + n = a + (m + n –1) d
And, am – n = a + (m – n –1) d
am = a + (m –1) d
Thus,
am + n + am – n = a + (m + n –1) d + a + (m – n –1) d
= 2a + (m + n –1 + m – n –1) d
= 2a + (2m – 2) d
= 2a + 2 (m – 1) d
=2 [a + (m – 1) d]
= 2am
Therefore, the sum of (m + n)th and (m – n)th terms of an A.P. is equal to twice the mth term
Example 2. If the sum of three numbers in A.P., is 24 and their product is 440, find the numbers.
Solution:
Let’s consider the three numbers in A.P. as a – d, a, and a + d.
Then, from the question we have
(a – d) + (a) + (a + d) = 24 … (i)
3a = 24
∴ a = 8
And,
(a – d) a (a + d) = 440 … (ii)
(8 – d) (8) (8 + d) = 440
(8 – d) (8 + d) = 55
64 – d2 = 55
d2 = 64 – 55 = 9
∴ d = ± 3
Thus,
When d = 3, the numbers are 5, 8, and 11 and
When d = –3, the numbers are 11, 8, and 5.
Therefore, the three numbers are 5, 8, and 11.
3. Geometric Progression (GP) and Geometric Mean (GM)
- Books Name
- AMARENDRA PATTANAYAK Mathmatics Book
- Publication
- KRISHNA PUBLICATIONS
- Course
- CBSE Class 11
- Subject
- Mathmatics
Geometric Progression (GP) and Geometric Mean (GM)
A sequence a1, a2, a3,…, an,… is called Geometric sequence or Geometric progression
if an + 1 / an = r, n Î N, where a1 is called the first term and the constant term r is called
the common ratio of the G.P.
It is represented by:
a, ar, ar2, ar3, ar4, and so on.
Where a is the first term and r is the common ratio.
The nth term of a GP is an=Tn = arn-1
a = the first term, r = the common ratio, l = the last term,
n = the numbers of terms,
Sn = the sum of first n terms.
Suppose a, ar, ar2, ar3,……arn-1 is the given Geometric Progression.
Then the sum of n terms of GP is given by:
Sn = a + ar + ar2 + ar3 +…+ arn-1
- Sn = a[(rn – 1)/(r – 1)] if r ≠ 1 and r > 1
Types of GP:
- Finite geometric progression (Finite GP)
- Infinite geometric progression (Infinite GP)
Finite Geometric Progression
The terms of a finite G.P. can be written as a, ar, ar2, ar3,……arn-1
a, ar, ar2, ar3,……arn-1 is called finite geometric series.
The sum of finite Geometric series is Sn = a[(rn – 1)/(r – 1)] if r ≠ 1 and r > 1
Infinite Geometric Progression
Terms of an infinite G.P. can be written as a, ar, ar2, ar3, ……arn-1,…….
a, ar, ar2, ar3, ……arn-1,……. is called infinite geometric series.
The sum of infinite geometric series is Sn =
Properties of GP:
- Three non-zero terms a, b, c are in GP if and only if b2 = ac
- In a GP,
Three consecutive terms can be taken as a/r, a, ar
Four consecutive terms can be taken as a/r3, a/r, ar, ar3
Five consecutive terms can be taken as a/r2, a/r, a, ar, ar2 - In a finite GP, the product of the terms equidistant from the beginning and the end is the same
That means, t1.tn = t2.tn-1 = t3.tn-2 = ….. - If each term of a GP is multiplied or divided by a non-zero constant, then the resulting sequence is also a GP with the same common ratio
- The product and quotient of two GP’s is again a GP
- If each term of a GP is raised to the power by the same non-zero quantity, the resultant sequence is also a GP
- If a1, a2, a3,… is a GP of positive terms then log a1, log a2, log a3,… is an AP (arithmetic progression) and vice versa
Geometric Mean (G .M.) : The geometric mean of two positive numbers a and b is the
Number √(ab).
Therefore, the geometric mean of 2 and 8 is 4.
We observe that the three numbers 2,4,8 are consecutive terms of a G.P.
Let G1, G2,…, Gn be n numbers between positive numbers a and b such that
a,G1,G2,G3,…,Gn,b is a G.P.
Thus, b being the (n + 2)th term,
we have b= arn+1
Example: Insert three numbers between 1 and 256 so that the resulting sequence
is a G.P.
Solution Let G1, G2,G3 be three numbers between 1 and 256 such that
1, G1,G2,G3 ,256 is a G.P.
Therefore 256 = r4 giving r = ± 4 (Taking real roots only)
For r = 4, we have G1 = ar = 4, G2 = ar2 = 16, G3 = ar3 = 64
Similarly, for r = – 4, numbers are – 4,16 and – 64.
Hence, we can insert 4, 16, 64 between 1 and 256 so that the resulting sequences are in G.P.
Relationship Between A.M. and G.M.:
Let A and G be A.M. and G.M. of two given positive real numbers a and b, respectively.
Thus, we have
- A – B ³ 0
we obtain the relationship A G.
Example 1. We know the sum of the interior angles of a triangle is 180°. Show that the sums of the interior angles of polygons with 3, 4, 5, 6, … sides form an arithmetic progression. Find the sum of the interior angles for a 21-sided polygon.
Solution:
Given the sum of interior angles of a polygon having ‘n’ sides is given by (n – 2) × 180°
Sum of angles with three sides that is n = 3 is (3 – 2) × 180° = 180°
Sum of angles with four sides that is n = 4 is (4 – 2) × 180° = 360°
Sum of angles with five sides that is n = 5 is (5 – 2) × 180° = 540°
Sum of angles with six sides that is n = 6 is (6 – 2) × 180° = 720°
As seen as the number of sides increases by 1 the sum of interior angles increases by 180°
Hence the sequence of sum of angles as number of sides’ increases is 180°, 360°, 540°, 720° …
The sequence is AP with first term as a = 180° and common difference as d = 180°
We have to find sum of angles of polygon with 21 sides
Using (n – 2) × 180°
⇒ Sum of angles of polygon having 21 sides = (21 – 2) × 180°
⇒ Sum of angles of polygon having 21 sides = 19 × 180°
⇒ Sum of angles of polygon having 21 sides = 3420°
Example 2. A side of an equilateral triangle is 20cm long. A second equilateral triangle is inscribed in it by joining the mid points of the sides of the first triangle. The process is continued as shown in the accompanying diagram. Find the perimeter of the sixth inscribed equilateral triangle.
Solution:
Let ABC be the triangle with AB = BC = AC = 20 cm
Let D, E and F be midpoints of AC, CB and AB respectively which are joined to form an equilateral triangle DEF
Now we have to find the length of side of ΔDEF
Consider ΔCDE
CD = CE = 10 cm … D and E are midpoints of AC and CB
Hence ΔCDE is isosceles
⇒ ∠CDE = ∠CED … base angles of isosceles triangle
But ∠DCE = 60° …∠ABC is equilateral
Hence ∠CDE = ∠CED = 60°
Hence ΔCDE is equilateral
Hence DE = 10 cm
Similarly, we can show that GH = 5 cm
Hence the series of sides of equilateral triangle will be 20, 10, 5 …
The series is GP with first term a = 20 and common ratio r = ½
To find the perimeter of 6th triangle inscribed we first have to find the side of 6th triangle that is the 6th term in the series
nth term in GP is given by tn = arn-1
⇒ t6 = (20) (1/2)6-1
⇒ t6 = 20/ 25
= 20/ (4 × 23)
⇒ t6 = 5/8
Hence the side of 6th equilateral triangle is 5/8 cm and hence its perimeter would be thrice its side length because it’s an equilateral triangle
Perimeter of 6th equilateral triangle inscribed is 3 × 5/8 = 15/8 cm
Example 3: Find the sum of GP: 10, 30, 90, 270 and 810, using formula.
Solution: Given GP is 10, 30, 90, 270 and 810
First term, a = 10
Common ratio, r = 30/10 = 3 > 1
Number of terms, n = 5
Sum of GP is given by;
Sn = a[(rn – 1)/(r – 1)]
S5 = 10[(35 – 1)/(3 – 1)]
= 10[(243 – 1)/2]
= 10[242/2]
= 10 × 121
= 1210
Check: 10 + 30 + 90 + 270 + 810 = 1210
Example 4: If 2, 4, 8,…., is the GP, then find its 10th term.
Solution: The nth term of GP is given by:
2, 4, 8,….
Here, a = 2 and r = 4/2 = 2
an = arn-1
Therefore,
a10 = 2 x 210 – 1
= 2 × 29
= 1024
Example 5. Find the sum of the following series up to n terms:
(i) 5 + 55 + 555 + … (ii) .6 + .66 + . 666 + …
Solution:
(i) Given, 5 + 55 + 555 + …
Let Sn = 5 + 55 + 555 + ….. up to n terms
(ii) Given, .6 + .66 + . 666 + …
Let Sn = 06. + 0.66 + 0.666 + … up to n terms
The Geometric Mean (G.M.) of a set of n observations is the nth root of their product. If x1, x2, ... , xn are n observations then
Example: If AM and HM of the data sets are 4 and 25 respectively, then find the GM.
Ans:
Given that, AM = 4
HM = 25.
We know that the relation between AM, GM and HM is
GM = √[ AM × HM]
Now, substitute AM and HM in the relation,
we get;
GM = √[4 × 25]
GM = √100 = 10
Hence, GM = 10.
4. Sum to n Terms of Special Series
- Books Name
- AMARENDRA PATTANAYAK Mathmatics Book
- Publication
- KRISHNA PUBLICATIONS
- Course
- CBSE Class 11
- Subject
- Mathmatics
Sum to n Terms of Special Series
sum of first n terms of some special series,
(i) 1 + 2 + 3 +… + n (sum of first n natural numbers)
(ii) 12 + 22 + 32 +… + n2 (sum of squares of the first n natural numbers)
(iii) 13 + 23 + 33 +… + n3 (sum of cubes of the first n natural numbers).
Proof:
(i) 1 + 2 + 3 +… + n (sum of first n natural numbers)
Sum of First n Natural Numbers
Natural numbers are: 1, 2, 3, 4,….
Sum of these natural numbers can be written as: 1 + 2 + 3 + 4 +….
This is an AP with first term 1 and common difference 1.
i.e. a = 1 and d = 2 – 1 = 1
Sum of first n terms of an AP = n/2 [2a + (n – 1)d]
Now,
Sn = 1 + 2 + 3 + 4 + ….. + n
Sn = n/2 [2a + (n – 1)d]
Substituting a = 1 and d = 1,
Sn = (n/2) [2(1) + (n – 1)(1)]
= (n/2) [2 + n – 1]
= n(n + 1)/2
Therefore, the sum of first n natural numbers = n(n + 1)/2
Sum of Squares of The First n Natural Numbers
(ii) 12 + 22 + 32 +… + n2 (sum of squares of the first n natural numbers)
The squares of natural numbers are: 12, 22, 32, 42,…
Or
1, 4, 9, 16, ….
We can express the sum of n terms as: 12 + 22 + 32 +…+ n2
This is neither AP nor GP since either the difference between two consecutive numbers is not constant or the ratio of two consecutive numbers is constant.
Let’s find the sum of this series by considering an expression given below:
k3 – (k – 1)3 = 3k2 – 3k + 1
Substituting k = 1,
13 – (1 – 1)3 = 3(1)2 – 3(1) + 1
13 – 03 = 3(1)2 – 3(1) + 1….(i)
Substituting k = 2,
23 – (2 – 1)3 = 3(2)2 – 3(2) + 1
23 – 13 = 3(2)2 – 3(2) + 1….(ii)
Substituting k = 3,
33 – (3 – 1)3 = 3(3)2 – 3(3) + 1
33 – 23 = 3(3)2 – 3(3) + 1….(iii)
Substituting k = 4,
43 – (4 – 1)3 = 3(4)2 – 3(4) + 1
43 – 33 = 3(4)2 – 3(4) + 1….(iv)
Substituting k = n,
n3 – (n – 1)3 = 3(n)2 – 3(n) + 1
Now, adding both sides of these equations together, we get;
13 – 03 + 23 – 13 + 33 – 23 + … + n3 – (n – 1)3 = 3(12 + 22 + 32 + 42 + … + n2) – 3(1 + 2 + 3 + 4 + … + n) + n(1)
n3 – 03 = 3(12 + 22+ 32 + 42 + … + n2) – 3(1 + 2 + 3 + 4 + … + n) + n
Here,
represents the sum of first n natural numbers and is equal to n(n + 1)/2.
So,
Rearranging the terms,
= (1/6) (2n3 + 3n2 + n)
= (1/6) [n(2n2 + 3n + 1)]
= (1/6)[n(n + 1)(2n + 1)]
Therefore, the sum of squares of first n natural numbers = [n(n + 1)(2n + 1)]/6
Sum of Cubes of The First n Natural Numbers
(iii) 13 + 23 + 33 +… + n3 (sum of cubes of the first n natural numbers).
The squares of natural numbers are: 13, 23, 33, 43,…
Or
1, 8, 27, 64,….
We can express the sum of n terms as: 13 + 23 + 33 +…+ n3
This is neither AP nor GP since either the difference between two consecutive numbers is not constant or the ratio of two consecutive numbers is constant.
Let’s find the sum of this series by considering an expression given below:
(k + 1)4 – k4 = 4k3 + 6k2 + 4k + 1
Substituting k = 1, 2, 3, …, n
24 – 14 = 4(1)3 + 6(1)2 + 4(1) + 1
34 – 24 = 4(2)3 + 6(2)2 + 4(2) + 1
44 – 34 = 4(3)3 + 6(3)2 + 4(3) + 1
(n – 1)4 – (n – 2)4 = 4(n – 2)3 + 6(n – 2)2 + 4(n – 2) + 1
n4 – (n – 1)4 = 4(n – 1)3 + 6(n – 1)2 + 4(n – 1) + 1
(n + 1)4 – n4 = 4n3 + 6n2 + 4n + 1
By adding both sides of these equations, we get;
24 – 14 + 34 – 24 + 44 – 34 + …. + (n + 1)4 – n4 = 4(1)3 + 6(1)2 + 4(1) + 1 + 4(2)3 + 6(2)2 + 4(2) + 1 + 4(3)3 + 6(3)2 + 4(3) + 1 + …. + 4n3 + 6n2 + 4n + 1
(n + 1)4 – 14 = 4(13 + 23 + 33 +…+ n3) + 6(12 + 22 + 32 + …+ n2) + 4(1 + 2 + 3 +…+ n) + n
We know that,
And
Thus,
By rearranging the terms,
= (1/4) [n4 + 4n3 + 6n2 + 4n – 2n3 – 3n2 – n – 2n2 – 2n – n]
= (1/4) [n4 + 2n3 + n2]
= (1/4)[n2(n2 + 2n + 1)]
= (1/4)[n2(n + 1)2]
Therefore, the sum of cubes of first n natural numbers = [n(n + 1)]2/4
Question:
Find the sum to n terms of the series: 2 + 5 + 14 + 41 +….
Solution:
2 + 5 + 14 + 41 +….
The difference between two consecutive terms of this series is: 3, 9, 27, ….
Let Sn be the sum of its n terms and an be its nth term. Then,
Sn =2 + 5 + 14 + 41 + … + an….(i)
And
Sn = 2 + 5+ 14 + 41 + … + an – 1 + an….(ii)
Subtracting equation (ii) from (i), we get
0 = 2 + [3 + 9 + 27 + … + (n – 1) term] – an
⇒ an = 2 + [3 + 9 + 27 +… + (n- 1) term]
Here, 3 + 9 + 27 + … is a geometric series.
So, an = 2 + [3(3n-1 – 1)/2]
= [4 + 3n – 3]/2
= (1 + 3n)/2
Now, we need to find the sum of the series whose general term is (1 + 3n)/2
Sn = [(1 + 3)/2] + [(1 + 32)/2] + [(1 + 33)/2] + …. + (1 + 3n)/2
= (1/2) [(3 + 32 + 33 + …. + 3n) + (1 + 1 + 1 + … + n)]
= (1/2) {[3(3n – 1)/(3 – 1)] + n}
= (1/2) [(3/2)(3n – 1) + n]
= [(3n+1 – 3 + 2n)/4]
Therefore, the sum of the given series = (3n+1 + 2n – 3)/4
5. Infinite GM and its sum
- Books Name
- AMARENDRA PATTANAYAK Mathmatics Book
- Publication
- KRISHNA PUBLICATIONS
- Course
- CBSE Class 11
- Subject
- Mathmatics
Infinite GM and its sum
The sum of an infinite Geometric Progression whose first term 'a' and common ratio 'r' (-1 < r < 1 i.e., |r| < 1) is
S = a/(1−r)
Proof:
A series of the form a + ar + ar22 + ...... + arnn + ............... ∞ is called an infinite geometric series.
Let us consider an infinite Geometric Progression with first term a and common ratio r, where -1 < r < 1 i.e., |r| < 1. Therefore, the sum of n terms of this Geometric Progression in given by
Example: Find the sum to infinity of the Geometric Progression
Example:. If the pth and qth terms of a G.P. are q and p respectively, show that its (p + q)th term is
Solution:
Hence the proof.