## 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 a_{n}.

The different numbers occurring in any particular sequence are known as terms. The terms of a sequence are denoted by

a_{1}, a_{2}, a_{3},….,a_{n}

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 *a*1, *a*2, *a*3,…,*an*, be a given sequence. Then, the expression

*a*1 + *a*2 + *a*3 +,…*+ 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 *a*1 *+ a*2 + *a*3 + ... + *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 *= 2*n *+ 5,

Ans.: (i) Here *an = *2*n + *5

Substituting *n *= 1, 2, 3, we get

*a*1 = 2(1) + 5 = 7, *a*2 = 9, *a*3 = 11

Therefore, the required terms are 7, 9 and 11.

**Example**** **Let the sequence *an *be defined as follows:

*a*1 = 1, *an *= *an *– 1 + 2 for *n *2.

Find first five terms and write corresponding series.

**Solution**** **We have

*a*1 = 1, *a*2 = *a*1 + 2 = 1 + 2 = 3, *a*3 = *a*2 + 2 = 3 + 2 = 5,

*a*4 = *a*3 + 2 = 5 + 2 = 7, *a*5 = *a*4 + 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 a _{n }= 2^{n}, then find the first five terms of the series.**

Solution: Given: a_{n }= 2^{n}

On substituting n = 1, 2, 3, 4, 5, we get

a_{1} = 2^{1} = 2

a_{2} = 2^{2} = 4

a_{3} = 2^{3} = 8

a_{4} = 2^{4} = 16

a_{5} = 2^{5} = 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,

S_{n} = n/2 [2a + (n-1)d]

S_{n} = 1001/2[2×1+(1001-1)2]

S_{n} = 1001/2[2+1000×2]

S_{n} = 1001/2[2002]

S_{n} = 1001 x 1001

S_{n} = 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 *a*1*, a*2, *a*3*,…*, *an,*… is called *arithmetic sequence or arithmetic progression *

if *an *+ 1 = *an *+ *d*, *n *Î **N**, where *a*1 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 *+ 2*d*, ...

The nth term of an AP is given by

**n ^{th} term = a_{n}= a + (n-1) d.**

*a *= the first term, *l *= the last term, *d *= common difference,

*n *= the number of terms.

**Sum of nth term:**

S*n*= the sum to *n *terms of A.P.

Let *a, a + d, a + *2*d, …, 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. A_{1,} A_{2,} A_{3,……,}A_{n} be n numbers between a and b such that a, A_{1 ,} A_{2 ,} A_{3,……,}A_{n}, 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, A_{1}= a+d= a+((b-a)/(n+1))

A_{2}= a+2d = a + ((2(b-a)/(n+1))

A_{n} = a+nd= a + ((n(b-a)/(n+1))}

The nth term of a geometric progression is given by a_{n }= ar^{n-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 *+ 2*d *= 3 + 2 × 3 = 9;

A3 = *a *+ 3*d *= 3 + 3 × 3 = 12; A4 = *a *+ 4*d *= 3 + 4 × 3 = 15;

A5 = *a *+ 5*d *= 3 + 5 × 3 = 18; A6 = *a *+ 6*d *= 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

**m**^{t}

^{h}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 *k*^{th} term of an A. P. is given by

*a _{k}* =

*a*+ (

*k*–1)

*d*

So, *a _{m}*

_{ + n}=

*a*+ (

*m*+

*n*–1)

*d*

And, *a _{m}*

_{ – n}=

*a*+ (

*m*–

*n*–1)

*d*

*a _{m}*

_{ }=

*a*+ (

*m*–1)

*d*

Thus,

*a _{m}*

_{ + n}+

*a*

_{m}_{ – n}=

*a*+ (

*m*+

*n*–1)

*d*+

*a*+ (

*m*–

*n*–1)

*d*

= 2*a* + (*m* + *n* –1 + *m* – *n* –1) *d*

= 2*a* + (2*m* – 2) *d*

= 2*a* + 2 (*m* – 1) *d*

*=*2 [*a* + (*m* – 1) *d*]

= 2*a _{m}*

Therefore, the sum of (*m* + *n*)^{th} and (*m* – *n*)^{th} terms of an A.P. is equal to twice the *m ^{t}*

^{h}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)

3*a* = 24

∴ *a* = 8

And,

(*a* – *d*) *a* (*a* + *d*) = 440 … (ii)

(8 – *d*) (8) (8 + *d*) = 440

(8 – *d*) (8 + *d*) = 55

64 – *d*^{2} = 55

*d*^{2} = 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 *a*1*, a*2, *a*3*,…*, *an,*… is called *Geometric sequence or Geometric progression *

if *an *+ 1 / *an *= r, *n *Î **N**, where *a*1 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, ar^{2}, ar^{3}, ar^{4}, and so on.

Where a is the first term and r is the common ratio.

The nth term of a GP is a_{n}=T_{n} = ar^{n-1}

*a *= the first term, *r *= the common ratio, *l *= the last term,

*n *= the numbers of terms,

S* _{n} *= the sum of first

*n*terms.

Suppose a, ar, ar^{2}, ar^{3},……ar^{n-1} is the given Geometric Progression.

Then the sum of n terms of GP is given by:

S_{n} = a + ar + ar^{2 }+ ar^{3 }+…+ ar^{n-1}

**S**_{n}= a[(r^{n }– 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, ar^{2}, ar^{3},……ar^{n-1}

a, ar, ar^{2}, ar^{3},……ar^{n-1} is called finite geometric series.

The sum of finite Geometric series is S_{n} = a[(r^{n }– 1)/(r – 1)] if r ≠ 1 and r > 1

Infinite Geometric Progression

Terms of an infinite G.P. can be written as a, ar, ar^{2}, ar^{3}, ……ar^{n-1},…….

a, ar, ar^{2}, ar^{3}, ……ar^{n-1},……. is called infinite geometric series.

The sum of infinite geometric series is S_{n} =

**Properties** of GP:

- Three non-zero terms a, b, c are in GP if and only if b
^{2}= ac - In a GP,

Three consecutive terms can be taken as a/r, a, ar

Four consecutive terms can be taken as a/r^{3}, a/r, ar, ar^{3}

Five consecutive terms can be taken as a/r^{2}, a/r, a, ar, ar^{2} - In a finite GP, the product of the terms equidistant from the beginning and the end is the same

That means, t_{1}.t_{n}= t_{2}.t_{n-1}= t_{3}.t_{n-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 a
_{1}, a_{2}, a_{3},… is a GP of positive terms then log a_{1}, log a_{2}, log a_{3},… 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,*…*, G*n *be *n *numbers between positive numbers *a *and *b *such that

*a*,G1,G2,G3,*…*,G*n*,*b *is a G.P.

Thus, *b *being the (*n *+ 2)th term,

we have *b= ar ^{n+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 = *r*^{4} giving *r *= ± 4 (Taking real roots only)

For *r *= 4, we have G1 = *ar *= 4, G2 = *ar*^{2} = 16, G3 = *ar*^{3} = 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 6^{th} triangle inscribed we first have to find the side of 6^{th} triangle that is the 6^{th} term in the series

n^{th} term in GP is given by t_{n} = ar^{n-1}

⇒ t_{6} = (20) (1/2)^{6-1}

⇒ t_{6} = 20/ 2^{5}

= 20/ (4 × 2^{3})

⇒ t_{6} = 5/8

Hence the side of 6^{th} equilateral triangle is 5/8 cm and hence its perimeter would be thrice its side length because it’s an equilateral triangle

Perimeter of 6^{th} 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;

S_{n} = a[(r^{n }– 1)/(r – 1)]

S_{5} = 10[(3^{5 }– 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

a_{n} = ar^{n-1}

Therefore,

a_{10} = 2 x 2^{10 – 1}

= 2 × 2^{9}

= 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 S* _{n}* = 5 + 55 + 555 + ….. up to

*n*terms

(ii) Given, .6 + .66 + . 666 + …

Let S* _{n}* = 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 *x*1, *x*2, ... , *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) 1^{2} + 2^{2} + 3^{2} +… + *n*^{2} (sum of squares of the first *n *natural numbers)

(iii) 1^{3 }+ 2^{3 } + 3^{3 } +… + *n*^{3 }(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,

S_{n} = 1 + 2 + 3 + 4 + ….. + n

S_{n} = n/2 [2a + (n – 1)d]

Substituting a = 1 and d = 1,

S_{n} = (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) 1^{2} + 2^{2} + 3^{2} +… + *n*^{2} (sum of squares of the first *n *natural numbers)

The squares of natural numbers are: 1^{2}, 2^{2}, 3^{2}, 4^{2},…

Or

1, 4, 9, 16, ….

We can express the sum of n terms as: 1^{2} + 2^{2} + 3^{2} +…+ n^{2}

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^{3} – (k – 1)^{3} = 3k^{2} – 3k + 1

Substituting k = 1,

1^{3} – (1 – 1)^{3} = 3(1)^{2} – 3(1) + 1

1^{3} – 0^{3} = 3(1)^{2} – 3(1) + 1….(i)

Substituting k = 2,

2^{3} – (2 – 1)^{3} = 3(2)^{2} – 3(2) + 1

2^{3} – 1^{3} = 3(2)^{2} – 3(2) + 1….(ii)

Substituting k = 3,

3^{3} – (3 – 1)^{3} = 3(3)^{2} – 3(3) + 1

3^{3} – 2^{3} = 3(3)^{2} – 3(3) + 1….(iii)

Substituting k = 4,

4^{3} – (4 – 1)^{3} = 3(4)^{2} – 3(4) + 1

4^{3} – 3^{3} = 3(4)^{2} – 3(4) + 1….(iv)

Substituting k = n,

n^{3} – (n – 1)^{3} = 3(n)^{2} – 3(n) + 1

Now, adding both sides of these equations together, we get;

1^{3} – 0^{3} + 2^{3} – 1^{3} + 3^{3} – 2^{3} + … + n^{3} – (n – 1)^{3} = 3(1^{2} + 2^{2 }+ 3^{2} + 4^{2} + … + n^{2}) – 3(1 + 2 + 3 + 4 + … + n) + n(1)

n^{3} – 0^{3} = 3(1^{2} + 2^{2}+ 3^{2} + 4^{2} + … + n^{2}) – 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) (2n^{3} + 3n^{2} + n)

= (1/6) [n(2n^{2} + 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) 1^{3 }+ 2^{3 } + 3^{3 } +… + *n*^{3 }(sum of cubes of the first *n *natural numbers).

The squares of natural numbers are: 1^{3}, 2^{3}, 3^{3}, 4^{3},…

Or

1, 8, 27, 64,….

We can express the sum of n terms as: 1^{3} + 2^{3} + 3^{3} +…+ n^{3}

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} – k^{4} = 4k^{3} + 6k^{2} + 4k + 1

Substituting k = 1, 2, 3, …, n

2^{4} – 1^{4} = 4(1)^{3} + 6(1)^{2} + 4(1) + 1

3^{4} – 2^{4} = 4(2)^{3} + 6(2)^{2} + 4(2) + 1

4^{4} – 3^{4} = 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

n^{4} – (n – 1)^{4} = 4(n – 1)^{3} + 6(n – 1)^{2} + 4(n – 1) + 1

(n + 1)^{4} – n^{4} = 4n^{3} + 6n^{2} + 4n + 1

By adding both sides of these equations, we get;

2^{4} – 1^{4} + 3^{4} – 2^{4} + 4^{4} – 3^{4} + …. + (n + 1)^{4} – n^{4} = 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 + …. + 4n^{3} + 6n^{2} + 4n + 1

(n + 1)^{4} – 1^{4} = 4(1^{3} + 2^{3} + 3^{3} +…+ n^{3}) + 6(1^{2} + 2^{2} + 3^{2} + …+ n^{2}) + 4(1 + 2 + 3 +…+ n) + n

We know that,

And

Thus,

By rearranging the terms,

= (1/4) [n^{4} + 4n^{3} + 6n^{2} + 4n – 2n^{3} – 3n^{2} – n – 2n^{2} – 2n – n]

= (1/4) [n^{4} + 2n^{3} + n^{2}]

= (1/4)[n^{2}(n^{2} + 2n + 1)]

= (1/4)[n^{2}(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 S_{n} be the sum of its n terms and an be its nth term. Then,

S_{n} =2 + 5 + 14 + 41 + … + a_{n}….(i)

And

S_{n} = 2 + 5+ 14 + 41 + … + a_{n – 1} + a_{n}….(ii)

Subtracting equation (ii) from (i), we get

0 = 2 + [3 + 9 + 27 + … + (n – 1) term] – a_{n}

⇒ a_{n} = 2 + [3 + 9 + 27 +… + (n- 1) term]

Here, 3 + 9 + 27 + … is a geometric series.

So, a_{n} = 2 + [3(3^{n-1} – 1)/2]

= [4 + 3^{n} – 3]/2

= (1 + 3^{n})/2

Now, we need to find the sum of the series whose general term is (1 + 3^{n})/2

Sn = [(1 + 3)/2] + [(1 + 3^{2})/2] + [(1 + 3^{3})/2] + …. + (1 + 3^{n})/2

= (1/2) [(3 + 3^{2} + 3^{3} + …. + 3^{n}) + (1 + 1 + 1 + … + n)]

= (1/2) {[3(3^{n} – 1)/(3 – 1)] + n}

= (1/2) [(3/2)(3^{n} – 1) + n]

= [(3^{n+1} – 3 + 2n)/4]

Therefore, the sum of the given series = (3^{n+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 p ^{th} and q^{th} terms of a G.P. are q and p respectively, show that its (p + q)^{th} term is **

**Solution:**

Hence the proof.