Introduction

Sequence and series help you to understand the various correlations and connections between different mathematical patterns. This can help you further solve complex concepts. This chapter is divided into arithmetic mean and progression, geometric mean and advance, and the sum of different series. For a strong foundation of this chapter, practicing enough problems related to Class 11 Maths Chapter 9 NCERT is necessary. 

In general terms, a sequence is a set of numbers that have been placed in a particular pattern and have something that connects them. The connecting factor between the terms of a sequence can be different in all cases. E.g.:- a₁, a₂, a₃, a₄, ……. is a sequence and can be infinite or finite.

On the other hand, a series is the sum of all the terms in that sequence. E.g.:- 

S_n = a₁+a₂+a₃ + .. + a_n

An AP is a sequence in which every term is created by adding or subtracting a definite number to the preceding digit is an arithmetic sequence.

Let us consider any two numbers, a and b. We insert a number A so that a, A, b is an AP. This means that A is the arithmetic mean of a and b. We can say that:-

A-a=b-A

A=(a+b)/2

A GP is a sequence in which every term is obtained by multiplying or dividing a definite number with the preceding digit is known as a geometric sequence.

Let us consider an as the first term of a GP and r as the common ratio. The GP becomes a, ar, ar², and so on. It may be noticed that this sequence follows a pattern, which is, an=ar^(n-1). This is the general term of a GP.

Let us consider a_n as the first term of a GP and r as the common ratio. S_n is the sum of all terms of the GP. 

S_n=a+ar+ar²……

Multiplying by r,

rS_n=ar+ar²+ar³……

Subtracting both, we get

S_n=a(rⁿ-1)/r-1

If we have two numbers, a and b, the GM can be calculated as G=sqrt(ab). For eg:- GM for 4 and 9 is sqrt(4x9)=6.

We know the formulas of AM and GM. Subtracting both, we get:

AM-GM= (a+b)/2 -sqrt(ab)

     = {(sqrt(a)-sqrt(b))}²/2 >=0

This implies that AM>=GM.

We can find the sum of some special series using the concepts above. Some of the formulas in this context include:-

• • sum of first n natural numbers = n(n + 1)/2

  • sum of squares of first n natural numbers = [n(n + 1)(2n + 1)]/6

  • sum of cubes of first n natural numbers = [n(n + 1)]²/4

1. Find the sum of all natural numbers lying between 100 and 1000, multiples of 5.

Solution:

The natural numbers between 100 and 1000, multiples of 5, are 105, 110, … 995.

It forms a sequence in A.P.

Where, a = 10, d = 5

Now,

a + (n -1)d = 995

105 + (n – 1)(5) = 995

5n = 995 – 105 + 5 = 895

n = 895/5=179

We know,

S_n = n/2 [2a + (n-1)d]

Therefore, the sum of all natural numbers lying between 100 and 1000, multiples of 5, is 98450.

 

2. If the sum of a certain number of terms of the A.P. 25, 22, 19, … is 116. Find the last term

Solution:

Given A.P.=25, 22, 19, …

Here, a = 25, d = 22 – 25 = -3

S_n= 116

The number of terms be n

So, we have

S_n = n/2 [2a + (n-1)d] = 116

116 = n/2 [2(25) + (n-1)(-3)]

232 = 53n – 3n²

3n² – 53n + 232 = 0

3n(n – 8) – 29(n – 8) = 0

(3n – 29) (n – 8) = 0

Hence,

n = 29/3 or n = 8

As n can only be an integral value, n = 8

Thus, 8th term is the last term of the A.P.

a₈ = 25 + (8 – 1)(-3)

= 25 – 21

= 4

 

3. Insert five numbers between 8 and 26 such that the resulting sequence is an A.P.

Solution:

Let’s assume A1, A2, A3, A4, and A5 to be five numbers between 8 and 26 such that 8, A1, A2, A3, A4, A5, 26 are in an A.P.

Here we have,

a = 8, b = 26, n = 7

So,

26 = 8 + (7 – 1) d

6d = 26 – 8 = 18

d = 3

Now,

A1 = a + d = 8 + 3 = 11

A2 = a + 2d = 8 + 2 × 3 = 8 + 6 = 14

A3 = a + 3d = 8 + 3 × 3 = 8 + 9 = 17

A4 = a + 4d = 8 + 4 × 3 = 8 + 12 = 20

A5 = a + 5d = 8 + 5 × 3 = 8 + 15 = 23

Therefore, the required five numbers between 8 and 26 are 11, 14, 17, 20, and 23.

 

4. A man starts repaying a loan as the first installment of Rs. 100. If he increases the installment by Rs 5 every month, what will he pay in the 30th installment?

Solution:

Given,

The first installment of the loan is Rs 100.

The second installment of the loan is Rs 105 as the installment increases by Rs 5 every month.

Thus, the amount that the man repays every month forms an A.P.

And the A.P. is 100, 105, 110, …

Where, the first term, a = 100

Common difference, d = 5

So, the 30th term in this A.P. will be

A₃₀  = a + (30 – 1)d

= 100 + (29) (5)

= 100 + 145

= 245

Therefore, the amount to be paid in the 30th installment will be Rs 245.

 

5. If a, b, c and d are in G.P. show that (a2 + b2 + c2)(b2 + c2 + d2) = (ab + bc + cd)2.

Solution:

Given, a, b, c, d are in G.P.

So, we have

bc = ad  … (1)

b² = ac  … (2)

c² = bd  … (3)

Taking the R.H.S. we have

R.H.S.

= (ab + bc + cd)²

= (ab + ad + cd)²  [Using (1)]

= [ab + d (a + c)]²

= a²b² + 2abd (a + c) + d² (a + c)²

= a²b² +2a²bd + 2acbd + d²(a² + 2ac + c²)

= a²b² + 2a²c² + 2b²c² + d²a² + 2d²b² + d²c²  [Using (1) and (2)]

= a²b²+ a²c² + a²c² + b²c² + b²c² + d²a² + d²b² + d²b² + d²c²

= a²b²+ a²c² + a²d² + b² × b² + b²c² + b²d² + c²b² + c² × c² + c²d²

[Using (2) and (3) and rearranging terms]

= a²(b² + c² + d²) + b² (b² + c² + d²) + c² (b²+ c² + d²)

= (a² + b² + c²) (b² + c² + d²)

= L.H.S.

Thus, L.H.S. = R.H.S.

Therefore,

(a² + b² + c²)(b² + c² + d²) = (ab + bc + cd)²

1. How to get Full Marks in Chapter 9 Sequence and Series?

Thorough knowledge of the topics in this chapter and good practice of questions from each case can indeed fetch you good marks.

2. What is the concept of Arithmetic Progression?

AP is one of the most important topics of class 11 Maths. It involves a sequence of numbers where the difference between the adjacent numbers is constant.

3. What are the Topics Covered in NCERT Solutions Class 11 Maths Chapter 9?

As mentioned above, the main topics covered in Chapter 9 include AP, GP, Arithmetic, and Geometric Mean and the relationship between them, along with sum to n terms of special series or GP.