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NCERT Class 11 Maths Chapter 8 Notes: Sequences and Series

Sequences and Series is a topic in Class 11 Maths that deals with ordered lists of numbers following a specific pattern. A sequence is a set of numbers arranged in a particular order, while a series is the sum of the terms of a sequence. This chapter covers important types like arithmetic and geometric progressions.

Sequence

The general definition is arranging something in a particular order.

Generally, we write the terms of a sequence by $a_1, a_2, a_3, \ldots \ldots \ldots a_n, \ldots \ldots$, etc., the subscripts define the position of the term. The nth term defines the nth position of the sequence.
From the above sentences, we can say that a function whose domain is the set of natural numbers or a subset of it can be defined as a sequence. A functional notation can be used for $a_n$ is a(n)

Series

Let $a_1, a_2, a_3, \ldots \ldots \ldots a_n, \ldots \ldots$, be a given sequence. Then, the expression $a_1+a_2+a_3+\ldots \ldots .+a_n$ is known as a series.

The sum of the following series is denoted by ∑a_n is a(n).

Arithmetic Progression (A.P)

Let the sequence be

$a_1, a_2, a_3, \ldots \ldots a_n$ called an AP when the difference between $a_n$ and $a_n+1$ is $d$
Thus $a_n$ can be written as $a_n=a+(n-1) d$
Let the sequence $a_1+a_2+\ldots. .+a_n$
Then the summation is $S_n=\left(\frac{n}{2}\right)[2 a+(n-1) d]$

Example:

If the 1,5,9,13… sequence is an A.P., what will be the nth term in the sequence?

Solution:

Given, sequence 1,5,9,13 . . . . . . .

Difference between the numbers, d =$(13-9)=4$

The nth term will be,

$\begin{aligned} & a_n=1+(n-1) 4 \\ & \Rightarrow a_n=1+4 n-4 \\ & \Rightarrow a_n=4 n-3\end{aligned}$

Arithmetic Mean

If the two given numbers are p and q. Insert the number F between p and q so that p, F, and q are in arithmetic progression. If F is the arithmetic mean of p and q numbers. Then we will have,

$\begin{aligned} & F-p=q-F \\ & \Rightarrow F+F=p+q \\ & \Rightarrow 2 F=p+q \\ & \Rightarrow F=\frac{p+q}{2}\end{aligned}$

If the terms of an A.P. are increased, decreased, multiplied, or divided by the same constant, they remain in A.P.
If $a_1, a_2, a_3 \ldots$ are in A.P. with common difference $d$, then
(i) $a_1 \pm k, a_2 \pm k, a_3 \pm k, \ldots$ are also in A.P with common difference $d$.
(ii) $a_1 k, a_2 k, a_3 k, \ldots$ are also in A.P with common difference $d k(k \neq 0)$. and $\frac{a_1}{k}, \frac{a_2}{k}, \frac{a_3}{k} \ldots$ are also in A.P. with common difference $\frac{d}{k}(k \neq 0)$.
If $a_1, a_2, a_3 \ldots$ and $b_1, b_2, b_3 \ldots$ are two A.P., then
(i) $a_1 \pm b_1, a_2 \pm b_2, a_3 \pm b_3, \ldots$ are also in A.P
(ii) $a_1 b_1, a_2 b_2, a_3 b_3, \ldots$ and $\frac{a_1}{b_1}, \frac{a_2}{b_2}, \frac{a_3}{b_3}, \ldots$ are not in A.P.

If $a_1, a_2, a_3 \ldots$ and $a_n$ are in A.Ps, then
(i) $a_1+a_n=a_2+a_{n-1}=a_3+a_{n-2}=\ldots$
(ii) $a_r=\frac{a_{r-k}+a_{r+k}}{2} \forall k, 0 \leq k \leq n-r$
(iii) If $n^{\text {th }}$ term of any sequence is linear expression in $n$, then the sequence is an A.P.
(iv) If sum of $n$ terms of any sequence is a quadratic expression in $n$, then sequence is an A.P.

Geometric Progression (G.P)

Let us consider the following sequences:

2, 4, 8, 16, .…..

1/9, -1/27, 1/81, -1/243, ......

Let the sequence be $a_1, a_2, a_3, \ldots \ldots . a_n$ called a GP when the difference in the ratio between $a_n$ and $a_{n+1}$ is $r$ by letting $a_1=a$, we obtain a geometric progression,$a$, ar, ar ${ }^2$, ar ${ }^3, \ldots .$.

General Term Of A G.P

The general term of GP is $a_n=a r^{n-1}$

Sum to $n$ terms of a G.P

$\begin{aligned} & S_n=a+a r+a r^2+\ldots \ldots+a r^{n-1} \\ & \Rightarrow S_n=a \frac{\left(1-r^n\right)}{(1-r)}\end{aligned}$

Geometric Mean (G.M.)

The formula for GM is

G=√ab

(i) If the terms of a G.P. are multiplied or divided by the same non-zero constant $(k \neq 0)$, they remain in G.P.

If $a_1, a_2, a_3, \ldots$, are in G.P., then $a_1 k, a_2 k, a_3 k, \ldots$ and $\frac{a_1}{k}, \frac{a_2}{k}, \frac{a_3}{k}, \ldots$ are also in G.P. with same common ratio, in particularly if $a_1, a_2, a_3, \ldots$ are in G.P., then $\frac{1}{a_1}, \frac{1}{a_2}, \frac{1}{a_3}, \ldots$ are also in G.P.
(ii) If $a_1, a_2, a_3, \ldots$ and $b_1, b_2, b_3, \ldots$ are two G.P.s, then $a_1 b_1, a_2 b_2, a_3 b_3, \ldots$ and $\frac{a_1}{b_1}, \frac{a_2}{b_2}, \frac{a_3}{b_3}, \ldots$ are also in G.P.
(iii) If $a_1, a_2, a_3, \ldots$ are in A.P. $\left(a_i>0 \forall i\right)$, then $x^{a_1}, x^{a_2}, x^{a_3}, \ldots$, are in G.P. $(\forall x>0)$

(iv) If $a_1, a_2, a_3, \ldots, a_n$ are in G.P., then $a_1 a_n=a_2 a_{n-1}=a_3 a_{n-2}=\ldots$

Relationship Between A.M. and G.M

A=(a+b)/2 and G=√(ab)

The relation is A ≥ G.

Sum To n Terms Of Special Series

$\begin{aligned} & 1+2+3+\ldots .+n \quad \text { (sum of first } \mathrm{n} \text { natural numbers) } \\ & k=\frac{n(n+1)}{2} \\ & 1^2+2^2+3^2+\ldots .+n^2 \quad \text { (sum of squares of the first } \mathrm{n} \text { natural numbers) } \\ & k=\frac{n(n+1)(2 n+1)}{6} \\ & 1^3+2^3+3^3+\ldots \ldots+n^3 \quad \text { (sum of cubes of the first } \mathrm{n} \text { natural numbers) } \\ & k=\left[\frac{n(n+1)}{2}\right]^2\end{aligned}$

Sequences and Series: Previous Year Question and Answer

Given below are some previous year question answers of various examinations from the NCERT class 11 chapter 8, Sequences and Series:

Question 1: If the sum of the second, fourth and sixth terms of a G.P. of positive terms is 21 and the sum of its eighth, tenth and twelfth terms is 15309, then the sum of its first nine terms is:

Solution:
Given:
$\begin{aligned}
& a r+a r^3+a r^5=21 ..............(i) \\
& a r^7+a r^9+a r^{11}=15309.............(ii)
\end{aligned}$
Dividing equation (ii) by equation (i), we get,
$\frac{r^7}{r}=729 \Rightarrow r=3 \quad(r>0)$
Using (i)
$\begin{aligned}
& a(3+27+243)=21 \\
& \Rightarrow a=\frac{21}{273}=\frac{1}{13} \\
& S_9=\frac{\frac{1}{13}\left(3^9-1\right)}{3-1}
\end{aligned}$
$\begin{aligned} & =\frac{1}{26}\left(3^9-1\right) \\ & =\frac{1}{26}\left(3^3-1\right)\left(3^6+1+3^3\right) \\ & =3^6+3^3+1 \\ & =757\end{aligned}$
Hence, the correct answer is $757$.

Question 2: If the sum of the first 10 terms of the series $\frac{4.1}{1+4.1^{4}}+\frac{4.2}{1+4.2^{4}}+\frac{4.3}{1+4.3^{4}}+\ldots$ is $\frac{m}{n}$, where $\operatorname{gcd}(m, n)=1$, then $m+n$ is equal to______.

Solution:
Given:
$T_r = \frac{4r}{1 + 4r^4}$
$= \frac{4r}{(2r^2 + 2r + 1)(2r^2 - 2r + 1)}$
$= \frac{(2r^2 + 2r + 1) - (2r^2 - 2r + 1)}{(2r^2 + 2r + 1)(2r^2 - 2r + 1)}$
$= \frac{1}{2r^2 - 2r + 1} - \frac{1}{2r^2 + 2r + 1}$
Now, sum of first 10 terms $S_{10} = \sum_{r=1}^{10}$ $T_r= \sum_{r=1}^{10} \left(\frac{1}{2r^2 - 2r + 1} - \frac{1}{2r^2 + 2r + 1}\right)$
$= \left(\frac{1}{1} - \frac{1}{5}\right) + \left(\frac{1}{5} - \frac{1}{13}\right) + \cdots + \left(\frac{1}{181} - \frac{1}{221}\right)$
$= 1 - \frac{1}{221} = \frac{220}{221}$
$m = 220, \quad n = 221$
$m + n = 441$
Hence, the correct answer is $441$.

Question 3: Let $\mathrm{a}_{\mathrm{n}}$ be the $\mathrm{n}^{\text {th }}$ term of an A. P.
If $\mathrm{S}_{\mathrm{n}}=\mathrm{a}_1+\mathrm{a}_2+\mathrm{a}_3+\ldots+\mathrm{a}_{\mathrm{n}}=700, \mathrm{a}_6=7$ and $S_7=7$, then $a_n$ is equal to :

Solution:
Given:
$\begin{aligned}
& a_6=7=a+5 d=7 \\
& S_7=\frac{7}{2}(2 a+6 d)=7(a+3 d)=7 \\
& \Rightarrow a+3 d=1
\end{aligned}$
Using (i) & (ii)
$\begin{aligned}
& \Rightarrow \quad 2 d=6 \Rightarrow d=3, a_1=-8 \\
& S_n=\frac{n}{2}(2 a+(n-1) d)=700 \\
& =\frac{n}{2}(-16+(n-1) 3)=700 \\
& =n(3 n-19)=1400, n>0 \Rightarrow n=25 \\
& \Rightarrow \quad a_{25}=(-8)+24 \cdot(3)=72-8=64
\end{aligned}$
Hence, the correct answer is $64$.

Importance of NCERT Class 11 Math Chapter 8 Notes

The Sequence and Series Class 11 notes cover all important topics of the chapter. Class 11 Math chapter 8 notes are based on the Class 11 CBSE Maths Syllabus. So, it will help students to get a detailed and compact knowledge of the chapter.

• NCERT notes are very important to strengthen your key concepts in order to perform well in exams. Students must try to solve all the NCERT problems, including miscellaneous exercises, and if needed, refer to the NCERT Solutions for Class 11 Maths Chapter 8 Sequence and Series
• Students are advised to go through the NCERT Class 12 Maths Chapter 8 Notes before solving the questions.
• These NCERT notes are very useful for boosting your exam preparation and quick revision.

NCERT Class 11 Notes Chapter Wise

All the links to chapter-wise notes for NCERT class 11 maths are given below: