Sequence And Series

In mathematics, a sequence is an ordered list of objects (which can be numbers, letters, names, etc.). Like a set, it contains members (also called elements or terms), and the position of each member is fixed. Unlike a set, in a sequence, the order of elements matters, and the same element can appear multiple times at different positions. In real life, we can use sequence and series to determine the number of people sitting around a table, the rate of change in population, etc.

This Story also Contains

1. What is Sequence?
2. Difference between Sequence and Series
3. Solved Examples Based on Sequence and Series

In this article, we will cover the concept of the sequence and series. This is a crucial Chapter in class 11 Mathematics. It is not only essential for board exams but also for competitive exams like the Joint Entrance Examination(JEE Main) and other entrance exams such as SRMJEE, BITSAT, WBJEE, BCECE, and more. Questions based on this topic have been asked frequently in JEE Mains.

What is Sequence?

A sequence is formed when terms are written in order such that they follow a particular pattern. A sequence can have any number of terms which can be finite or infinite.

The total number of terms is called the length of the sequence.

E.g., 1, 2, 3, 4, 5,.....

1, 4, 9, 16, ......

1/3, 1/4, 1/5, 1/6, ......

Based on the number of terms, there are two types of sequences:

1. Finite sequence
2. Infinite Sequence

Finite Sequence

If the sequence has only a finite number of terms, then the sequence is called a finite sequence.

Eg, 2, 4, 6, 8

Infinite Sequence

If a sequence has three dots at the end, then it indicates the list never ends. It has an infinite number of terms. Then, the sequence is said to be an infinite sequence.

$
\text { E.g., } 2,4,6,8,10, \ldots
$

Terms in a few sequences, follow a certain pattern which can be expressed mathematically as an algebraic expression. This expression can be used as a formula to obtain the $\mathrm{n}^{\text {th }}$ term (or any term) of the sequence without looking at the sequence.
$\mathrm{n}^{\text {th }}$ term of the Sequence

In sequences $\mathrm{n}^{\text {th }}$ term is usually denoted by $a_n$ or $t_n$ or $T_n$

So, in sequence $2,4,6,8, \ldots$
$a_1=2, a_2=4, a_3=6$, and so on

We can write it in compact form as $a_n=2 n$
$a_n$ or $n^{\text {th }}$ term is also called the General term of the sequence.

So a sequence can be written as $a_1, a_2, a_3$

a_n or n^th term is also called the General term of the sequence.

$a_n$ or $\mathrm{n}^{\text {th }}$ term is also called the General term of the sequence.

So a sequence can be written as $a_1, a_2, a_3$ $\qquad$

Conversely, if the general term of a sequence is given, we can find any term of that sequence.

Eg, If $T_n=2^n$, then fourth term can be obtained by putting $n=4$

So, $T_4=2^4=16$

If we add or subtract all the terms of a sequence we will get an expression, which is called a series. It is denoted by S_n.

If the sequence is $a_1, a_2, a_3, \ldots \ldots, a_n$, then it's sum i.e. $a_1+a_2+a_3+\ldots \ldots \ldots . .+a_n$ is a series.
$
\mathrm{S}_{\mathrm{n}}=a_1+a_2+a_3+\ldots \ldots \ldots . .+a_n=\displaystyle \sum_{\mathrm{r}=1}^{\mathrm{n}} a_r=\sum a_r
$

Then,
$
\begin{aligned}
& \mathrm{S}_{\mathrm{n}}-\mathrm{S}_{\mathrm{n}-1}=\left(a_1+a_2+a_3+\ldots \ldots \ldots . .+a_{n-1}+a_n\right) -\left(a_1+a_2+a_3+\ldots \ldots \ldots . .+a_{n-1}\right)
\end{aligned}
$

Thus, $a_n=\mathrm{S}_{\mathrm{n}}-\mathrm{S}_{\mathrm{n}-1}$

This is the formula for finding the general term of a sequence if the sum of n terms is given.

Some of the most common examples of sequences are:

• Arithmetic Sequence
• Geometric Sequence
• Harmonic Sequence

Arithmetic Progression

An arithmetic sequence is a sequence in which each term increases or decreases by a constant term or fixed number. This fixed number is called the common difference of an AP and is generally denoted by ‘d’.

Geometric progression

A geometric sequence is a sequence where the first term is non-zero and the ratio between consecutive terms is always constant. The ‘constant factor’ is called the common ratio and is denoted by ‘r’. r is also a non-zero number.

Harmonic Progression

A sequence $a_1, a_2, a_3, \ldots ., a_n, \ldots$ of non-zero numbers is called a harmonic progression if the sequence $\frac{1}{a_1}, \frac{1}{a_2}, \frac{1}{a_3}, \ldots ., \frac{1}{a_n}, \ldots$ is an arithmetic progression.

Or

Reciprocals of arithmetic progression is a Harmonic progression.

E.g. is an HP because their reciprocals 2, 5, 8, 11,... form an A.P.

Difference between Sequence and Series

A sequence is formed when terms are written in order such that they follow a particular pattern. A sequence can have any number of terms which can be finite or infinite.

E.g., 1, 2, 3, 4, 5,.....

1, 4, 9, 16, ......

1/3, 1/4, 1/5, 1/6, ......

If we add or subtract all the terms of a sequence we will get an expression, which is called a series.

If the sequence is $a_1, a_2, a_3, \ldots . ., a_n$, then it's sum i.e. $a_1+a_2+a_3+\ldots \ldots \ldots . .+a_n$ is a series.

Recommended Video Based on Sequence and Series

Solved Examples Based on Sequence and Series

Example 1: $\frac{2^3-1^3}{1 \times 7}+\frac{4^3-3^3+2^3-1^3}{2 \times 11}+\frac{6^3-5^3+4^3-3^3+2^3-1^3}{3 \times 15}+\cdots+\frac{30^3-29^3+28^3-27^3+\ldots+2^3-1^3}{15 \times 63}$ is equal to [JEE MAINS 2022]

Solution:

$
\begin{aligned}
& 2^3-1^3=7 \\
& 4^3-3^3+2^3-1^3=37+7=44 \\
& 6^3-5^3+4^3-3^3+2^3-1^3=91+44=135
\end{aligned}
$

Required expression
$
\begin{aligned}
& =1+\frac{4}{2}+\frac{9}{3}+\cdots-\cdots \\
& =1+2+3+\ldots+15 \\
& =120
\end{aligned}
$
Hence, the required answer is 120.

Example 2: Let $\mathrm{a}_{\mathrm{n}}=\int_{-1}^{\mathrm{n}}\left(1+\frac{\mathrm{x}}{2}+\frac{\mathrm{x}^2}{3}+\ldots \ldots+\frac{\mathrm{x}^{\mathrm{n}-1}}{\mathrm{n}}\right) \mathrm{dx}$ for every $\mathrm{n} \in \mathrm{N}$. Then the sum of all the elements of the set $\left\{\mathrm{n} \in \mathrm{N}: \mathrm{a}_{\mathrm{n}} \in(2,30)\right\}$ is $\qquad$ [JEE MAINS 2022]

Solution:

$\begin{aligned} & a_n=x+\frac{x^2}{4}+\frac{x^3}{9}+\cdots+\left.\frac{x^n}{n^2}\right|_{-1} ^n \\ & =(n+1)+\frac{n^2-1}{4}+\frac{n^3+1}{9}+\cdots \cdots+\frac{n^n-(-1)^n}{n^2} \\ & \therefore \quad a_1=2 \\ & a_2=2+1+\frac{4-1}{4}=\frac{15}{4} \\ & a_3=4+\frac{8}{4}+\frac{28}{9}=\frac{100}{9} \\ & a_4=5+\frac{15}{4}+\frac{65}{9}+\cdots>30 \\ & \therefore \text { Sum }=2+3=5\end{aligned}$

Hence, the required answer is 5.

Example 3: If the sum of the first ten terms of the series $\frac{1}{5}+\frac{2}{65}+\frac{3}{325}+\frac{4}{1025}+\frac{5}{2501}+\ldots$ is $\frac{\mathrm{m}}{\mathrm{n}}$, where $\mathrm{m}$ and $\mathrm{n}$ are co-prime numbers, then $\mathrm{m}+\mathrm{n}$ is equal [JEE MAINS 2022]

Solution:

$\begin{aligned} & \mathrm{S}=\frac{1}{5}+\frac{2}{65}+\frac{3}{325}+\frac{4}{1025}+\frac{5}{2501}+\cdots=\frac{1}{4 \times 1^4+1}+\frac{2}{4 \times 2^4+1}+\frac{3}{4 \times 3^4+1}+\frac{4}{4 \times 5^4}+\frac{5}{4 \times 5^4+1}+\cdots \\ & \mathrm{T}_{\mathrm{r}}=\frac{\mathrm{r}}{4 \mathrm{r}^4+1} \\ & \mathrm{~S}_{10}=\displaystyle \sum_{\mathrm{r}=1}^{10} \frac{\mathrm{r}}{4 \mathrm{r}^4+1}=\displaystyle \sum_{\mathrm{r}=1}^{10} \frac{\mathrm{r}}{4 \mathrm{r}^4+4 \mathrm{r}^2+1-4 \mathrm{r}^2} \\ & =\displaystyle \sum_{\mathrm{r}=1}^{10} \frac{\mathrm{r}}{\left(2 \mathrm{r}^2+1\right)^2-(2 \mathrm{r})^2} \\ & =\displaystyle \sum_{\mathrm{r}=1}^{10} \frac{\mathrm{r}}{\left(2 \mathrm{r}^2+2 \mathrm{r}+1\right)\left(2 \mathrm{r}^2-2 \mathrm{r}+1\right)}=\frac{1}{4} \displaystyle \sum_{\mathrm{r}=1}^{10}\left(\frac{1}{2 \mathrm{r}^2-2 \mathrm{r}+1}-\frac{1}{2 \mathrm{r}^2+2 \lambda+1}\right) \\ & =\frac{1}{4}\left[\left(1-\frac{1}{5}\right)+\left(\frac{1}{5}-\frac{1}{13}\right)+\left(\frac{1}{13}-\frac{1}{25}\right)+\cdots\left(\frac{1}{2 \times 10^2-2 \times 10+1}-\frac{1}{2 \times 10^2+2 \times 10+1}\right)\right] \\ & =\frac{1}{4}+\left[1-\frac{1}{221}\right]=\frac{1}{4} \times \frac{220}{221}=\frac{55}{221}=\frac{\mathrm{m}}{\mathrm{n}} \\ & \mathrm{m}+\mathrm{n}: 55+221=276\end{aligned}$

Hence, the required answer is 276.

Example 4: The series of positive multiples of 3 is divided into sets: $\{3\},\{6,9,12\},\{15,18,21,24,27\}, \cdots$ Then the sum of the elements in the $11^{\text {th }}$ set is equal to [JEE MAINS 2022]

Solution
$
\begin{aligned}
\mathrm{S}_{11} & =3[101+102+\cdots+121] \\
& =\frac{3}{2}(222) \times 21 \\
& =6993
\end{aligned}
$

Hence, the required answer is 6993.

Example 5: Example 5: Let $\left\{\mathrm{a}_{\mathrm{n}}\right\}_{\mathrm{n}=0}^{\infty}$ be a sequence such that $\mathrm{a}_0=\mathrm{a}_1=0$ and $\mathrm{a}_{\mathrm{n}+2}=2 \mathrm{a}_{\mathrm{n}+1}-\mathrm{a}_{\mathrm{n}}+1$ for all $n \geqslant 0$. Then, $\displaystyle \sum_{n=2}^{\infty} \frac{a_n}{7^n}$ is equal to: [JEE MAINS 2022]

Solution:

$\begin{aligned} & a_0=a_1=0 \\ & a_{n+2}=2 a_{n+1}-a_n+1 \\ & \Rightarrow \frac{a_{n+2}}{7^{n+2}}=2 \frac{a_{n+1}}{7^{n+2}}-\frac{a_n}{7^{n+2}}+\frac{1}{7^{n+2}} \\ & \Rightarrow \frac{a_{n+2}}{7^{n+2}}=\frac{2}{7} \cdot\left(\frac{a_{n+1}}{7^{n+1}}\right)-\frac{1}{7^2}\left(\frac{a_n}{7^n}\right)+\frac{1}{7^{n+2}}\end{aligned}$

Applying summation from $\mathrm{n}=0$ to infinity,
$
\Rightarrow \sum_{n=0}^{\infty} \frac{a_{n+2}}{7^{n+2}}=\frac{2}{7} \sum_0^{\infty} \frac{a_{n+1}}{7^{n+1}}-\frac{1}{49} \sum_0^{\infty} \frac{a_n}{7^n}+\sum_0^{\infty} \frac{1}{7^{n+2}}
$

Now

$\begin{aligned} & \sum_0^{\infty} \frac{\mathrm{a}_{\mathrm{n}+2}}{7^{\mathrm{n}+2}}=\frac{\mathrm{a}_2}{7^2}+\frac{\mathrm{a}_3}{7^3}+\cdots=\mathrm{S} \text { (let) } \\ & \text { And } \sum_0^{\infty} \frac{a_n+1}{7^{\mathrm{n}+1}}=\frac{\mathrm{a}_1}{7}+\frac{\mathrm{a}_2}{7^2}+\frac{\mathrm{a}_3}{7^3}+\cdots \\ & =\frac{\mathrm{a}_2}{7^2}+\frac{\mathrm{a}_3}{7^3}+\cdots\left(\text { as } \mathrm{a}_1=0\right) \\ & =\mathrm{S} \\ & \end{aligned}$

And $\begin{aligned} \sum_0^{\infty} \frac{\mathrm{a}_{\mathrm{n}}}{7^{\mathrm{n}}}= & \frac{\mathrm{a}_0}{1}+\frac{\mathrm{a}_1}{7}+\frac{\mathrm{a}_2}{7^2}+\cdots \\ & =\frac{\mathrm{a}_2}{7^2}+\frac{\mathrm{a}_3}{7^3}+\cdots=\mathrm{S}\end{aligned}$
$\begin{aligned} & \text { From (i) } \\ & \Rightarrow \mathrm{S}=\frac{2}{7} \mathrm{~S}-\frac{1}{49} \mathrm{~S}+\left(\frac{1}{7^2}+\frac{1}{7^3}+\cdots\right) \\ & \Rightarrow \mathrm{S}-\frac{2}{7} \mathrm{~S}+\frac{1}{49} \mathrm{~S}=\frac{1}{49} \cdot \frac{1}{1-\frac{1}{7}} \\ & \Rightarrow \frac{(49-14+1) \mathrm{S}}{49}=\frac{1}{7 \cdot 6} \\ & \Rightarrow \mathrm{S}=\frac{49}{7 \cdot 6 \cdot 36} \\ & \Rightarrow \mathrm{S}=\frac{7}{216}\end{aligned}$

Hence, the required answer is \frac{7}{216}