Sequences: Definition, Types, General Form, Convergence & Examples 

1.0Introduction to Sequences

Sequences are one of the most basic ideas in higher-level mathematics. For JEE (Main and Advanced), sequences and series are essential because they cover arithmetic progression, geometric progression, limits, calculus and even probability. Mastering sequences gives us a strong footing to build concepts for advanced topics, such as series summation and infinite series, as well as concepts like mathematical induction. 

In simple words, a sequence is an ordered list of numbers according to some rule. A sequence is not just a random collection of numbers but rather a logical arrangement of numbers.

For example:

• 2,4,6,8,… is a sequence of even numbers.
• $1,\frac{1}{2},\frac{1}{3},\frac{1}{4},\dots$ is a sequence of reciprocals of natural numbers.

2.0Definition of a Sequence

A sequence is a function whose domain is the set of natural numbers (or a subset of it) and whose range is a set of real numbers.

If the nth term of a sequence is denoted as an, then the sequence can be written as:

$a_1,a_2,a_3,\dots ,a_n,\dots$

Where:

• $a_1$ = first term
• $a_2$ = second term
• $a_n$ = general term (nth term)

Example: If $a_n=2n+1$, then the sequence is 3,5,7,9,….

3.0Types of Sequences

Arithmetic Sequence (Arithmetic Progression – AP)

• Definition: A sequence where the difference between consecutive terms is constant.
• General form: $a,a+d,a+2d,a+3d,\dots$
• nth term:
  $a_n=a+(n-1)d$
• Example: 2,5,8,11,… where a=2,d=3.

Geometric Sequence (Geometric Progression – GP)

• Definition: A sequence where the ratio between consecutive terms is constant.
• General form: $a,ar,a{r}^2,a{r}^3,\dots$
• nth term: $a_n=a{r}^{n-1}$
• Example: 3,6,12,24,… where a=3,r=2.

3.3 Harmonic Sequence (HP)

• Definition: A sequence whose reciprocals form an AP.
• General form:$\frac{1}{a},\frac{1}{a+d},\frac{1}{a+2d},\dots$
• $Example:1,\frac{1}{2},\frac{1}{3},\frac{1}{4},\dots$

3.4 Fibonacci Sequence

• Definition: A sequence where each term is the sum of the two preceding terms.
• General form:

$F_n=F_{n-1}+F_{n-2}]with(F_1=1,F_2=1).$

• Sequence: 1, 1, 2, 3, 5, 8, 13,….

Special Sequences in JEE

• Recurrence sequences
• Sequence defined by limits $a_n=\frac{n}{n+1}$
• Alternating sequences $(-1{)}^n$
• Nested sequences

4.0General Form of a Sequence

If a sequence is defined by a formula:

$a_n=f(n)$

then the sequence is determined by substituting n=1,2,3,….

Examples:

1. $a_n=n^2⟹\text{Sequence: }1,4,9,16,\dots$
2. $a_n=\frac{1}{n}⟹\text{Sequence: }1,\frac{1}{2},\frac{1}{3},\dots$

5.0Convergence and Divergence of Sequences

A sequence is said to converge if its terms approach a finite limit as n→∞.

• $\text{Example: }{a}_n=\frac{1}{n}\to \text{converges to }0$
• $\text{Example: }{a}_n=n\to \text{diverges to infinity}$

6.0Solved Examples on Sequences

Example 1: Arithmetic Progression (AP)

Question:
The 3rd term of an AP is 12 and the 7th term is 24. Find the first term and common difference. Also, find the 10th term.

Solution:
Let the first term be ( a ) and the common difference be ( d ).

• 3rd term: $(a_3=a+2d=12)$
• 7th term: $(a_7=a+6d=24)$

Subtract the first equation from the second:

$$\begin{gathered} (a+6d)-(a+2d)=24-12 \\ 4d=12⟹d=3 \end{gathered}$$

Now, substitute ( d ) into the first equation:

$a+2\times 3=12⟹a=12-6=6$

10th term:

$a_{10}=a+9d=6+9\times 3=6+27=33$

Answer:
First term ( a = 6 ), Common difference ( d = 3 ), 10th term = 33.

Example 2: Geometric Progression (GP)

Question:
Find the 5th term and the sum of the first 6 terms of a GP whose first term is 2 and common ratio is 3.

Solution:
Given: ( a = 2 ), ( r = 3 ).

5th term:

$a_5=a\cdot r^4=2\cdot 3^4=2\cdot 81=162$

Sum of first 6 terms:

$S_6=a\frac{r^6-1}{r-1}=2\frac{3^6-1}{3-1}=2\frac{729-1}{2}=2\cdot 364=728$

Answer:
5th term = 162, Sum of first 6 terms = 728.

Example 3: Harmonic Progression (HP)

Question:
If the 2nd and 4th terms of an HP are 1 and $(\frac{1}{3})$ respectively, find the first term.

Solution:
Let the HP: $(a_1,a_2,a_3,a_4,\dots )$
The reciprocals form an AP: $(b_1,b_2,b_3,b_4,...)$

Given:

$(a_2=1⟹b_2=1)$

$(a_4=\frac{1}{3}⟹b_4=3)$

The AP: ( b_1, 1, b_3, 3, ... )

The difference in AP:

$(b_4=b_1+3d=3)$

$(b_2=b_1+d=1)$

Subtract:

$(b_1+3d)-(b_1+d)=3-1⟹2d=2⟹d=1$

Now, $(b_1+d=1⟹b_1=0)$

So, ( $a_1=\frac{1}{b_1})$, but $(b_1=0)$ is not possible (since 1/0 is undefined).
Check the problem again:

Let’s try again with the correct approach:

Let $(a_2=\frac{1}{A+d}=1⟹A+d=1⟹d=1-A)$

$(a_4=\frac{1}{A+3d}=\frac{1}{3}⟹A+3d=3)$

Now substitute ( d ) from above:

$$\begin{gathered} A+3(1-A)=3 \\ A+3-3A=3 \\ -2A+3=3 \\ -2A=0⟹A=0 \end{gathered}$$
So, first term $(a_1=\frac{1}{A}=\text{undefined})$.

Conclusion:
The given values lead to an undefined first term, which is not possible for a real HP. Please check for a different HP question, or the terms in the question.

Example 4: Finding the General Term

Question:
Find the general (nth) term for the sequence: 5, 8, 13, 20, 29, ...

Solution:
Let's check the pattern:

Find the difference between consecutive terms:

• 8 - 5 = 3
• 13 - 8 = 5
• 20 - 13 = 7
• 29 - 20 = 9

The differences are: 3, 5, 7, 9 (which form an AP with d = 2).

Let’s try to represent the nth term as a quadratic function: $(a_n=a{n}^2+bn+c)$.

Let’s use the first three terms:

• $(n=1:a_1=a+b+c=5)$
• $(n=2:a_2=4a+2b+c=8)$
• $(n=3:a_3=9a+3b+c=13)$

Now solve:

1. ( a + b + c = 5 )
2. ( 4a + 2b + c = 8 )
3. ( 9a + 3b + c = 13 )

Subtract 1 from 2:

$((4a+2b+c)-(a+b+c)=8-5⟹3a+b=3)(Eqn4)$

Subtract 2 from 3:

$((9a+3b+c)-(4a+2b+c)=13-8⟹5a+b=5)$

Now, subtract Eqn 4 from this:

$((5a+b)-(3a+b)=5-3⟹2a=2⟹a=1)$

From Eqn 4: $(3a+b=3⟹3\cdot 1+b=3⟹b=0)$

From the first equation: $(a+b+c=5⟹1+0+c=5⟹c=4)$

So, the nth term is:

$a_n=n^2+4$

Answer:
General term: $(a_n=n^2+4)$

Example 5: Sequence in a Word Problem

Question:
A student saves Rs 10 in the first week, Rs 15 in the second week, Rs 20 in the third week, and so on. How much will the student save in the 20th week? What is the total savings after 20 weeks?

Solution:
This is an AP with ( a = 10 ), ( d = 5 ).

20th term:

$a_{20}=a+(20-1)d=10+19\times 5=10+95=105$
So, in the 20th week, Rs 105 will be saved.

Total savings after 20 weeks:

$S_{20}=\frac{20}{2}[2a+(20-1)d]=10[2\times 10+19\times 5]=10[20+95]=10\times 115=1150$

Answer:
20th week savings = Rs 105; Total savings after 20 weeks = Rs 1150.

Example 6: Convergence of a Sequence

Question:
Does the sequence $(a_n=\frac{2n}{n+1})$ converge? If yes, what is its limit?

Solution:

${\lim }_{n\to \infty }\frac{2n}{n+1}={\lim }_{n\to \infty }\frac{2}{1+\frac{1}{n}}=\frac{2}{1+0}=2$

Answer:
Yes, the sequence converges to 2.