Sequences and Series Formulas For JEE 2026
Sequences and Series is an important chapter in JEE Mathematics and is known for being quite scoring, with around 2–3 questions asked every year. In this topic, students learn about Arithmetic Progression (AP), Geometric Progression (GP), and Harmonic Progression (HP). These concepts help in understanding number patterns and make it easier to solve sequence-based problems.
The chapter also includes topics like AGP, special series, AM-GM-HM inequality, and telescoping series, which are commonly asked in exams. Since many questions are based on formulas and standard results, regular practice is very important. With proper revision, this chapter can become a strong scoring area. For quick revision, students can also use a well-organized JEE Mains Maths Formula PDF to go through important formulas and concepts easily.
Sequences and Series
Basic Definitions
A sequence is an ordered list of numbers following a definite pattern. A series is what you get when you add up the terms of a sequence.
Sequence: An ordered list of numbers $$a_1, a_2, a_3, \ldots$$ where each term is determined by a rule. The $$n$$th term is often written as $$a_n$$.
Series: The sum of terms of a sequence: $$S_n = a_1 + a_2 + a_3 + \cdots + a_n$$. Here $$S_n$$ is called the sum of the first $$n$$ terms (or partial sum).
Note: The $$n$$th term can be recovered from the partial sums: $$a_n = S_n - S_{n-1}$$ for $$n \geq 2$$, and $$a_1 = S_1$$.
Arithmetic Progression (AP) Formula
An arithmetic progression is a sequence where the difference between consecutive terms is constant. For example, $$2, 5, 8, 11, 14, \ldots$$ is an AP where each term is 3 more than the previous one.
Common Difference ($$d$$): The constant difference between consecutive terms: $$d = a_{n+1} - a_n$$ for all $$n$$.
AP Formulas
For an AP with first term $$a$$ and common difference $$d$$:
| Quantity | Formula |
|---|---|
| $$n$$th term | $$a_n = a + (n-1)d$$ |
| Sum of $$n$$ terms | $$S_n = \dfrac{n}{2}[2a + (n-1)d]$$ |
| Sum (alternate form) | $$S_n = \dfrac{n}{2}(a + l)$$ where $$l = a_n$$ is the last term |
Worked Example
Find the 20th term and the sum of the first 20 terms of the AP: $$3, 7, 11, 15, \ldots$$
$$a = 3$$, $$d = 7 - 3 = 4$$
$$a_{20} = a + (20-1)d = 3 + 19 \times 4 = 3 + 76 =$$ 79
$$S_{20} = \frac{20}{2}(a + a_{20}) = 10(3 + 79) = 10 \times 82 =$$ 820
Worked Example
The sum of the first $$n$$ terms of an AP is $$S_n = 3n^2 + 5n$$. Find the common difference.
$$a_n = S_n - S_{n-1} = (3n^2 + 5n) - [3(n-1)^2 + 5(n-1)]$$
$$= 3n^2 + 5n - 3n^2 + 6n - 3 - 5n + 5 = 6n + 2$$
So $$a_n = 6n + 2$$, which gives $$a_1 = 8$$, $$a_2 = 14$$, $$d = 14 - 8 =$$ 6.
Arithmetic Mean Formulas
AM of two numbers $$a$$ and $$b$$:
$$$A = \frac{a + b}{2}$$$
$$n$$ arithmetic means between $$a$$ and $$b$$:
If $$A_1, A_2, \ldots, A_n$$ are inserted between $$a$$ and $$b$$ to form an AP, then:
$$$d = \frac{b - a}{n + 1}, \quad A_k = a + \frac{k(b - a)}{n + 1}$$$
Worked Example
Insert 3 arithmetic means between 4 and 20.
Here $$a = 4$$, $$b = 20$$, $$n = 3$$.
$$d = \frac{20 - 4}{3 + 1} = \frac{16}{4} = 4$$
$$A_1 = 4 + 4 = 8$$, $$A_2 = 4 + 8 = 12$$, $$A_3 = 4 + 12 = 16$$
The sequence is: $$4,$$ $$8, 12, 16,$$ $$20$$ ✓
Properties of AP
- If $$a$$, $$b$$, $$c$$ are in AP, then $$2b = a + c$$ (the middle term is the average of the other two)
- If each term is increased/decreased by a constant, the result is still an AP with the same $$d$$
- If each term is multiplied/divided by a non-zero constant $$k$$, the result is an AP with common difference $$kd$$ (or $$d/k$$)
- In an AP with $$n$$ terms, $$a_k + a_{n+1-k} = a_1 + a_n$$ for all $$k$$ (sum of terms equidistant from the ends is constant)
- Sum of $$n$$ terms of an AP is always a quadratic in $$n$$ (i.e., $$S_n = An^2 + Bn$$ for some constants $$A, B$$)
Tip: When three numbers are in AP, take them as $$a - d$$, $$a$$, $$a + d$$. When four numbers are in AP, take them as $$a - 3d$$, $$a - d$$, $$a + d$$, $$a + 3d$$. This simplifies calculations because the sum is easy to compute.
Geometric Progression (GP) Formula
A geometric progression is a sequence where each term is obtained by multiplying the previous term by a fixed number called the common ratio. For example, $$2, 6, 18, 54, \ldots$$ is a GP where each term is 3 times the previous one.
Common Ratio ($$r$$): The constant ratio between consecutive terms: $$r = \dfrac{a_{n+1}}{a_n}$$ for all $$n$$ (provided terms are non-zero).
GP Formulas
For a GP with first term $$a$$ and common ratio $$r$$ ($$r \neq 0$$):
| Quantity | Formula |
|---|---|
| $$n$$th term | $$a_n = ar^{n-1}$$ |
| Sum of $$n$$ terms ($$r \neq 1$$) | $$S_n = a\cdot\dfrac{r^n - 1}{r - 1} = a\cdot\dfrac{1 - r^n}{1 - r}$$ |
| Sum of $$n$$ terms ($$r = 1$$) | $$S_n = na$$ |
Worked Example
Find the 8th term and sum of first 8 terms of the GP: $$3, 6, 12, 24, \ldots$$
$$a = 3$$, $$r = \frac{6}{3} = 2$$
$$a_8 = 3 \times 2^{8-1} = 3 \times 128 =$$ 384
$$S_8 = 3 \times \frac{2^8 - 1}{2 - 1} = 3 \times \frac{256 - 1}{1} = 3 \times 255 =$$ 765
Sum to Infinity of a GP
Infinite GP Sum Formula
If $$|r| < 1$$:
$$$S_\infty = \frac{a}{1 - r}$$$
If $$|r| \geq 1$$, the infinite sum does not exist (the series diverges).
Worked Example
Find the sum of the infinite GP: $$1 + \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8} + \cdots$$
$$a = 1$$, $$r = \frac{1}{2}$$
Since $$|r| = \frac{1}{2} < 1$$:
$$S_\infty = \frac{1}{1 - \frac{1}{2}} = \frac{1}{\frac{1}{2}} =$$ 2
Worked Example
Express $$0.7\overline{2} = 0.7222\ldots$$ as a fraction.
$$0.7222\ldots = 0.7 + 0.02(1 + 0.1 + 0.01 + \cdots)$$
The GP in the bracket has $$a = 1$$, $$r = 0.1$$: $$S_\infty = \frac{1}{0.9} = \frac{10}{9}$$
$$0.7222\ldots = \frac{7}{10} + 0.02 \times \frac{10}{9} = \frac{7}{10} + \frac{2}{90} = \frac{63}{90} + \frac{2}{90} = \frac{65}{90} =$$ $$\dfrac{13}{18}$$
Geometric Mean Formulas
GM of two positive numbers $$a$$ and $$b$$:
$$$G = \sqrt{ab}$$$
$$n$$ geometric means between $$a$$ and $$b$$:
If $$G_1, G_2, \ldots, G_n$$ are inserted between $$a$$ and $$b$$ to form a GP, then:
$$$r = \left(\frac{b}{a}\right)^{\!\frac{1}{n+1}}, \quad G_k = a \cdot r^k = a\left(\frac{b}{a}\right)^{\!\frac{k}{n+1}}$$$
Worked Example
Insert 2 geometric means between 3 and 192.
$$a = 3$$, $$b = 192$$, $$n = 2$$.
$$r = \left(\frac{192}{3}\right)^{\!\frac{1}{3}} = (64)^{1/3} = 4$$
$$G_1 = 3 \times 4 = 12$$, $$G_2 = 3 \times 16 = 48$$
The sequence is: $$3,$$ $$12, 48,$$ $$192$$ ✓
Properties of GP
- If $$a$$, $$b$$, $$c$$ are in GP, then $$b^2 = ac$$
- If each term is multiplied/divided by a non-zero constant, the result is still a GP with the same $$r$$
- The product of terms equidistant from the beginning and end is constant: $$a_k \cdot a_{n+1-k} = a_1 \cdot a_n$$
- The reciprocals of a GP form a GP with common ratio $$\frac{1}{r}$$
- If $$a_1, a_2, a_3, \ldots$$ is a GP, then $$\log a_1, \log a_2, \log a_3, \ldots$$ is an AP (provided all terms are positive)
Tip: When three numbers are in GP, take them as $$\frac{a}{r}$$, $$a$$, $$ar$$. Their product is $$a^3$$ — this often simplifies problems immediately.
Harmonic Progression (HP) Formulas
A harmonic progression is a sequence whose reciprocals form an arithmetic progression. There is no direct formula for the sum of an HP — instead, we convert to AP, solve, and convert back.
Harmonic Progression: A sequence $$a_1, a_2, a_3, \ldots$$ is an HP if $$\frac{1}{a_1}, \frac{1}{a_2}, \frac{1}{a_3}, \ldots$$ is an AP.
Harmonic Mean Formula
HM of two numbers $$a$$ and $$b$$:
$$$H = \frac{2ab}{a + b}$$$
$$n$$th term of HP: Find the $$n$$th term of the corresponding AP and take its reciprocal.
Worked Example
Find the 5th term of the HP: $$\dfrac{1}{3}, \dfrac{1}{5}, \dfrac{1}{7}, \ldots$$
The corresponding AP is: $$3, 5, 7, \ldots$$ with $$a = 3$$, $$d = 2$$.
5th term of AP: $$a_5 = 3 + 4 \times 2 = 11$$
5th term of HP: $$\dfrac{1}{11}$$
Arithmetico-Geometric Progression (AGP) Formulas
An AGP is formed by multiplying corresponding terms of an AP and a GP. The general term is of the form (linear in $$n$$) $$\times$$ (exponential in $$n$$).
AGP: A sequence whose $$n$$th term is $$a_n = [a + (n-1)d] \cdot r^{n-1}$$, where $$a, d$$ are the AP parameters and $$r$$ is the GP common ratio.
Sum of an AGP
For $$S_n = a + (a+d)r + (a+2d)r^2 + \cdots + [a+(n-1)d]r^{n-1}$$:
Method: Multiply $$S_n$$ by $$r$$, subtract, and simplify.
Infinite sum ($$|r| < 1$$):
$$$S_\infty = \frac{a}{1 - r} + \frac{dr}{(1 - r)^2}$$$
Worked Example
Find the sum to infinity: $$S = 1 + 3x + 5x^2 + 7x^3 + \cdots$$ where $$|x| < 1$$.
This is an AGP with $$a = 1$$, $$d = 2$$, $$r = x$$.
$$S_\infty = \frac{1}{1 - x} + \frac{2x}{(1 - x)^2} = \frac{(1 - x) + 2x}{(1 - x)^2} = \frac{1 + x}{(1 - x)^2}$$
$$S =$$ $$\dfrac{1 + x}{(1 - x)^2}$$
Tip: The multiply-and-subtract trick (also called the "$$rS$$ method") works for any AGP. Write $$S$$, write $$rS$$ (shifted by one position), and subtract to turn the AGP into a GP. JEE frequently tests this technique.
Special Series Summation Formula
| Sum | Formula |
|---|---|
| $$\displaystyle\sum_{k=1}^n k = 1 + 2 + \cdots + n$$ | $$\dfrac{n(n+1)}{2}$$ |
| $$\displaystyle\sum_{k=1}^n k^2 = 1^2 + 2^2 + \cdots + n^2$$ | $$\dfrac{n(n+1)(2n+1)}{6}$$ |
| $$\displaystyle\sum_{k=1}^n k^3 = 1^3 + 2^3 + \cdots + n^3$$ | $$\left[\dfrac{n(n+1)}{2}\right]^2$$ |
Note: $$\sum k^3 = \left(\sum k\right)^2$$ — the sum of cubes equals the square of the sum of natural numbers.
Worked Example
Find $$1^2 + 2^2 + 3^2 + \cdots + 15^2$$.
$$\sum_{k=1}^{15} k^2 = \frac{15 \times 16 \times 31}{6} = \frac{7440}{6} =$$ 1240
Worked Example
Find the sum $$1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + \cdots + n(n+1)$$.
The general term is $$k(k+1) = k^2 + k$$.
$$\sum_{k=1}^n k(k+1) = \sum k^2 + \sum k = \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2}$$
$$= \frac{n(n+1)}{6}[(2n+1) + 3] = \frac{n(n+1)(2n+4)}{6} =$$ $$\dfrac{n(n+1)(n+2)}{3}$$
Worked Example
Find $$1^3 + 2^3 + 3^3 + \cdots + 10^3$$.
$$\sum_{k=1}^{10} k^3 = \left[\frac{10 \times 11}{2}\right]^2 = (55)^2 =$$ 3025
AM-GM-HM Inequality Formula
The AM-GM-HM inequality is one of the most important inequalities in mathematics and is extensively used in optimization problems in JEE.
AM-GM-HM Inequality
For positive real numbers $$a_1, a_2, \ldots, a_n$$:
$$$\text{AM} \geq \text{GM} \geq \text{HM}$$$
Explicitly:
$$$\frac{a_1 + a_2 + \cdots + a_n}{n} \geq \sqrt[n]{a_1 a_2 \cdots a_n} \geq \frac{n}{\frac{1}{a_1} + \frac{1}{a_2} + \cdots + \frac{1}{a_n}}$$$
Equality holds if and only if $$a_1 = a_2 = \cdots = a_n$$ (all numbers are equal).
For two positive numbers $$a$$ and $$b$$:
$$$\frac{a + b}{2} \geq \sqrt{ab} \geq \frac{2ab}{a + b}$$$
Also: $$\text{AM} \times \text{HM} = \text{GM}^2$$, i.e., $$A \cdot H = G^2$$.
Worked Example
Find the minimum value of $$x + \dfrac{1}{x}$$ for $$x > 0$$.
By AM-GM inequality:
$$\frac{x + \frac{1}{x}}{2} \geq \sqrt{x \cdot \frac{1}{x}} = \sqrt{1} = 1$$
$$x + \frac{1}{x} \geq 2$$
Equality when $$x = \frac{1}{x}$$, i.e., $$x = 1$$.
Minimum value $$=$$ 2 (at $$x = 1$$).
Worked Example
If $$a + b = 10$$ and $$a, b > 0$$, find the maximum value of $$ab$$.
By AM-GM: $$\frac{a + b}{2} \geq \sqrt{ab}$$
$$5 \geq \sqrt{ab}$$, so $$ab \leq 25$$
Maximum value of $$ab =$$ 25 (when $$a = b = 5$$).
Worked Example
The AM of two positive numbers is 10 and their GM is 8. Find their HM.
Using $$G^2 = A \times H$$:
$$H = \frac{G^2}{A} = \frac{64}{10} =$$ 6.4
Tip: AM-GM is the go-to tool when JEE asks for "minimum value of $$f(x)$$" and $$f(x)$$ is a sum of terms whose product is constant (or vice versa). The minimum of a sum occurs when all terms are equal.
Method of Differences and Telescoping Series
When the general term of a series is not obvious, but the differences between consecutive terms form a recognizable pattern, we can use the method of differences to find the $$n$$th term and the sum.
Method of Differences
Given a sequence $$a_1, a_2, a_3, \ldots$$:
Step 1: Compute the first differences $$d_k = a_{k+1} - a_k$$.
Step 2: If $$d_1, d_2, d_3, \ldots$$ form an AP or GP, then:
$$$a_n = a_1 + \sum_{k=1}^{n-1} d_k$$$
Step 3: Sum the series using the known formula for $$\sum d_k$$.
Worked Example
Find the $$n$$th term and sum of: $$1, 3, 7, 13, 21, \ldots$$
Step 1: Differences: $$2, 4, 6, 8, \ldots$$ — an AP with first term 2 and $$d = 2$$.
Step 2: $$a_n = 1 + \sum_{k=1}^{n-1} 2k = 1 + 2 \cdot \frac{(n-1)n}{2} = 1 + n(n-1) = n^2 - n + 1$$
Step 3: $$S_n = \sum_{k=1}^n (k^2 - k + 1) = \frac{n(n+1)(2n+1)}{6} - \frac{n(n+1)}{2} + n =$$ $$\dfrac{n(n^2 + 2)}{3}$$
Telescoping Series Formulas
Telescoping Technique
If $$a_k = f(k) - f(k+1)$$, then:
$$$\sum_{k=1}^n a_k = f(1) - f(n+1)$$$
Most intermediate terms cancel, leaving only the first and last.
Worked Example
Find $$\displaystyle\sum_{k=1}^n \frac{1}{k(k+1)}$$.
Use partial fractions: $$\frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1}$$
$$\sum_{k=1}^n \frac{1}{k(k+1)} = \left(\frac{1}{1} - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \cdots + \left(\frac{1}{n} - \frac{1}{n+1}\right)$$
$$= 1 - \frac{1}{n+1} =$$ $$\dfrac{n}{n+1}$$
Worked Example
Find $$\displaystyle\sum_{k=1}^n \frac{1}{k(k+1)(k+2)}$$.
Partial fractions: $$\frac{1}{k(k+1)(k+2)} = \frac{1}{2}\left[\frac{1}{k(k+1)} - \frac{1}{(k+1)(k+2)}\right]$$
This telescopes to:
$$\sum_{k=1}^n \frac{1}{k(k+1)(k+2)} = \frac{1}{2}\left[\frac{1}{1 \cdot 2} - \frac{1}{(n+1)(n+2)}\right] =$$ $$\dfrac{n(n+3)}{4(n+1)(n+2)}$$
Tip: When you see $$\frac{1}{k(k+1)}$$, $$\frac{1}{k(k+2)}$$, or similar terms in a JEE sum, immediately try partial fractions — chances are it will telescope. Also, $$\frac{1}{(2k-1)(2k+1)} = \frac{1}{2}\left(\frac{1}{2k-1} - \frac{1}{2k+1}\right)$$ is a common pattern
Sequences and Series Formulas For JEE 2026: Conclusion
Sequences and series play a crucial role in building a strong mathematical foundation for competitive exams. Concepts like arithmetic progression, geometric progression, and harmonic progression help in understanding number patterns and solving a wide range of problems efficiently. With regular practice and clarity in formulas, students can easily tackle both basic and advanced questions from this chapter.
To score well, consistent revision and application of concepts are key. Topics like AGP, special series, inequalities, and telescoping methods require conceptual understanding along with practice. By focusing on important formulas and solving a variety of questions, students can improve their speed, accuracy, and overall performance in the exam.
