Mathematics Sequences and Series

Sequences and Series Formula Sheet

All key Sequences and Series formulas for JEE Main & Advanced: AP, GP, HP, AGP, AM-GM-HM, special summations, and telescoping series quick revision.

7 min read Updated Jun 2026 #formula sheet#quick revision#jee-main

Every must-know formula for Sequences and Series in one scannable sheet, grouped by progression type. Use it for last-minute revision before JEE Main and Advanced.

Arithmetic Progression (AP)

General form: $a,\ a+d,\ a+2d,\ a+3d,\ \ldots$ where $a$ is the first term and $d$ is the common difference.

$$\boxed{T_n = a + (n-1)d}$$$$\boxed{S_n = \frac{n}{2}\big[2a + (n-1)d\big] = \frac{n}{2}(a + l)}$$

where $l = T_n$ is the last term.

QuantityFormulaNotes
$n$th term$T_n = a + (n-1)d$Not $a + nd$ — first term is $a$
Sum of $n$ terms$S_n = \frac{n}{2}[2a + (n-1)d]$Note the $2a$, not $a$
Sum (last term known)$S_n = \frac{n}{2}(a + l)$Count $\times$ average of ends
AP test$T_{n+1} - T_n = $ constantOr $2b = a + c$ for $a,b,c$
$n$th term from sum$T_n = S_n - S_{n-1}$Valid for $n \geq 2$

Key Properties

  • Three terms in AP: $$\boxed{2b = a + c}$$
  • Sum of equidistant terms: $T_k + T_{n-k+1} = a + l$ (constant)
  • Arithmetic Mean of $a$ and $b$: $A = \dfrac{a+b}{2}$
  • Inserting $n$ AMs between $a$ and $b$: common difference $d = \dfrac{b-a}{n+1}$
  • $S_n,\ S_{2n} - S_n,\ S_{3n} - S_{2n},\ \ldots$ are themselves in AP

Smart Selection of Terms

TermsChoose asSum
3 terms$a-d,\ a,\ a+d$$3a$
4 terms$a-3d,\ a-d,\ a+d,\ a+3d$$4a$
5 terms$a-2d,\ a-d,\ a,\ a+d,\ a+2d$$5a$
High-Yield Shortcut
If $S_n$ is quadratic in $n$, the sequence is an AP. Recover the AP using $T_n = S_n - S_{n-1}$.

Geometric Progression (GP)

General form: $a,\ ar,\ ar^2,\ ar^3,\ \ldots$ with first term $a \neq 0$ and common ratio $r \neq 0$.

$$\boxed{T_n = ar^{n-1}}$$$$\boxed{S_n = a\frac{r^n - 1}{r - 1} = a\frac{1 - r^n}{1 - r}\quad (r \neq 1)}$$
QuantityFormulaNotes
$n$th term$T_n = ar^{n-1}$Not $ar^n$
Sum ($r > 1$)$S_n = a\frac{r^n - 1}{r - 1}$Positive denominator
Sum ($r < 1$)$S_n = a\frac{1 - r^n}{1 - r}$Positive denominator
Sum ($r = 1$)$S_n = na$Formula above is undefined
Common ratio$r = \frac{T_{n+1}}{T_n}$Constant for a GP

Key Properties

  • Three terms in GP: $$\boxed{b^2 = ac}$$
  • Product of equidistant terms: $T_k \cdot T_{n-k+1} = a \cdot l$ (constant)
  • Geometric Mean of $a$ and $b$: $G = \sqrt{ab}$
  • Inserting $n$ GMs between $a$ and $b$: ratio $r = \left(\dfrac{b}{a}\right)^{\frac{1}{n+1}}$
  • Reciprocals of GP terms form a GP with ratio $\frac{1}{r}$
  • $\frac{T_8}{T_5} = r^3$ — divide terms to get $r$ raised to the index gap

Smart Selection of Terms

TermsChoose asProduct
3 terms$\frac{a}{r},\ a,\ ar$$a^3$
4 terms$\frac{a}{r^3},\ \frac{a}{r},\ ar,\ ar^3$$a^4$
5 terms$\frac{a}{r^2},\ \frac{a}{r},\ a,\ ar,\ ar^2$$a^5$
Watch the Sign
For negative ratios, track even/odd powers: $(-2)^4 = +16$ but $(-2)^5 = -32$. Both sum forms are valid — pick the one with cleaner arithmetic.

Infinite GP

Converges if and only if $|r| < 1$.

$$\boxed{S_\infty = \frac{a}{1-r}\quad (|r| < 1)}$$
QuantityFormulaNotes
Infinite sum$S_\infty = \frac{a}{1-r}$Only when $\lvert r\rvert < 1$
Convergence$\lvert r\rvert < 1$Use absolute value; $1-r$ in denominator, not $1-\lvert r\rvert$
Pure recurring$0.\overline{a_1\ldots a_k} = \frac{a_1\ldots a_k}{\underbrace{99\ldots9}_{k}}$$k$ digits $\to k$ nines
Mixed recurring$0.a_1\ldots a_m\overline{b_1\ldots b_k} = \frac{a_1\ldots a_m b_1\ldots b_k - a_1\ldots a_m}{\underbrace{9\ldots9}_{k}\underbrace{0\ldots0}_{m}}$$k$ nines then $m$ zeros

Standard infinite-series results (for $|x| < 1$):

$$1 - x + x^2 - x^3 + \cdots = \frac{1}{1+x}\qquad (\text{ratio } -x)$$$$0.\overline{9} = 1 \quad\text{(exactly)}$$
Convergence First
Always verify $|r| < 1$ before applying $S_\infty = \frac{a}{1-r}$. For $r=2$, “$\frac{1}{1-2}=-1$” is meaningless — the series diverges.

Harmonic Progression (HP)

A sequence whose reciprocals form an AP. If the reciprocal AP has first term $a$ and common difference $d$:

$$\boxed{H_n = \frac{1}{a + (n-1)d}}$$
QuantityFormulaNotes
$n$th term$H_n = \frac{1}{a+(n-1)d}$Work with reciprocal AP
Reciprocal relation$\frac{1}{H_n} = \frac{1}{H_1} + (n-1)d$$d = \frac{1}{H_2} - \frac{1}{H_1}$
Three terms in HP$\frac{2}{b} = \frac{1}{a} + \frac{1}{c}$Equivalent to $b = \frac{2ac}{a+c}$
Harmonic Mean (2 nos.)$H = \frac{2ab}{a+b}$Not $\frac{a+b}{2ab}$
Harmonic Mean ($n$ nos.)$H = \frac{n}{\frac{1}{a_1} + \cdots + \frac{1}{a_n}}$Reciprocal of mean of reciprocals
  • Three terms in HP: $$\boxed{b = \frac{2ac}{a+c}}$$
  • Selection of 3 terms: $\frac{1}{a-d},\ \frac{1}{a},\ \frac{1}{a+d}$ (reciprocals of an AP)
No Sum Formula
HP has no simple sum formula. Convert to the reciprocal AP, find terms, then add as fractions. All terms should keep the same sign.

AM, GM, HM and the Mean Inequality

For positive numbers $a$ and $b$:

MeanTwo numbers$n$ numbers
Arithmetic (AM)$\frac{a+b}{2}$$\frac{a_1 + \cdots + a_n}{n}$
Geometric (GM)$\sqrt{ab}$$\sqrt[n]{a_1 a_2 \cdots a_n}$
Harmonic (HM)$\frac{2ab}{a+b}$$\frac{n}{\frac{1}{a_1} + \cdots + \frac{1}{a_n}}$
$$\boxed{\text{AM} \geq \text{GM} \geq \text{HM}}$$$$\boxed{\frac{a+b}{2} \geq \sqrt{ab} \geq \frac{2ab}{a+b}}$$

Equality holds if and only if all numbers are equal ($a = b$).

Key Relations

$$\boxed{\text{GM}^2 = \text{AM} \times \text{HM}}$$
  • Standard optimization result: $$\boxed{x + \frac{k}{x} \geq 2\sqrt{k}\quad (x > 0)}$$(so $x + \frac{1}{x} \geq 2$)
  • Weighted AM-GM: $\dfrac{ma + nb}{m+n} \geq \sqrt[m+n]{a^m b^n}$, equality when $a = b$

Advanced Results (JEE Advanced)

  • Power Mean Inequality (for $r > s$): $$\left(\frac{a_1^r + \cdots + a_n^r}{n}\right)^{1/r} \geq \left(\frac{a_1^s + \cdots + a_n^s}{n}\right)^{1/s}$$ ($r=1, s=0$ gives AM $\geq$ GM; $r=0, s=-1$ gives GM $\geq$ HM)
  • Cauchy-Schwarz Inequality: $$(a_1^2 + \cdots + a_n^2)(b_1^2 + \cdots + b_n^2) \geq (a_1 b_1 + \cdots + a_n b_n)^2$$
Positivity Required
AM-GM-HM applies only to positive numbers. For negatives the inequality can reverse: with $a=-2,\ b=-3$, GM $\approx 2.45 >$ AM $=-2.5$.

Relationship Between AP, GP, HP

For a number $b$ between $a$ and $c$:

graph LR
    A["a, b, c in AP
b = (a+c)/2"] --> M["AM ≥ GM ≥ HM
(positive a, c)"] G["a, b, c in GP
b = √(ac)"] --> M H["a, b, c in HP
b = 2ac/(a+c)"] --> M

Arithmetico-Geometric Progression (AGP)

Each term is the product of an AP term and a GP term: $a,\ (a+d)r,\ (a+2d)r^2,\ \ldots$

$$\boxed{T_n = \big[a + (n-1)d\big]\, r^{n-1}}$$$$\boxed{S_n = \frac{a}{1-r} + \frac{dr\,(1 - r^{n-1})}{(1-r)^2} - \frac{[a+(n-1)d]\,r^n}{1-r}\quad (r \neq 1)}$$

When $r = 1$, the AGP reduces to an AP: $S_n = na + \frac{n(n-1)d}{2}$.

Infinite AGP ($|r| < 1$)

$$\boxed{S_\infty = \frac{a}{1-r} + \frac{dr}{(1-r)^2} = \frac{a + r(d-a)}{(1-r)^2}}$$

Must-Know Standard Sums ($|x| < 1$)

SeriesSum
$1 + 2x + 3x^2 + 4x^3 + \cdots$$\frac{1}{(1-x)^2}$
$x + 2x^2 + 3x^3 + 4x^4 + \cdots$$\frac{x}{(1-x)^2}$
$1 + 4x + 9x^2 + 16x^3 + \cdots$$\frac{1+x}{(1-x)^3}$
$a + (a+d)r + (a+2d)r^2 + \cdots$$\frac{a + r(d-a)}{(1-r)^2}$
Method: Multiply and Subtract
To sum an AGP, compute $S - rS$. The middle becomes a plain GP. This is often faster than plugging into the full formula.

Special Series and Standard Summations

$$\boxed{\sum_{k=1}^{n} k = \frac{n(n+1)}{2}}$$$$\boxed{\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}}$$$$\boxed{\sum_{k=1}^{n} k^3 = \left[\frac{n(n+1)}{2}\right]^2}$$
SeriesSumNotes
$\sum k$$\frac{n(n+1)}{2}$Natural numbers
$\sum k^2$$\frac{n(n+1)(2n+1)}{6}$Three factors, $\div 6$
$\sum k^3$$\left[\frac{n(n+1)}{2}\right]^2$Square of $\sum k$; $\div 4$
$\sum 2k$$n(n+1)$First $n$ even numbers
$\sum (2k-1)$$n^2$First $n$ odd numbers
$\sum k(k+1)$$\frac{n(n+1)(n+2)}{3}$Product of 2 consecutive
$\sum k(k+1)(k+2)$$\frac{n(n+1)(n+2)(n+3)}{4}$Product of 3 consecutive
$\sum k^4$$\frac{n(n+1)(2n+1)(3n^2+3n-1)}{30}$Rare in JEE

Useful Combinations and Patterns

  • Linear combination: $\sum (ak + b) = a\dfrac{n(n+1)}{2} + bn$
  • General product rule: $$\boxed{\sum_{k=1}^{n} k(k+1)\cdots(k+r-1) = \frac{n(n+1)\cdots(n+r)}{r+1}}$$
  • Range adjustment: $\displaystyle\sum_{k=a}^{b} f(k) = \sum_{k=1}^{b} f(k) - \sum_{k=1}^{a-1} f(k)$
  • Alternating sum: $1 - 2 + 3 - 4 + \cdots + (-1)^{n-1}n = \begin{cases} \frac{n+1}{2} & n \text{ odd} \\ -\frac{n}{2} & n \text{ even} \end{cases}$
Memory Hooks
Sum of first $n$ odd numbers $= n^2$. Sum of cubes $=$ (sum of naturals)$^2$. Verify any formula by checking $n = 1, 2, 3$.

Method of Differences (Telescoping)

If $T_n = f(n+1) - f(n)$, then almost everything cancels:

$$\boxed{S_n = \sum_{k=1}^{n} \big[f(k+1) - f(k)\big] = f(n+1) - f(1)}$$

Standard Telescoping Forms

SeriesSumKey step
$\sum \frac{1}{k(k+1)}$$\frac{n}{n+1}$$\frac{1}{k} - \frac{1}{k+1}$
$\sum \frac{1}{k(k+2)}$$\frac{3}{4} - \frac{2n+3}{2(n+1)(n+2)}$$\frac{1}{2}\left(\frac{1}{k} - \frac{1}{k+2}\right)$
$\sum \frac{1}{k(k+1)(k+2)}$$\frac{1}{4} - \frac{1}{2(n+1)(n+2)}$Difference of products
$\sum \frac{1}{(2k-1)(2k+1)}$$\frac{n}{2n+1}$$\frac{1}{2}\left(\frac{1}{2k-1} - \frac{1}{2k+1}\right)$
$\sum \frac{1}{\sqrt{k} + \sqrt{k+1}}$$\sqrt{n+1} - 1$Rationalize: $\sqrt{k+1} - \sqrt{k}$
$\sum k\cdot k!$$(n+1)! - 1$$k\cdot k! = (k+1)! - k!$
$\sum \frac{k}{(k+1)!}$$1 - \frac{1}{(n+1)!}$$\frac{1}{k!} - \frac{1}{(k+1)!}$
Don't Forget the First Term
After cancellation, write $f(n+1) - f(1)$ — a common slip is to keep $f(n+1)$ but drop the $-f(1)$. For gap-2 forms, the first two and last two terms survive.