Sequence & Series

• Sequence & Series
  • Formula List
  • Arithmetic Progression (AP)
    • General Term
    • Sum of n Terms
    • nth term from Sum
    • Properties of AP
    • Considering number in AP
    • Common AP
    • Arithmetic Mean
  • Geometric Progression (GP)
    • General Term
    • Sum of n Terms
    • Sum of infinite Terms
    • Properties of GP
    • Considering number in AP
    • Geometric Mean
  • Harmonic Progression (HP)
    • Harmonic Mean
  • Relation between AM, GM, HM
    • Inequality between A, G, H
    • For positive Distinct Numbers
    • For Negative Distince Numbers
    • For Equal Numbers
  • Arithmetic Geometric Progression (AGP)
    • Sum of n Terms
  • Important Summations
  • Difference Method
  • Licensing and Links

tags: Mathematics

All Mathematics Formula by Abhyas here

Please see README if this is the first time you are here.

Formula List

Arithmatic Progression
General Term	$T_n=a+(n-1)d$
Sum of $n$ terms	$S_n=\frac{n}{2}[2a+(n-1)d]=\frac{n}{2}[a+l]$
$a,b,c\to$ AP	$2b=a+c$
AM B/w two terms $a$ and $b$	$b=\frac{a+c}{2}$
AM of $n$ terms	$\frac{a_1+a_2{+}_3+...+a_n}{n}$
Inserting $n$ AM between $a$ and $b$	$A_n=a+nd$

Geometric Progression
General Term	$T_n=a{r}^{n-1}$
Sum of $n$ terms	$S_n=\frac{a(r^n-1)}{r-1}$
Sum of $\mathrm{\infty }$ terms	$S_{\mathrm{\infty }}=\frac{a}{(1-r)}(\left|\begin{array}{c}r\end{array}\right|<1)$
$a,b,c\to$ GP	$b^2=ac$
GM B/w two terms $a$ and $b$	$$\begin{gathered} b=\sqrt{ac};a\mathrm{\&}b>0 \\ b=-\sqrt{ac};a\mathrm{\&}b<0 \end{gathered}$$
GM of $n$ terms	$(a_1.a_2.a_3....a_n{)}^{\frac{1}{n}}$
Inserting $n$ GM between $a$ and $b$	$G_n=a{r}^n$

Harmonic Progression
HM B/w two terms $a$ and $b$	$b=\frac{2ac}{a+c}$
HM of $n$ terms	$\frac{n}{\frac{1}{a_1}+\frac{1}{a_2}+\frac{1}{a_3}+...+\frac{1}{a_n}}$
Inserting $n$ HM between $a$ and $b$	$(\frac{1}{H_n}{)}^{-1}=(\frac{1}{a}+nd{)}^{-1}$

Relation B/w A, G & H
Relation	$G^2=AH$
If $a,b\in {\mathbb{R}}^{\mathbb{+}}$	$AM>GM>HM$
If $a,b\in {\mathbb{R}}^{\mathbb{-}}$	$AM<GM<HM$
If $a,b\in \mathbb{R}$ and $a=b$	$AM=GM=HM$

Arithmetic Geometric Progression
Sum of $n$ terms	Put $S=$ sum; take $rS$; Subtract $S-rS$; Solve for $S$

Important Summations
$\sum 1=n$
$\sum n=\frac{n(n+1)}{2}$
$\sum n^2=\frac{n(n+1)(2n+1)}{6}$
$\sum n^3={\left(\frac{n(n+1)}{2}\right)}^2$

Difference Method
$n$th term from sum	$T_n=S_n-S_{n-1}$

Arithmetic Progression (AP)

AP is a sequence whose terms increase or decrease by a fixed number. This fixed number is called the common difference.

eg.

$1,4,7,10,13,...$

General Term

If

$a$ is the first term and

$d$ the common difference then series is:

$a,(a+d),(a+2d),(a+3d),...$

General term =

$T_n=a+(n-1)d$

Sum of n Terms

$S_n=\frac{n}{2}[2a+(n-1)d]=\frac{n}{2}[a+l]$
where $l$ is the last term

Proof:

$S_n=T_1+T_2+T_3+...+T_n$

$S_n=a+(a+d)+(a+2d)+...+\{a+(n-1)d\}$ –-(1)

$S_n=\{a+(n-1)d\}+\{a+(n-2)d\}+...+a$ –-(2) By reversing (1)
Adding (1) and (2)

$\therefore 2S_n=\{2a+(n-1)d\}+\{2a+(n-1)d\}+...+\{2a+(n-1)d\}$

$\therefore 2S_n=n[2a+(n-1)d]$

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

n^th term from Sum

$T_n=S_n-S_{n-1}$

Proof:

$S_5=T_1+T_2+T_3+T_4+T_5$

$S_4=T_1+T_2+T_3+T_4$

$\therefore S_5-S_4=T_5$

Properties of AP

1. The common difference can be zero , positive or negative.
2. If
   $a,b,c$ in AP then

   $2b=a+c$
   Proof:
   $a,b,c$ are in AP
   $\therefore (b-a)=(c-b)\
   b+b=a+c\
   2b=a+c$

3. The sum of the terms of an AP equidistant from the beginning & end is constant and equal to the sum of first & last terms
4. If each term of an AP is increased, decreased, multiplied or divided by the same non-zero number, then the resulting sequence is also an AP.

Considering number in AP

3 Numbers: $(a-d),a,(a+d)$

4 Numbers: $(a-2d),(a-d),(a+d),(a+2d)$

5 Numbers: $(a-2d),(a-d),a,(a+d),(a+2d)$

Common AP

1. Sum of First n natural numbers

   $\frac{n(n-1)}{2}$

Arithmetic Mean

AM between two terms

If three terms are in AP then the middle term is called AM between the other two, so if

$a,b,c$ are in AP.

$b$ is AM of

$a$ and

$c$

$b=\frac{a+c}{2}$

AM for n terms

AM for any

$n$ positive number

$a_1,a_2,a_3,...,a_n$ is

$A=\frac{a_1+a_2+a_3,...,a_n}{n}$

Inserting n AM between two terms

Inserting n AM between

$a$ and

$b$. The new AP is

$a,A_1,A_2,A_3,...,A_n,b$

First Term

$=a=T_1$
Second Term

$=A_1=T_2$
…
Last Term

$=b=T_{n+2}$

Use

$T_{n+2}=b=a+(n+1)d$ to find

$d$

All inserted means can be found using

$a$ and

$d$ as

$$\begin{gathered} A_1=a+d \\ {A}_2=a+2d \\ {A}_n=a+nd \end{gathered}$$

Geometric Progression (GP)

GP. is a sequence of numbers whose first term is non zero & each of the succeeding terms is equal to the proceeding terms multiplied by a constant.

eg.

$1,2,4,8,16,...$

General Term

If

$a$ if the first term and

$r$ the common ratios then series is:

$a,ar,a{r}^2,a^3,../$

General term =

$T_n=a{r}^{n-1}$

Sum of n Terms

$S_n=\frac{a(1-r^n)}{1-r}$
where $r\ne 1$

Proof:

$S_n=a+ar+a{r}^2+...+a{r}^{n-1}$ –-(1)

$r{S}_n=ar+a{r}^2+...+a{r}^{n-1}+a{r}^n$ –-(2)
Taking (1)-(2)

$S_n-r{S}_n=a-a{r}^n$

$S_n(1-r)=a(1-r^n)$

$S_n=\frac{a(1-r^n)}{1-r}$

Sum of infinite Terms

$S_{\mathrm{\infty }}=\frac{a}{(1-r)}(\left|\begin{array}{c}r\end{array}\right|<1)$

Proof:

$1,\frac{1}{2},\frac{1}{4},\frac{1}{8},\frac{1}{16},\frac{1}{32},...$ here

$r=\frac{1}{2}$

If

$|r|<1$ then it is an infinite series

$\because S_n=\frac{a(1-r^n)}{(1-r)}$

$\therefore S_{\mathrm{\infty }}=\frac{a(1-r^{\mathrm{\infty }})}{(1-r)}$

$\because r$ is a fraction

$\therefore r^{\mathrm{\infty }}$ is negligible

$\therefore S_{\mathrm{\infty }}=\frac{a}{(1-r)}$

Properties of GP

1. The common ratio can be positive or negative but not zero.
2. If a, b, c in GP then

   $\frac{b}{a}=\frac{c}{b}$

3. The product if the terms of a GP equidistant from the beginning & end is constant and equal to the product of first & last term.
4. If each term of a GP is multiplied or divided or raised to the power by the same non zero number, then the resulting sequence is also a GP.
5. If
   $a_1,a_2,a_3,...$ and
   $b_1,b_2,b_3,...$ are two GP with common ratio
   $r_1$ and
   $r_2$ respectively then the sequence
   $a_1{b}_2,a_2{b}_2,a_3{b}_3,..$ are also a GP with common ratio
   $r_1,r_2$
6. if
   $a_1,a_2,a_3,...$ are in GP, where
   $a_i>0$, then
   $\log a_1,\log a_2,\log a_3,...$ are in AP and its converse is also true

Considering number in AP

3 Numbers: $\frac{a}{r},a,ar$

4 Numbers: $\frac{a}{r^2},\frac{a}{r}.ar,a{r}^2$

5 Numbers: $\frac{a}{r^2},\frac{a}{r},a,ar,a{r}^2$

Geometric Mean

GM between two terms

if

$a,b,c$ are in GP, b is the GM between

$a$ and

$c$

$$\begin{gathered} b=\sqrt{ac};a\mathrm{\&}b>0 \\ b=-\sqrt{ac};a\mathrm{\&}b<0 \end{gathered}$$

GM for n terms

GM for any

$n$ positive number,

$a_1,a_2,a_3,...,a_n$ is

$(a_1.a_2.a_3.....a_n{)}^{\frac{1}{n}}$

Inserting n GM between two terms

Inserting n GM beween

$a$ and

$b$. The new GP is

$a,G_1,G_2,G_3,...,G_n,b$

First Term

$=a=T_1$
Second Term

$=G_1=T_2$
…
Last Term

$=b=T_{n+2}$

Use

$T_{n+2}=\frac{a(1-r^{n+1})}{1-r}$ to find

$r$

All inserted means can be found using

$a$ and

$r$ as

$$\begin{gathered} G_1=ar \\ {G}_2=a{r}^2 \\ ... \\ {G}_n=a{r}^n \end{gathered}$$

Harmonic Progression (HP)

A sequence is said to be in HP, if the reciprocals of its terms are in AP.

$1,\frac{1}{2},\frac{1}{3},\frac{1}{4},...$

Harmonic Mean

HM between two terms

if

$a,b,c$ are in HP, b is the HM between

$a$ and

$c$

$b=\frac{2ac}{a+c}$

HM for n terms

$\frac{n}{\frac{1}{a_1}+\frac{1}{a_2}+\frac{1}{a_3}+...+\frac{1}{a_n}}$

Method:

1. Take reciprocal for each term to form an AP
2. Add n terms and divide by n
3. Take reciprocal of the fraction

Inserting n GM between two terms

1. Take reciprocal for each term to form an AP
2. Find
   $d$
3. Find
   $n$ AM between
   $\frac{1}{a}$ and
   $\frac{1}{b}$
4. Take reciprocal for each term to from the HP

Relation between AM, GM, HM

$\because A=\frac{a+b}{2}$,

$G=\sqrt{ab}$ and

$H=\frac{2ab}{a+b}$

$\therefore G^2=AH$

Inequality between A, G, H

For positive Distinct Numbers

If

$a,b\in {\mathbb{R}}^{\mathbb{+}}$

$AM>GM>HM$

Taking,

$$\begin{gathered} A-G \\ =\frac{a+b}{2}-\sqrt{ab} \\ =\frac{a+b-2\sqrt{ab}}{2} \\ =\frac{(\sqrt{a}-=\sqrt{b}{)}^2}{2}>0 \\ \therefore A-G>0 \\ \therefore A>G \end{gathered}$$

Again,

$$\begin{gathered} G-H \\ =\sqrt{ab}-\frac{2ab}{a+b} \\ =\frac{(a+b)\sqrt{ab}-2ab}{a+b} \\ =(\sqrt{ab})(1-\frac{2\sqrt{ab}}{a+b}) \\ =\frac{(\sqrt{ab})(\sqrt{a}-\sqrt{b}{)}^2}{a+b}>0 \\ \therefore G-H>0 \\ \therefore G>H \end{gathered}$$

For Negative Distince Numbers

If

$a,b\in {\mathbb{R}}^{\mathbb{-}}$

$AM<GM<HM$

For Equal Numbers

If

$a,b\in \mathbb{R}$ and

$a=b$

$AM=GM=HM$

Arithmetic Geometric Progression (AGP)

$a,(a+d),(a+2d),(a+3d),...\to$ AP

$b,br,b{r}^2,b{r}^3,...\to$ GP

$ab,br(a+d),b{r}^2(a+2d),...\to$ AGP

$1(x)+3(x^2)+5(x^3)+7(x^4)+...\to$ AGP

Sum of n Terms

$$\begin{gathered} S=1(x)+3(x^2)+5(x^3)+7(x^4)+... \\ xS=1(x^2)+3(x^3)+5(x^4)+7(x^5)+... \\ S-xS=1(x)+2(x^2)+2(x^3)+... \\ (1-x)S=x+2x^2(1+x+x^2+...) \end{gathered}$$

Important Summations

$\sum _{r=1}^{n}1=n$

$\sum _{r=1}^{n}1=1+1+1+1+1+...$ upto n terms

$=n$

$\sum _{r=1}^{n}r=\frac{n(n+1)}{2}$

$\sum _{r=1}^{n}r=1+2+3+4+...$ upto n terms

$=\frac{n(n+1)}{2}$

$\sum _{r=1}^n{r}^2=\frac{n(n+1)(2n+1)}{6}$

$\sum _{r=1}^n{r}^2=1^2+2^2+3^2+4^2+...$ upto

$n$ terms=

$\frac{n(n+1)(2n+1)}{6}$

$\sum _{r=1}^n{r}^3={\left(\frac{n(n+1)}{2}\right)}^2$

$\sum _{r=1}^n{r}^3=1^3+2^3+3^3+4^3+...$ upto

$n$ terms=

${\left(\frac{n(n+1)}{2}\right)}^2$

Difference Method

$T_n=S_n-S_{n-1}$

Licensing and Links

All Mathematics Formula by Abhays here

This work is licensed under a Creative Commons Attribution-NonCommercial 4.0 International License.