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

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Formula List

Arithmatic Progression
General Term	\(T_n=a+(n-1)d\)
Sum of \(n\) terms	\(S_n=\frac{n}{2}[2a+(n-1)d]=\frac{n}{2}[a+l]\)
\(a, b, c \rightarrow\) 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	\(T_n = ar^{n-1}\)
Sum of \(n\) terms	\(S_n=\frac{a(r^n-1)}{r-1}\)
Sum of \(\infty\) terms	\(S_\infty=\frac{a}{(1-r)}(\begin{vmatrix}r\end{vmatrix}<1)\)
\(a, b, c \rightarrow\) GP	\(b^2=ac\)
GM B/w two terms \(a\) and \(b\)	\(b=\sqrt{ac}; a \,\&\, b>0\\b=-\sqrt{ac}; a \,\&\, b<0\)
GM of \(n\) terms	\((a_1.a_2.a_3....a_n)^{\frac{1}{n}}\)
Inserting \(n\) GM between \(a\) and \(b\)	\(G_n=ar^n\)

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 \(a, b \in \mathbb{R^+}\)	\(AM>GM>HM\)
If \(a, b \in \mathbb{R^-}\)	\(AM<GM<HM\)
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)
\(\therefore2S_n=\{2a+(n-1)d\}+\{2a+(n-1)d\}+...+\{2a+(n-1)d\}\)
\(\therefore2S_n=n[2a+(n-1)d]\)
\(\therefore S_n=\frac{n}{2}[2a+(n-1)d]=\frac{n}{2}[a+l]\)

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
\(A_1=a+d\\ A_2=a+2d\\ A_n=a+nd\)

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:
\(a, ar, ar^2, a^3,../\)

General term =

\(T_n=ar^{n-1}\)

Sum of n Terms

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

Proof:
\(\,\,\,\,S_n=a+ar+ar^2+...+ar^{n-1}\) –-(1)
\(rS_n=\quad\quad ar+ar^2+...+ar^{n-1}+ar^n\) –-(2)
Taking (1)-(2)
\(S_n-rS_n= a-ar^n\)
\(S_n(1-r)=a(1-r^n)\)
\(S_n=\frac{a(1-r^n)}{1-r}\)

Sum of infinite Terms

\(S_\infty=\frac{a}{(1-r)}(\begin{vmatrix}r\end{vmatrix}<1)\)

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

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

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

\(\because r\) is a fraction \(\,\therefore r^\infty\) is negligible
\(\therefore S_\infty=\frac{a}{(1-r)}\)

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 \(a_1b_2, a_2b_2, a_3b_3,..\) 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: \(\frac{a}{r^2}, \frac{a}{r}. ar,ar^2\)

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

Geometric Mean

GM between two terms

if \(a, b, c\) are in GP, b is the GM between \(a\) and \(c\)

\(b=\sqrt{ac}; a \,\&\, b>0\\b=-\sqrt{ac}; a \,\&\, b<0\)

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

\(G_1=ar\\ G_2=ar^2\\ ...\\ G_n=ar^n\)

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 \(a, b \in \mathbb{R^+}\)

\(AM>GM>HM\)

Taking, \(A-G\\ =\frac{a+b}{2}-\sqrt{ab}\\ =\frac{a+b-2\sqrt{ab}}{2}\\ =\frac{(\sqrt{a}-=\sqrt{b})^2}{2}>0\\ \therefore A-G>0\\ \therefore A>G\)

Again, \(G-H\\ =\sqrt{ab}-\frac{2ab}{a+b}\\ =\frac{(a+b)\sqrt{ab}-2ab}{a+b}\\ =(\sqrt{ab})(1-\frac{2\sqrt{ab}}{a+b})\\ =\frac{(\sqrt{ab})(\sqrt{a}-\sqrt{b})^2}{a+b}>0\\ \therefore G-H>0\\ \therefore G>H\)

For Negative Distince Numbers

If \(a, b \in \mathbb{R^-}\)

\(AM<GM<HM\)

For Equal Numbers

If \(a, b \in \mathbb{R}\) and \(a=b\)

\(AM=GM=HM\)

Arithmetic Geometric Progression (AGP)

\(a, (a+d), (a+2d), (a+3d),...\rightarrow\) AP
\(b, br, br^2, br^3,...\rightarrow\) GP

\(ab, br(a+d),br^2(a+2d),...\rightarrow\) AGP

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

Sum of n Terms

\(S=1(x)+3(x^2)+5(x^3)+7(x^4)+...\\ xS=1(x^2)+3(x^3)+5(x^4)+7(x^5)+...\\ S-xS=1(x)+2(x^2)+2(x^3)+...\\ (1-x)S=x+2x^2(1+x+x^2+...)\)

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

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