Notes - Mathematics Olympiads - Sequence Series

**Category : **11th Class

** Sequence and series**

**Sequence:**A sequence is a mapping of or function whose domain is the set of natural number and its range is the set of real number or complex number. But we will study the real number sequence only.

**e.g.** **(a)** \[{{x}_{1}},\,{{x}_{2}},{{x}_{3}}.....{{x}_{n}}.\]

**(b)** 2, 4, 6, 8, 10.....

**(c)** 1, 4, 7, 10, 13....... etc.

These above example is a sequence of real no.

**Note:** Sequence is also said to be progression.

Generally there are three type of sequence

**(a)** Arithmatic progression or sequence (A.P.)

**(b)** Geometric progression (G.P.)

**(c)** Harmonic prograssion (H.P.)

**Arithmetic sequence:**A sequence (Sn) is said to be an arithmetic sequence if difference between any term and its proceeding term give the constant quantity.

**e.g.** Sn: 3,8,13,18, 23.....

we choose any term

\[13-5=5\]

\[23-18=5\] etc.

The constant quantity is said to common difference (c.d.) in A.P.

**Remember Some Points about A.P.**

** **

**(a)** \[{{n}^{th}}\] term whose 1st term and common difference is given in A.P. is written as \[{{t}_{n}}=a+(n-1).d\] where a = 1st term and d = comman difference n = no. of terms in the A.P.

** **

**(b)** Sum of the \[{{n}^{th}}\] term in A.P. be

\[Sn=\frac{n}{2}\{2a+(n-1).d\}=\frac{n}{2}\{a+a+(n-1).d\}\]

\[=\frac{n}{2}\{a+{{t}_{n}}\}=\frac{n}{2}\{1st\,\,term+\,last\,\,term\}\]

**(c)** Single arithmetic mean between two given quantity is \[A.M=\frac{a+b}{2}\]

**(d)** For n arithmetic mean between two terms a and b.

\[\therefore \] So, the total no. of terms = n+ 2

Last term \[=b=a+(n+2-1).d=a+(n+1).d\]

Where d = comman difference in A.P.

** **

**(e)** For choosing five term in A.P. We take them as \[a-2b,a-b,a+b\]and \[(a+2b)\] etc.

**Some very useful results:**If the last term, \[{{t}_{n}}\] is in the linear form. i.e. \[{{t}_{n}}=an+b.\] Here, n is considered as variable then the series so formed is said to be in A.P. Similarly if last tern, \[{{t}_{n}}=a{{n}^{2}}+bn+c\] (i.e. in quadratic form), then the series so formed is said to be in A.P.

If \[{{a}_{1}},\,{{a}_{2}},\,{{a}_{3}},\,{{a}_{4}}....{{a}_{n}}\]is said to be in A.P.

Then

**(a)** \[{{a}_{1}}\pm \text{k},\,{{a}_{2}}\pm \text{k},....\]will be in A.P.. Where k be any const quantity

**(b)** Even \[\text{k}.{{a}_{2}},\text{k}.{{a}_{2}},\text{k}.{{a}_{3}}.....\]will be said to be in A.P.

**(c)** \[\frac{{{a}_{1}}}{\text{k}},\frac{{{a}_{2}}}{\text{k}},\frac{{{a}_{3}}}{\text{k}}...\]is said to be in A.P.

**(d)** If \[{{a}_{1}},\,{{a}_{2}},\,{{a}_{3}},\,{{a}_{4}},\,{{a}_{5}}....\]be in A.P. and also \[{{b}_{1}},\,{{b}_{2}},\,{{b}_{3}},\,{{b}_{4}},\,{{b}_{5}}....\]be in A.P. then the resulting sequence whose elements are corresponding **element addition** or multiplication of given sequences is said to be in A.P.

\[\Rightarrow (b-a)=(n+1).d\]

\[d=\frac{b-a}{(n+1)}\]

Hence, 1st A.M. between a and b

\[=a+d\,\,\,\,\,=a+\left( \frac{b-a}{n+1} \right)\]

2nd A.M. between a and \[=a+2d\,\,\,\,\,=a+2.\left( \frac{b-a}{n+1} \right)\]

2nd A.M. between a and b

\[{{n}^{th}}\] A.M. between a and \[b=a+n\frac{(b-a)}{n+1}\]

**(e)** Sum of the n-natural number is written as

\[\text{Sn=}\frac{\text{n}(n+1)}{2},\] where \[n\in N\]

**(f) ** Sum of the square of first n-natural number is written as

\[\text{S}{{\text{n}}^{\text{2}}}\text{=}\frac{\text{n}(n+1)(2n+1)}{6}\]

** **

**(g)** Sum of the cube of the 1st n-natural number be

\[\text{S}{{\text{n}}^{3}}\text{=}{{\left( \frac{n(n+1)}{2} \right)}^{2}}\]

**(h) **(a) For choosing three terms in A.P.. We choose thenas, \[a-b,\,\,b,\,\,a+b\]

**(b**) For choosing four terms in A.P., we take \[a-3b,a-b,a+3b\]

**Geometric Sequence:**A sequence {Sn} is said to be geometric sequence if the ratio of any term to its proceeding or succeeding term gives the constant term. The const term/number is said to becomman ratio.

** **

**e.g.** **(1)** 2, 4, 8, 16, 32, 64.....

**(2)** \[\text{Sn:}\,\,\frac{1}{9},\frac{1}{27},\frac{1}{81},\frac{1}{243}...\]

The comman ratio \[=\frac{{{\text{a}}_{\text{k}}}}{{{\text{a}}_{\text{k-1}}}}\] or \[\frac{{{\text{a}}_{\text{k}}}}{{{\text{a}}_{\text{k+1}}}}\] and same

**General term of a G.P.:**If the 1st term comman ratio of a G.P. is given be a and r respective The Last term of G.P, \[{{t}_{\text{n}}}=\text{a}\text{.}{{\text{r}}^{\text{G-1}}}\] If \[\text{r}\,\,\text{1}\]

Sum to n terms of G.P.

i.e. \[\text{Sn=}\frac{a(1-{{\text{r}}^{4}})}{1-\text{r}}\] if \[\text{r}\,\,<\text{1}\]

\[\text{Sn=}\frac{a({{r}^{4}}-1)}{\text{r}-1}\] if \[\text{r}\,\,>\text{1}\]

Sum of the infinite terms \[({{\text{S}}_{\infty }})\]

\[{{\text{S}}_{\infty }}=\frac{a}{1-\text{r}}\] if \[\text{r}\,\,<\text{1}\]

When \[\text{r}\,\,>\text{1,}\], then sum of this series is not possible because this series is divergent. If terms of the G.P, be given and its product is known, then we can choose the terms by following way.

**(a)** for selecting the three terms in G.P. then these can be \[\frac{a}{\text{r}}\], a, ar

**(b)** for selecting the four terms in G.P. then these can be \[-\frac{a}{{{\text{r}}^{3}}}\], a, ar, ar^{2 }etc.

If each terms of a G.P. is multiplied or divided by non-zero quantity, the resulting progression will be in G.P.

If \[{{a}_{1}},\,{{a}_{2}},\,{{a}_{3}},\,.....{{a}_{n}}\] and \[{{b}_{1}},\,{{b}_{2}},\,{{b}_{3}},\,.....{{b}_{n}}\] be two sequence in G.P. Then the corresponding product or division of each terms form a new sequence and this new sequence must be in G.P.

** **

**i.e.** \[{{a}_{1}}{{b}_{1}},\,{{a}_{2}}{{b}_{2}},\,{{a}_{3}}{{b}_{3}}.....\] will be in G.P.

and \[\frac{{{a}_{1}}}{{{b}_{1}}},\,\frac{{{a}_{2}}}{{{b}_{2}}},\,\frac{{{a}_{3}}}{{{b}_{3}}}....\] will be in G.P.

If \[{{a}_{1}},\,{{a}_{2}},\,{{a}_{3}},{{a}_{4}}....\] are in G.P. provided

\[\forall {{a}_{\text{i}}}\ge 0,\text{i}=1,\,2,3,....\text{n}.\] Then \[\log {{a}_{1}},\,\log {{a}_{2}},\,lig\,{{a}_{3}}\] will be in A.P. and vice versa..

** **

**i.e. \[{{a}_{1}}+{{b}_{1}},\,\,{{b}_{1}}+{{b}_{2}},\,\,{{a}_{3}}+{{b}_{3}}.....\] **be in A.P.

\[{{a}_{1}}{{a}_{1}},\,\,{{a}_{2}}{{b}_{2}},\,\,{{a}_{3}}{{b}_{3}}.....\] be in A.P.

** **

**Harmonic Sequence (Progression) i.e. H.P.**

A sequence is said to be in Harmonic progression; If and only if the reciprocal of its terms form the arithmetic sequence and vice versa.

**e.g.** \[(1)\frac{1}{2},\frac{1}{4},\frac{1}{16}.....\]are in H.P.

then 2, 4, 6, 8....... will be in A.P.

(2) If 5, 8, 11, 14, 17,......are in A.P. then \[\frac{1}{5},\frac{1}{8},\frac{1}{11},\frac{1}{14}.....\]will be in H. P.

**To find the nth term in H.P.**

For this 1st of all, convert the H.P. into A.P. then find the nth term of this changeable A.P. The inverse of this nth terms of A.P. will be the nth term of the given H.P.

**Some useful results:**

If a, b are 1st two term of in H.P.

\[\therefore \] So then \[\frac{1}{a},\frac{1}{b}\]will be the 1st two ten of

\[{{t}_{n}}={{a}_{n}}+(n-1).\left( \frac{1}{b}-\frac{1}{a} \right)\,\,\,\left( \because \,d=\frac{1}{b}-\frac{1}{a} \right)\]

So, nth term of H.P. be

\[{{t}_{n}}=\frac{1}{{{a}_{n}}+\left( \frac{1}{b}-\frac{1}{a} \right)(n-1)}.\]

Harmonic mean H of any two numbers a and b be \[\text{H=}\frac{2}{\frac{1}{a}+\frac{1}{b}}=\frac{2ab}{a+b}\]

Some useful Resultants about A.M., G.M. and H.M.

Arithmetic nean A of any two numbers a and b. Then \[A=\frac{a+b}{2}\]

**Geometric mean (G.M.)**

\[G=\sqrt{a.b}\]is the geometric mean of two positive numbers a and b.

If \[a,{{A}_{1}},\,{{A}_{2}},\,{{A}_{3}}....{{b}_{3}}\] are in A.P. then \[{{A}_{1}},\,{{A}_{2}},\,{{A}_{3}}....{{A}_{n}}\] are said to be n A.M. between two terms a and b and d, the comman difference, is given.

Thenn = no. of terms in \[\text{A}\text{.P}\text{.=n+2}\]

\[\therefore \,\,\,{{t}_{n}}=\]last term \[=b=a+(n+2-1).d\]

\[b-a=+(n+1).d\]

\[\therefore \,\,\,d=\left( \frac{b-a}{n+1} \right).\]

So, 1^{st} A.M. \[{{A}_{1}}=a+d=a+\frac{b-a}{n+2}\]

2^{nd} A.M. \[{{A}_{2}}=a+2d=a+2\left( \frac{b-a}{n+2} \right)\]

2nd A.M. \[{{A}_{2}}=a+2d=a+2\left( \frac{b-a}{n+2} \right)\]

\[Ai=a+id=a+id....\]

** **

**Note:** The sum of all arithmetic means.

i.e. \[{{A}_{1}}+{{A}_{2}}+{{A}_{3}}+.....+{{A}_{n}}\]is written as \[\frac{n}{2}(a+b)\]

**Geometric Means:**If a and b be two terms in G.P. and \[{{G}_{1}},\,{{G}_{2}},\,{{G}_{3}}.....{{G}_{n}}\] be n G.P. between a and b. and its comman ratio is r. then \[{{t}_{n}}=b=\text{a}\text{.}{{\text{r}}^{\text{n-1}}}=a.{{r}^{n+2-1}}=a.{{r}^{(n+1)}}\]

Here no. of term \[n=n+2\]

\[\Rightarrow \,\,r={{\left( \frac{b}{a} \right)}^{\frac{1}{n+1}}}\]

\[\therefore \,\,\,{{G}_{1}}=ar=a.\left( \frac{b}{a} \right)\frac{c}{n+1}={{a}^{n+1-1}}.{{b}^{\frac{1}{n+1}}}\]

\[i=1,\,2,3,....n={{a}^{n}}.{{b}^{\frac{1}{n+1}}}\]

\[\therefore \] Similarly \[{{G}_{i}}=a.{{\left( \frac{b}{a} \right)}^{\frac{i}{n+1}}}\]

Hence, the product of n G.M

i.e. \[{{G}_{1}},\,{{G}_{2}},{{G}_{3}}........{{G}_{n}}=({{\sqrt{ab)}}^{n}}.\]

**Harmonic Mean:**For H.M. basically there is no general method to finding of the terms. For operation in H.M. or H.P., AU time, strike the mind to convert into in A.P. or A.M. and find its reciprocal term.

**Relation between A.P., G.P. and H.P.**

If A, G and H be denoted the A.M., G.M. and H.M. respectively. Then \[A.M.\ge G.M.\ge H.M.\]

If no. of terms is two i.e. n = 2

\[\therefore \,\,\,{{G}^{2}}=A\times H\]

\[\therefore \,\,\,\frac{A}{G}=\frac{G}{H}\]

**Arithmetic-Geometric Series:**A series whose each term is formed by multiplying the corresponding terms of an A.P. and G.P. is called an Arithmetic-geometric series.

** **

**e.g.** \[1+3x+6{{x}^{2}}+9{{x}^{2}}+12{{x}^{2}}+.....\]

Summation of n terms of Arithmetic-Geometric series.

Let \[{{S}_{n}}=a+(a+d)x+(a+2d)x+.....\{a+(n-1)d\}{{x}^{n-1}}\] (1)

\[d\ne 0,x\ne 1\]

multiplying by x both sides we have

\[x.Sn=ax+(a+d){{x}^{2}}+(a+2d){{x}^{3}}+......\{a+(n-1)d\}{{x}^{3}}\] (2)

Subtracting (2) by (1), we have

\[xSn-Sn=a+d(x+{{x}^{2}}+{{x}^{3}}+.....{{x}^{n-1}})\]

\[+(a+(n-1)d){{x}^{n}}\]

\[Sn(x-1)=a+\frac{dxx.(1-{{x}^{n-1}})}{1-x}\]

\[-(a+(n-1).d){{x}^{n}}\]

\[\therefore \,\,Sn=\frac{a}{x-1}+\frac{d.x.(1-{{x}^{n-1}})}{{{(1-x)}^{2}}}-\frac{(a+(n-1).d){{x}^{n}}}{1-x}\]

Here the result

For summation of infinite series, if \[\left| \,\,r\,\, \right|<1\]

Then \[{{S}_{\infty }}=S=\frac{a}{1-x}+\frac{dx}{{{(1-x)}^{2}}}\]

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