Notes - Mathematics Olympiads - Sequence Series
Category : 11th Class
Sequence and series
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.)
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.
(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.
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 \[a-3b,a-b,a+3b\]
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
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, ar2 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.
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.
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 \[A=\frac{a+b}{2}\]
\[G=\sqrt{a.b}\]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, 1st A.M. \[{{A}_{1}}=a+d=a+\frac{b-a}{n+2}\]
2nd 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 \[\frac{n}{2}(a+b)\]
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. \[{{G}_{1}},\,{{G}_{2}},{{G}_{3}}........{{G}_{n}}=({{\sqrt{ab)}}^{n}}.\]
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
\[\therefore \,\,\,{{G}^{2}}=A\times H\]
\[\therefore \,\,\,\frac{A}{G}=\frac{G}{H}\]
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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