To insert a given number of arithmetic mean between two give quantities:-
let a and b be the two given quantities , n the number of means. Including the extremes the number of terms will be (n+2) ; so that we have find a series of(n+2) terms in A.P. of which a is the first term and b is the last or (n+2) term.
Let d be the common difference
b= the (n+2)th term = a + (n+1)d
b-a = (n+1)d;
(b-a)/(n+1) = d;
so the required means are a+(b-a)/(n+1) , a+2(b-a)/(n+1).....................
Q: Find the five arithmetic means between 5 and 25.
Ans: Here given that first term a=5 last term b = 25 and number of arithmetic mean n=5
Let d be the common difference of the required A.P.
so , b= a + (n+1)d;
putting the values of a ,b and n in the equation,we get
25= 5 + (5+1)d;
20= 6d;
d= 10/3;
Now the arithmetic means are 5+10/3,5+2*10/3,5+30/3,5+40/3 and 5+ 50/3
i.e. 25/3,35/3,45/3,55/3 and 65/3 are arithmetic means between 5 and 25.
If you any problem then mail me at er.abhilashthakur@gmail.com
indian mathematics
July 31, 2009
July 25, 2009
syllbus of IMO
number systems, arithmetic of integers, geometry, quadratic equations and expressions, trigonometry, co-ordinate geometry, systems of linear equations, permutations and combinations, factorisation of polynomials, inequalities, elementary combinatorics, probability theory, number theory, infinite series, complex numbers and elementary graph theory. The syllabus does not include calculus and statistics. The typical areas for problems are: number theory, geometry, algebra and combinatorics. The syllabus is in a sense spread over class IX to class XII levels.
Mathematics Olympiad proplem 1995
Here are the IMO 1995 problems as posted by Abhilash thakur
1. Let A, B, C and D be four distinct points on a line, in that order. The
circles with diameters AC and BD intersect at the points X and Y. The
line XY meets BC at the point Z. Let P be a point on the line XY different
from Z. The line CP intersects the circle with diameter AC at the points
C and M, and the line BP intersects the circle with diameter BD at the
points B and N. Prove that the lines AM, DN and XY are concurrent.
2. Let a, b and c be positive real numbers such that a*b*c=1. Prove that
1 1 1 3
---------- + ---------- + ---------- >= -
(a^3)(b+c) (b^3)(c+a) (c^3)(a+b) 2
3. Determine all integers n>3 for which there exist n points A1, A2, ..., An
in the plane, and real numbers r1, r2, ..., rn satisfying the following two
conditions:
(i) no three of the points A1, A2, ..., An lie on a line;
(ii) for each triple i, j, k (1 <= i < j < k <= n) the triangle AiAjAk
has area equal to ri+rj+rk.
4. Find the maximum value of x[0] for which there exists a sequence of positive
real numbers x[0], x[1], ..., x[1995] satisfying the two conditions:
(i) x[0]=x[1995];
(ii) x[i-1] + 2/(x[i-1]) = 2x[i] + 1/x[i] for each i = 1, 2, ..., 1995.
[Note: x[i] means x subscript i]
5. Let ABCDEF be a convex hexagon with AB=BC=CD and DE=EF=FA, and
angle BCD=angle EFA=60 degrees. Let G and H be two points in the interior of
the hexagon such that angle AGB=andgle DHE=120 degrees. Prove that
AG+GB+GH+DH+HE >= CF
6. Let p be an odd prime number. Find the number of subsets of A of the set
{1, 2, ..., 2p} such that
(i) A has exactly p elements, and
(ii) the sum of all the elements in A is divisible by p.
1. Let A, B, C and D be four distinct points on a line, in that order. The
circles with diameters AC and BD intersect at the points X and Y. The
line XY meets BC at the point Z. Let P be a point on the line XY different
from Z. The line CP intersects the circle with diameter AC at the points
C and M, and the line BP intersects the circle with diameter BD at the
points B and N. Prove that the lines AM, DN and XY are concurrent.
2. Let a, b and c be positive real numbers such that a*b*c=1. Prove that
1 1 1 3
---------- + ---------- + ---------- >= -
(a^3)(b+c) (b^3)(c+a) (c^3)(a+b) 2
3. Determine all integers n>3 for which there exist n points A1, A2, ..., An
in the plane, and real numbers r1, r2, ..., rn satisfying the following two
conditions:
(i) no three of the points A1, A2, ..., An lie on a line;
(ii) for each triple i, j, k (1 <= i < j < k <= n) the triangle AiAjAk
has area equal to ri+rj+rk.
4. Find the maximum value of x[0] for which there exists a sequence of positive
real numbers x[0], x[1], ..., x[1995] satisfying the two conditions:
(i) x[0]=x[1995];
(ii) x[i-1] + 2/(x[i-1]) = 2x[i] + 1/x[i] for each i = 1, 2, ..., 1995.
[Note: x[i] means x subscript i]
5. Let ABCDEF be a convex hexagon with AB=BC=CD and DE=EF=FA, and
angle BCD=angle EFA=60 degrees. Let G and H be two points in the interior of
the hexagon such that angle AGB=andgle DHE=120 degrees. Prove that
AG+GB+GH+DH+HE >= CF
6. Let p be an odd prime number. Find the number of subsets of A of the set
{1, 2, ..., 2p} such that
(i) A has exactly p elements, and
(ii) the sum of all the elements in A is divisible by p.
July 18, 2009
Arithmetical Progression
Today i am solve some problem related to A.P.
1. find the 11th of a A.P. which first term is 3 and common difference is 5.
Ans: Give that first term a=3 and common difference is 5
so nth term = a+(n-1)d ;
putting the value of a, d and n in above equation , we get;
a=3; d=5 and n=11
11th term = 3+ (11-1)5
= 3+ 10*5
= 3+50
= 53 ans
Q 2. A arithmetical progression is as follows 10, 6,2,...............
Find the 10th term and the addition of first ten terms.
Ans: Here given series is 10,6,2,.........
so first term a=10 and common difference = 6- 10= -4 ( astounding to definition )
so the 10th term = 10+ (10-1)(-4)
= 10 + 9(-4)
= 10-36 [we know (+)*(-)= -]
= -26
Now the addition
first term a= 10, d= -4 and number of term n =10
First method,
summation of first n term s= n/2[2a+ (n-1)d]
Putting the value of a, d, and n, in above equation , we get
s= 10/2[2*10 +(10-1)*(-4)];
= 5[20 -36];
= 5(-16)= -80
so the summation of ten term is -80 .
Second method
sum of n term =n/2[first term + last term];
s= 10/2[ 10 +(-26)] [ here last term is -26]
= 5[10 - 26];
= 5( -16) = -80 .
NOTE: in this type of problem any three variable are give out of four variable such as a ,d ,n and s . and uyou have to find the other variable . put the all value in given equation of summation and find the unknow .
Thanks
1. find the 11th of a A.P. which first term is 3 and common difference is 5.
Ans: Give that first term a=3 and common difference is 5
so nth term = a+(n-1)d ;
putting the value of a, d and n in above equation , we get;
a=3; d=5 and n=11
11th term = 3+ (11-1)5
= 3+ 10*5
= 3+50
= 53 ans
Q 2. A arithmetical progression is as follows 10, 6,2,...............
Find the 10th term and the addition of first ten terms.
Ans: Here given series is 10,6,2,.........
so first term a=10 and common difference = 6- 10= -4 ( astounding to definition )
so the 10th term = 10+ (10-1)(-4)
= 10 + 9(-4)
= 10-36 [we know (+)*(-)= -]
= -26
Now the addition
first term a= 10, d= -4 and number of term n =10
First method,
summation of first n term s= n/2[2a+ (n-1)d]
Putting the value of a, d, and n, in above equation , we get
s= 10/2[2*10 +(10-1)*(-4)];
= 5[20 -36];
= 5(-16)= -80
so the summation of ten term is -80 .
Second method
sum of n term =n/2[first term + last term];
s= 10/2[ 10 +(-26)] [ here last term is -26]
= 5[10 - 26];
= 5( -16) = -80 .
NOTE: in this type of problem any three variable are give out of four variable such as a ,d ,n and s . and uyou have to find the other variable . put the all value in given equation of summation and find the unknow .
Thanks
HELP
HI all of my friend ,this is a new blog and need your help for development of this blog.
i am happy to help you people , if you give me chance to help you . This is a math blog . I requested all my friends,students,teachers, parent please post your problem and dot related to any branch of mathematics .E-mail your question and and dot on my E-mail ID er.abhilashthakur@gmail.com , and i give the step by step solution with logic.
I start a math tuition in kolkata . Here i teach math of class 5 to 12 and all the engineering entrance examination such as IIT,AIEEE,WBJEE etc . i also teach the B.TECH maths . contact with me for group and home tuition .
My mobile No is 09674185828
and E-mail is er.abhilashthakur@gmail.com
thanks
i am happy to help you people , if you give me chance to help you . This is a math blog . I requested all my friends,students,teachers, parent please post your problem and dot related to any branch of mathematics .E-mail your question and and dot on my E-mail ID er.abhilashthakur@gmail.com , and i give the step by step solution with logic.
I start a math tuition in kolkata . Here i teach math of class 5 to 12 and all the engineering entrance examination such as IIT,AIEEE,WBJEE etc . i also teach the B.TECH maths . contact with me for group and home tuition .
My mobile No is 09674185828
and E-mail is er.abhilashthakur@gmail.com
thanks
July 5, 2009
Arithmatical progression
Quantity are said to be in arithmetical progression when they increase or decrease by a common difference . OR
the difference between two consecutive terms of a series is content is called the A.P. For example 3,7,11,15,.....................
8,2,-4,-10,...................
a,a+r,a+2r,a+3r,................
the common difference is found by subtracting any term of the series that which follows it .
# If you examine the series , a,a+r,a+2r,a+3r,............
We notice that any term that coefficient of d is always less by one then that number of terms in the series.
so, nth term of the series = a+(n-1)d;
where a is the first term and d is common difference of the series
# Sum of the number of terms of A.P.
let a is the first term,d the common difference , n the number of terms ,l the last term and s the required sum
s=a+(a+d)+(a+2d)+(a+3d)+..............+(l-2d)+(l-d)+l
s=l+(l-d)+l-2d)+(l-3d)+...............+(a+2d)+a+d)+a
by adding, 2s=(a+l)+(a+l)+(a+l)+.........to n terms
2s=n(a+l)
s=n/2(a+l) (i)
and you know that l=a+(n-1)d
so s = n/2[2a+(n-1)d] (ii)
#When three quantities are in A.P the middle term is said to be arithmetic mean of other two.
# Arithmetic mean:- let a and b be the two quantities and A is the arithmetic mean then, a,A,b are in A.P. We must have
A-a=b-A
2A=a+b
A=(a+b)/2
cont. for next post
in next some of theory and problem related to A.P.
the difference between two consecutive terms of a series is content is called the A.P. For example 3,7,11,15,.....................
8,2,-4,-10,...................
a,a+r,a+2r,a+3r,................
the common difference is found by subtracting any term of the series that which follows it .
# If you examine the series , a,a+r,a+2r,a+3r,............
We notice that any term that coefficient of d is always less by one then that number of terms in the series.
so, nth term of the series = a+(n-1)d;
where a is the first term and d is common difference of the series
# Sum of the number of terms of A.P.
let a is the first term,d the common difference , n the number of terms ,l the last term and s the required sum
s=a+(a+d)+(a+2d)+(a+3d)+..............+(l-2d)+(l-d)+l
s=l+(l-d)+l-2d)+(l-3d)+...............+(a+2d)+a+d)+a
by adding, 2s=(a+l)+(a+l)+(a+l)+.........to n terms
2s=n(a+l)
s=n/2(a+l) (i)
and you know that l=a+(n-1)d
so s = n/2[2a+(n-1)d] (ii)
#When three quantities are in A.P the middle term is said to be arithmetic mean of other two.
# Arithmetic mean:- let a and b be the two quantities and A is the arithmetic mean then, a,A,b are in A.P. We must have
A-a=b-A
2A=a+b
A=(a+b)/2
cont. for next post
in next some of theory and problem related to A.P.
June 14, 2009
WELCOM
All of my friend i am happy to inform you that i am starting a now blog which is based upon the most interested subject mathematics. math is subject of game , this is my through because i like to spend my time with maths.
let me tall some thing about myself .... i am a B.TECH 3rd year student under WBUT university . my college name is AIEMD. I want to going in the field of research . my branch of study is electronics and communication . math is my most lover ,love or you can say it is my heart because it my best time pass friend.
i want to work in field of mathematics, people say maths is a very hard subject and do not able to handle it. so i am here to help those person.
so welcome all the math lover and those person who are not interested in maths.
thanks
let me tall some thing about myself .... i am a B.TECH 3rd year student under WBUT university . my college name is AIEMD. I want to going in the field of research . my branch of study is electronics and communication . math is my most lover ,love or you can say it is my heart because it my best time pass friend.
i want to work in field of mathematics, people say maths is a very hard subject and do not able to handle it. so i am here to help those person.
so welcome all the math lover and those person who are not interested in maths.
thanks
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