There are 153 people in total. The number of class one is 1.12 times that of class three. The number of class two is 3 less than that of class three. How many people are there in each of the three classes?

Class one 56. Class two 47. Class three 50

It is known that {an} and {BN} are two different arithmetic sequences. Is there two integers P and Q such that AP = BP and AQ = BQ? Explain the reason

If it exists: AP = BP, AQ = BQ,
Then AP AQ = BP BQ
Let the tolerances of an and BN be s and t respectively
ap-aq=(p-q)* s
bp-bq=(p-q)* t
So s = t
If s = t, then AP = a1 + (p-1) s, BP = B1 + (p-1) * t = B1 + (p-1) s
So A1 = B1, so the two sequences are the same, so they are contradictory
There are no such integers

Let {an} be the arithmetic sequence and {BN} be the proportional sequence, tangent A1 = B1 = 1. A2 + A4 = B3. B2b4 = A3
The sum S10 and T10 of the first 10 terms of {an} and {BN} are obtained respectively

2a3=b3,b3^2=a3,
So 4A3 ^ 2 = A3
A3 = 1 / 4 or 0, then B3 = 1 / 2,0
Because the common ratio of the equal ratio sequence is not zero
So A3 = 1 / 4, B3 = 1 / 2
an=-3n/8+11/8
BN = (radical 2 / 2) ^ (n-1)
The first n items and you can count them by yourself

A series problem,
In the sequence an, A1 = 3, an = 3A (n-1) - 2 (n-1, which is the numerator) denominator is a (n-1) (n ≥ 2, n is positive integer) 1. If the sequence BN satisfies BN = a (n-2) / 1-an (the numerator and denominator correspond respectively), it is proved that BN is equal to 2. Find the general term and maximum term of sequence an
I've offered all the rewards, and you can't help solving them. Speed, big brothers and big sisters
The second supplementary question
How could that be. Don't help me if I don't have any points. I'll give it a day

I won't help you if I don't have any points

A series problem
1. Definition: in the sequence {an}, if {an} ^ 2 - {an-1} ^ 2 = P, (n ≥ 2, n ∈ n , P is a constant), then {an} is called "equivariance sequence". The following is the judgment of "equivariance sequence": ① if {an} is "equivariance sequence", then {an2} is equal difference sequence; ② {(- 1) ^ n} is "equivariance sequence"; ③ if {an} is "equivariance sequence", then {AKN} (K ∈ n ), (4) if {an} is both {an} and {a} and {a}, then the sequence is a constant sequence
Note: {AKN} kn is the subscript {an-1} n-1 is the subscript
Who can prove the third one for me

The items in the sequence {an} are listed as: A1, A2,. AK, AK + 1, AK + 2,. A2k. A3k
The items in the sequence {AKN} are listed as: AK, a2k, a3k
Because AK + 1 ^ 2-ak ^ 2 = AK + 2 ^ 2-ak + 1 ^ 2 = AK + 3 ^ 2-ak + 2 ^ 2 =. = a2k ^ 2-a2k-1 ^ 2 = P
So (AK + 1 ^ 2-ak ^ 2) + (AK + 2 ^ 2-ak + 1 ^ 2) + (AK + 3 ^ 2-ak + 2 ^ 2) +... + (a2k ^ 2-a2k-1 ^ 2) = a2k ^ 2-ak ^ 2 = KP
Similarly, there are
(akn^2-akn-1^2)=(akn-1^2-akn-2^2)=.=(akn+3^2-akn+2^2)=akn+2^2-akn+1^2=akn+1^2-akn^2=p
Same as above, lianjiakede
akn+1^2-akn^2=kp
Therefore, the sequence {AKN} is an equivariance sequence

Find a sequence problem
Given the first term A13 of the sequence an, the general term an satisfies 2An = Sn * s (n-1) with the first n terms and Sn, (1) prove that 1 / Sn is an arithmetic sequence and find the tolerance, (2) find the general term formula of the sequence an, (3) whether there is a natural number k in the sequence an, so that the inequality AK is greater than a (K + 1) for any natural number greater than k or equal to k? If there is, find the minimum K, if not, explain the reason

1.
When n ≥ 2,
2an=2[Sn-S(n-1)]=2Sn-2S(n-1)
2Sn-2S(n-1)=SnS(n-1)
Divide both sides of the equation by 2SNS (n-1)
1/S(n-1)- 1/Sn=1/2
1/Sn -1/S(n-1)=-1/2
1 / S1 = 1 / A1 = 1 / 3, the sequence {1 / Sn} is an arithmetic sequence with 1 / 3 as the first term and - 1 / 2 as the tolerance, and the tolerance = - 1 / 2
two
1/Sn=(1/3)+(-1/2)(n-1)=(5-3n)/6
Sn=6/(5-3n)
When n ≥ 2, an = SN-S (n-1) = 6 / (5-3n) - 6 / [5-3 (n-1)] = 6 / (5-3n) - 6 / (8-3n)
When n = 1, A1 = 6 / (5-3) - 6 / (8-3) = 9 / 5 ≠ 3
The general formula of sequence {an} is
an=3 n=1
6/(5-3n)-6/(8-3n) n≥2
three
Suppose that there is a k-path satisfying the meaning of the problem
When k = 1,
a2=6/(5-3×2)-6/(8-3×2)=-9
When k ≥ 2,
ak>a(k+1)
6/(5-3k) -6/(8-3k)>6/[5-3(k+1)]-6/[8-3(k+1)]
1/(3k-8)+1/(3k-2)>2/(3k-5)
1/[(3k-2)(3k-8)]>1/(3k-5)²
On the right side of the inequality, 1 / (3k-5) ² constant > 0, if the inequality has a solution, (3K-2) (3k-8) > 0, k > 8 / 3, K is a positive integer, K ≥ 3
(3K-2) (3k-8) 0, the inequality holds, that is, when k ≥ 3, the inequality satisfies the problem
In conclusion, the minimum value of K is 3

A problem of sequence
1. In known sequence {an}, Sn is the sum of the first n terms, and an + Sn = 1,
(1) Finding the general term formula of sequence {an}
(2) If the sequence {BN} satisfies BN = 3 + log4an, let TN = | B1 | + | B2 | + |BN |, finding TN

（1）an+Sn=1,a(n+1)+S(n+1)=1
A (n + 1) - an + a (n + 1) = 0
Then an = 2A (n + 1), and A1 = S1, then A1 = 1 / 2
So an = 1 / 2 ^ n
（2） bn=3+log4an=3-n/2
When n ≤ 6, BN ≥ 0
When n > 6, BN < 0
So when n ≤ 6, TN = (B1 + BN) n / 2 = n (11-n) / 4
When n > 6, TN = (n-6) (N-5) / 4 + 15 / 2

A problem about sequence of numbers
It is known that {an} is equal ratio sequence, A2, (A3) + 1, A4 is equal difference sequence, and the general term formula is obtained
a1=2

From the question, A2 + A4 = 2 (A3 + 1)
Namely: a1q + a1q ^ 3 = 2 (a1q ^ 2 + 1)
Substituting A1 = 2 into
We get: Q ^ 3-2q ^ 2 + Q-2 = 0
That is, (Q ^ 2 + 1) (Q-2) = 0
The solution is: q = 2
So the general formula is: an = 2 ^ n

There is also a problem of sequence
It is known that both {an} and {BN} are arithmetic sequences. The sum of the first n terms of {an} is denoted as Sn, and the sum of the first n terms of {BN} is denoted as TN, and Sn / TN = (2n-3) / (4n-3). The solution is
（1）（a2000）/（b2000）
（2)[(b100）/（a1999+a1)]+[(b1900)/(a1900+a100)]

Sn = n (a1 + an) / 2tn = n (B1 + BN) / 2Sn / TN = (a1 + an) / (B1 + BN) = 2n-3 / 4n-3 (1): a2000 = 1 / 2 (a1 + a3999) b2000 = 1 / 2 (B1 + b3999) so a2000 / b2000 = (a1 + a3999) / (B1 + b3999) = (2 × 3999-3) / (4 × 3999-3) = 8995 / 15993 (2): a1999 + A1 = 2a1000a1900 + A10

Principles of Statistics: what is the distributive sequence?

The concept of distributive sequence
On the basis of statistical grouping, all units of the population are grouped and arranged to form the distribution of each unit in the population among each group, which is called distribution sequence, also known as distribution sequence or frequency distribution
The distributive sequence consists of two elements: one is the group of the whole according to a sign; the other is the total number of units occupied by each group
Distributive sequence is of great significance in statistical research. Distributive sequence is not only the main form of statistical grouping results, but also an important method of statistical analysis. It can show the distribution characteristics and structure of population units in each group, and further study the composition, average level and change regularity of indicators on this basis
2、 Types of distributive sequence
According to the nature of grouping marks, allocation sequence is divided into quality allocation sequence and variable allocation sequence
Variable sequence is divided into single value sequence and group distance sequence
1. Single valued sequence: refers to each group of values with only one specific variable value
Compilation conditions: variables are discrete variables; the number of different values of variables is small (both available)
[example] it is known that there are 24 workers in a workshop, and their daily output (pieces) are 20,23,20,24,23,21,22,25,26,20,21,21,22,22,23,22,24,25,21,22,21,24,23 respectively. It is required to compile variable series according to the above data
Daily output (piece) x number of workers (person) f
twenty
twenty-one
twenty-two
twenty-three
twenty-four
twenty-five
26 3
five
six
four
three
two
one
Total 24
2. Group distance sequence: it refers to the sequence of variable values of each group represented by an interval
Compilation conditions: variables are continuous variables; or: discrete variables with more total units and different values
Group distance sequence is divided into equidistance sequence and different distance sequence
Equidistant sequence: the length of variable value change interval is equal
Different distance sequence: variable value change interval length is not completely equal
Related concepts:
Group limit: it refers to the variable values representing the limits of each group at both ends of each group. The minimum value of each group is the low limit and the maximum value is the upper limit
Group distance: the length of the variable range of each group, which is the difference between the upper and lower limits
Group median value: the midpoint value of each variable value range
Group median = (upper limit + lower limit) / 2

There are 153 people in total. The number of class one is 1.12 times that of class three. The number of class two is 3 less than that of class three. How many people are there in each of the three classes?