Senior high school mathematics compulsory 5 arithmetic sequence of the first N and explore the problem to prove that Sn = PN ^ 2 + QN + R is arithmetic sequence and find out the tolerance

a(1)=s(1)=p+q+r
a(n+1)=s(n+1)-s(n)=p(2n+1)+q=2p(n+1)-p+q,
a(2)=2p*2-p+q=3p+q,a(2)-a(1)=3p+q-(p+q+r)=2p-r.
a(3)-a(2)=2p
If r = 0, then a (2) - A (1) = a (3) - A (2),
At this point,
a(n)=2np-p+q=p+q+(n-1)*2p,n=1,2,...
{a (n)} is an arithmetic sequence with the first term P + Q and the tolerance 2p

It is known that each item of an equal ratio sequence is not zero, and there is another equal difference sequence whose first item is zero and tolerance is not zero. Add the corresponding items of each sequence to form a new sequence 1,1,2. Find the general term formula of the original equal difference sequence and the equal ratio sequence

The first term of arithmetic sequence is zero
Then the first term of the equal ratio sequence is 1
Then d + q = 1
2d+q^2=2
Then q = 2, d = - 1
Then the arithmetic sequence is an = 0 + (n-1) * (- 1) = 1-n
The general term of equal ratio sequence is an = 1 * q ^ (n-1) = 2 ^ (n-1)

Is a sequence with zero tolerance an arithmetic sequence

A sequence with zero tolerance is an arithmetic sequence
This is a special arithmetic sequence d = 0
as
1,1,1、、,1
a,a,a,.a

Let A1, A2 , an is an n (n ≥ 4) arithmetic sequence with nonzero items, and the tolerance D ≠ 0. If the sequence (in the original order) obtained by deleting a certain item from the sequence is an equal ratio sequence, then the set composed of all number pairs (n, a1d) is______ ．

Let the tolerance of sequence {an} be D, then the items are: A1, a1 + D, a1 + 2D If the first term is removed, there is (a1 + D) (a1 + 3D) = (a1 + 2D) 2, and the solution is d = 0, which is not suitable for the problem; if the second term is removed, there is A1 (a1 + 3D) = (a1 + 2D) 2, and the reduction is 4D2 + a1d = 0, that is, D (4D + A1) = 0, and the solution is d = - A14. Because each item of the sequence is not zero, the fifth term (a1 + 4D = 0) will not appear in the sequence, so the number pair (n, a1d) = (4) If the third term is removed, there is A1 (a1 + 3D) = (a1 + D) 2, which is reduced to d2-a1d = 0, that is, D (d-a1) = 0, and the solution is d = A1, then the sequence is: A, 2a, 3a, 4a The fifth term will not appear in this sequence, because the fifth term does not appear in the sequence, so the number pair (n, a1d) = (4, 1); when the fourth term is removed, there is A1 (a1 + 2D) = (a1 + D) 2, which is reduced to D = 0, which is not the meaning of the problem; when the fifth term or further term is removed, there must be the situation when the first term and the fourth term are removed, that is, d = 0, which is not the meaning of the problem There are two pairs of numbers in meaning, and the set is {(4, - 4), (4,1)}. So the answer is: {(4, - 4), (4,1)}

Let {an} be an arithmetic sequence with nonzero tolerance,
SN is the sum of the first n terms, satisfying (A2) &# 178; + (A3) &# 178; = (A4) &# 178; + (A5) &# 178;, S7 = 7
(1) Find the general term formula of sequence and the first n terms and Sn
(2) Try to find all positive integers m so that [am × a (M + 1)] / a (M + 2) is the term in the sequence SN
This is a question for college entrance examination, especially the second question. I finished it differently from the answer. In addition, the am, a (M + 1), a (M + 2) in the second question represent the m-th item of the sequence a, (M + 1) and (M + 2)

(1) Because it's an arithmetic sequence
So the meaning of the title is: (a1 + D) ^ 2 + (a1 + 2D) ^ 2 = (a1 + 3D) ^ 2 + (a1 + 4D) ^ 2
The reduction is: (2 * a1 + 5d) d = 0
Because D ≠ 0, 2 * a1 + 5D = 0
That is: D = - 2 * A1 / 5
Because S7 = 7, that is, A4 = a1 + 3D = a1-6 * A1 / 5 = - A1 / 5 = 1
So A1 = - 5, d = 2
So an = a1 + (n-1) d = - 5 + 2 (n-1) = 2n-7
Sn=(a1+an)*n/2=[(-5)+(2n-7)]n/2=n(n-6)
(2) If am × a (M + 1)] / a (M + 2) is a term in Sn, then there is
(2m-7)(2m-5)/(2m-3)=n(n-6)
When m = 1, am × a (M + 1)] / a (M + 2) = - 15, can not be expressed as n (n-6), n ∈ n+
When m = 2, am × a (M + 1)] / a (M + 2) = 3, can not be expressed as n (n-6), n ∈ n+
When m = 3, am × a (M + 1)] / a (M + 2) = 1, can not be expressed as n (n-6), n ∈ n+
When m ≥ 4
Because: 2m-9 ＜ (2m-7) (2m-5) / (2m-3) ＜ 2m-7 ------ This can be seen by multiplying
Because (2m-7) (2m-5) / (2m-3) = n (n-1) is an integer
So: (2m-7) (2m-5) / (2m-3) = 2m-8
The solution is m = 5.5
It's impossible
So [am × a (M + 1)] / a (M + 2) is not a term in SN

In the arithmetic sequence an with nonzero tolerance
SN is the sum of the first n terms of the sequence an. Given S3 square = 9s2, S4 = 4s2, find the general term of an

The solution is as follows
(S3)^2=(3a1+3d)^2=9S2=9(a1+a2)=9(2a1+d)
That is: (3A1 + 3D) ^ 2 = 9 (2A1 + D)
And because: S4 = 4s2
We get: a1 + A2 + a3 + A4 = 4 (a1 + A2)
That is 4A1 + 6D = 8A1 + 4D
The relationship between A1 and D is obtained: D = 2A1
Substituting the relation into (3A1 + 3D) ^ 2 = 9 (2A1 + D)
Yes, (3A1 + 6A1) ^ 2 = 9 (2A1 + 2A1)
The result is: 9 (A1) ^ 2-4a1 = 0
Because the tolerance is not zero
So A1 is not zero, so solving the equation, A1 = 4 / 9
d=8/9
an=a1+(n-1)d=4/9（2n-1）
That's it,

Can the tolerance of arithmetic sequence be 0

Arithmetic sequence: if a sequence, starting from the second term, subtracts the previous term from each term, and the difference is equal to the same constant, the sequence is called arithmetic sequence, and the constant is called tolerance of arithmetic sequence, and the tolerance is usually expressed by the letter D
Because 0 is also a constant, 0 can be used as a tolerance

If the first term of an arithmetic sequence is 0, the tolerance is 2
If the first term of an arithmetic sequence is 0 and the tolerance is 2, what is the sum of the first 20 terms of the arithmetic sequence?

0,2,4,6
a1=0
a20=a1+19d=0+19x2=38
Sn=20x(a1+a20)/2=10x(0+38)=380

If A1, A2 If D ≠ 0, then ()
A. a1a8＞a4a5B. a1a8＜a4a5C. a1+a8＞a4+a5D. a1a8=a4a5

∵ 1 + 8 = 4 + 5 ∵ a1 + A8 = A4 + A5 ∵ exclude C; if an = n, then a1a8 = 1.8 < 20 = 4.5 = a4a5 ∵ exclude D, a

When the tolerance of arithmetic sequence is 0, does the sum of the first n terms have the maximum value
We also need to ask: 1. Find the square of the maximum value of the sum of the first n terms of the arithmetic sequence {an}. 2. How can an ≥ 0 and an + 1 determine the natural number n

D = 0, Sn = Na1, if A1 = 0, the maximum and minimum values are 0
If A1 is less than 0, the maximum value is S1 = A1 and the minimum value is negative infinity
If A1 is greater than 0, the maximum value is positive infinity and the minimum value is A1

Senior high school mathematics compulsory 5 arithmetic sequence of the first N and explore the problem to prove that Sn = PN ^ 2 + QN + R is arithmetic sequence and find out the tolerance