It is known that the sequence {an} is an arithmetic sequence with non-zero tolerance, and a 2 = 3, and a 4, a 5, a 8 are equal proportion sequences. (1) find the general term formula of the sequence {an}; (2) let Sn be the sum of the first n terms of the sequence {an}, find the values of all n that make an = Sn hold

(1) Since A4, A5 and A8 are equal ratio sequence, a52 = a4a8. Let the tolerance of sequence {an} be D, then (3 + 3D) 2 = (3 + 2D) (3 + 6D) is reduced to D2 + 2D = 0. ∵ D ≠ 0, ∵ d = - 2. Then an = A2 + (n-2) d = - 2n + 7, that is, the general term formula of sequence {an} is an = - 2n + 7; (2) Sn = n (5
If 5x-y = 51, X-Y = 3, then x =, y=
1-2 gives 4x = 48, x = 12, 12-y = 3, y = 9
So x = 12, y = 9
x=12，y=9
x=12 y=9
x=12 y=9
x=12，y=9
If the greatest common factor of a and B is 4 and the least common multiple is 84, then the sum of a and B is 88 or 84（
Because 4x () x () = 84, the brackets can be 1 and 21, or 3 and 7, so these two numbers can be used
It's 4 and 84, it can be 12 and 28, and their sum has two 88 or 40
It is known that the sequence {an} is an arithmetic sequence with non-zero tolerance, and a 2 = 3, and a 4, a 5, a 8 are equal proportion sequences. (1) find the general term formula of the sequence {an}; (2) let Sn be the sum of the first n terms of the sequence {an}, find the values of all n that make an = Sn hold
(1) Since A4, A5 and A8 are equal ratio sequence, a52 = a4a8. Let the tolerance of sequence {an} be D, then (3 + 3D) 2 = (3 + 2D) (3 + 6D) is reduced to D2 + 2D = 0. ∵ D ≠ 0, ∵ d = - 2. Then an = A2 + (n-2) d = - 2n + 7, that is, the general term formula of sequence {an} is an = - 2n + 7; (2) Sn = n (5
5X+（20-x)2=51
5X+（20-x)2=51
5x+40-2x=51
3x=51-40
3x=11
X = 3 and 2 / 3
5x+40-2x=51
3x=11
x=11/3
5X+40-2x=51
3x+40=51
3x=51-40
3x=11
x=11/3
Ask the teacher to solve a math problem
Two companies, company a and company B, are going to organize a tour to Shanghai. The number of first-class tickets, second-class tickets and the cost are as follows:
First class ticket / second class ticket / fee / yuan
Company a 25 1800
Company B 16 1600
Q: calculate the unit prices of class I and class II tickets respectively
(1) Suppose the unit price of class I tickets is x yuan / piece, and that of class II tickets is y yuan / piece,
Then there are 2x + 5Y = 1800x + 6y = 1600 &;,
The solution is: x = 400Y = 200 & # 8203;,
A: the unit price of first class tickets is 400 yuan / piece, and that of second class tickets is 200 yuan / piece
It is known that the tolerance of the arithmetic sequence {an} whose sum of the first n terms is SN is not zero, and A2 = 3, and A4, A5, a8 are equal proportion sequence
Is there a positive integer pair (n, K), yes Nan = KSN? Find all positive integer pairs (n, K)

X △ 4 = 2x-3.5 step by step
Here is a math problem, please help me!
A square billet with 0.8 m edge length is forged into a rectangular square steel with a cross-sectional area of 0.16 square meters. How long is the forged square steel? (use equation solution)!
0.16*X=0.8*0.8*0.8
X=32
Because the volume of a cube is equal to that of a cuboid, the equation is formulated by using this equivalence relation
Cross sectional area * height (here is length) = edge length × edge length × edge length
Let this steel be x meters long
0.16x=0.8×0.8×0.8
x=32
A: the steel is 32 meters long.
0.8^3/0.16=3.2
0.8*0.8*0.8=0.16x
x=3.2
0.16*X=0.8*0.8*0.8 X=32
0.8*0.8*0.8/0.16=3.2m
. set the length as X meters
0.16x=0.512
0.16x=0.512÷0.16
x=3.2
A: this square steel is 3.2 meters long
The tolerance of arithmetic sequence is 3. If A2, A4 and A8 are in the same ratio sequence, calculate A4
a2=a1+3,
a4=a1+9,
a8=a1+21
a4/a2=a8/a4
(a1+9)/(a1+3)=(a1+21)/(a1+9)
a1=3
a4=3+9=12
A2 = a4-6, a8 = A4 + 12. So A4 / A2 = A8 / A4, take A2 and A8 above and calculate A4 = 12
The tolerance is 3
a2=a4-6
a8=a4+12
A2, A4 and A8 are equal ratio series
therefore
a2*a8=a4^2
So A4 ^ 2 = (a4-6) (A4 + 12)
The solution is A4 = 12
A2 = a4-6, a8 = A4 + 12
If A2, A4 and A8 are in equal proportion sequence, then: (A4) ^ 2 = A2 * A8
(a4)^2 = (a4-6)*(a4+12)
(a4)^2 = (a4)^2 + 6a4 - 72
a4=12
Let A2 = a4-6
a8=a4+12
Because A2, A4 and A8 are equal ratio series
A4 * A4 = A2 * A8: A4 * A4 = (a4-6) (A4 + 12)
The solution is: 4 = 72
a4=12

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