Question 1 I won't write it.

The second question is similar. First of all, we should believe that only arithmetic progression can meet those two conditions at the same time, and then make bold guesses and prove the conclusion on this premise. The difficulty of the college entrance examination is usually relatively low, and the knowledge of middle school students is not much. I believe this conclusion is very simple.

Use conditions first.

N>3 point (s _ {n+3}-s _ {n})+(s _ {n}-s _ {n-3}) = 2s _ 3, that is.

a _ { n+3 }+a _ { n+2 }+a _ { n+ 1 }-a _ { n }-a _ { n- 1 }-a _ { n-2 } = 2S _ 3(*)

N+ 1 and this formula is obtained by subtracting n.

A _ {n+4}-2a _ {n+1}+a _ {n-2} = 0, that is, a _ {n+4}-a _ {n+1} = a _ {n+1}

In this way, we get three groups of arithmetic subsequences of the first kind with an interval of 3: A _ 1 = {A _ 2, A _ 5, ...}, A _ 2 = {A _ 3, A _ 6, ...}, A _ 3 = {A _ 4, A _ 7, ...}

Similarly, the condition of k=4

a _ { n+4 }+a _ { n+3 }+a _ { n+2 }+a _ { n+ 1 }-a _ { n }-a _ { n- 1 }-a _ { n-2 }-a _ { n-3 } = 2S _ 4(* *)

Four groups of arithmetic subsequences of the second kind can be obtained through one iteration, with the interval of 4, b _ 1 = {a _ 2, a _ 6, ...}, b _ 2 = {a _ 3, a _ 7, ...}, b _ 3 = {a _ 4, a _ 8, ...

And note that {a_n} anything except a_ 1 must belong to both A_u and a b _ v.

The next step is to prove that the tolerance of several arithmetic progression in each category is the same, because 3 and 4 are coprime, so we should believe that the conclusion must be correct.

A_{n+4}-a_{n-3}=2a_4 is obtained from (* *)-(*), that is to say, a kind of arithmetic subsequence with an interval of 7 is obtained. Assuming that the tolerance of A_u is d_u, then for any a_n belonging to A_u, 7d _ u = a _ {n+21}-a _ {n} = 6a _ 4, so d_u=6/7*a_4, that is, the tolerance of three groups of sequences of the first kind. Similarly, the four groups of sequence tolerances of the second category of a_{n+28}-a_{n} are the same, abbreviated as D, and its size is D=2a_4.

(If you didn't think of this step (* *)-(*), then you can examine a_{n+ 12}-a_{n}. Note that a_{n} can search all A_u and B_v, and you can get that d_u and D_v have nothing to do with U and V, but you can't get D directly.

The next step is very clear, that is, to prove that the whole {a_n} (except the first term) is arithmetic progression, and we also need to take several special points to solve the equation from the common points of the two series.

use

A8 = a2+2d = a4+D

a_ 10 = a_2+2D = a_4+2d

Solve d/3=D/4, and then substitute A _ {n+4} = A _ {n}+D = A _ {n+1}+D, that is, starting from a_2, {a_n} is arithmetic progression, and the tolerance is D-D.

Finally, combining the previous d = 6/7 * a _ 4 and d = 2a _ 4, we get D=8, d = 6, a_4=7, and then we get a_n=2n- 1, which is also true for 1.

(If you didn't expect this step (* *)-(*), you can change (*) to 3d=2S_3 and (* *) to 4D=2S_4, and you can get the same conclusion. In a word, the last step is to solve the linear equations without thinking.