The title of the third question is problematic. First of all, a 1 does not satisfy Sn=(an+2)? /8。

I think Sn=[(an+2)? -1]/8, as follows:

When n≥ 1, 8Sn=(an+2)? - 1，8S(n+ 1)=[a(n+ 1)+2]？ - 1,

The two expressions are subtracted to get 8a (n+1) = 8s (n+1)-8sn = [(a (n+1)+2]? -(an+2)？ ,

[a (n+1)+an] {a (n+1)-an-4] = 0, so a(n+ 1)-an-4=0 because an∈N*,

That is, a(n+ 1)-an=4, and a 1= 1 satisfies Sn=[(an+2)? - 1]/8,

So the sequence {an} is a arithmetic progression, the first term is 1, and the tolerance is 4, so an= 1+4(n- 1)=4n-3.

7. In arithmetic progression, a1+an = a2+a (n-1) = a3+a (n-2),

And because a 1+a2+a3 = 15, an+a (n- 1)+a (n-2) = 78,

So 3 (a1+an) =15+78 = 93, that is, a 1+an=3 1,

Because sn = [n (a1+an)]/2 = 31n/2 =155, n= 10.

8. In arithmetic progression, a9 and b9 are the arithmetic averages of a 1 and a 17, and b 1 and b 17, respectively.

So A9/B9 = [17 (a1+a17)/2]/[17 (b1+b17)/2] = s.