∫{ an} is arithmetic progression, a 1, a3 and a4 are geometric progression.

∴ (a 1+2d)？ =a 1*(a 1+3d)

a 1？ +4a 1*d+4d？ =a 1？ +3a 1*d

Simplify and get a 1*d=-4d?

∫d≠0

∴ a 1=-4d

∴(a 1+a5+a 17)/(a2+a6+a 18)

=(a 1+a 1+4d+a 1+ 16d)/(a 1+d+a 1+5d+a 1+ 17d)

=(3a 1+20d)/(3a 1+23d)

=(- 12d+20d)/(- 12d+23d)

= 8d/ 1 1d

= 8/ 1 1

the second question

The sequence {an} is monotonically decreasing, so for any n∈N, an > a (n+ 1).

North? +λn>-(n+ 1)？ +λ(n+ 1)

-n？ +λn>-n？ -2n- 1+λn+λ

Simplified to 2n+ 1 > λ.

∫n∈ natural number n

The minimum value of ∴ 2n+ 1 is 3, which means 2n+ 1≥3.

In order to keep 2n+ 1 > λ, λ < 3.

So the range of λ is λ < 3.

Third question

Sn=2^n-3 (n∈N)

When n≥2 and n ≥ n,

an= Sn-S(n- 1)

= (2^n-3)-(2^(n- 1)-3)

= 2^n-2^(n- 1)

= 2^(n- 1)*(2- 1)

= 2^(n- 1)

When n= 1, a1= s1= 2-3 =-1.

So the general formula of {an} is an=-1(n= 1).

2 (n- 1) (n ≥ 2 and n∈N)

The fourth question

For any m, n∈N, there is Sn+Sm=S(m+n).

Then when n= 1 and m=9, there is S 1+S9=S 10.

∴a 10 = s 10-S9 = s 1

∫s 1 = a 1 = 1

∴ a 10= 1

I hope you can adopt it. thank you