∫{ 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