Solution 1, sinb = sin (180-b) = sin (a+c) = sin3c.

a/sinA=b/sinB=c/sinC=2R

a+c = 2R(sin2C+sinC)= 8 = 2b = 4 rsin 3c

2 sinccosc+sinC = 6 sinC-8 sincsincsinc

CosC=3/4 cosC=- 1/2 (obsolete) sinC=√7/4.

sinA=sin2C=3√7/8

a=24/5 c= 16/5

2、 1 1(a 1+4d)= 5(a 1+7d)- 13

1 1a 1+44d = 5a 1+35d- 13 9d =-6a 1- 13

9d =-6 *(3)- 13 9d = 5d = 5/9

sn=na 1+n(n- 1)d/2=-3n+5n(n- 1)/ 18=(5n^2-59n)/ 18=(5/ 18)((n-59/ 10)^2-(5/ 18)(59/ 10)^2

When n=6, the minimum value is Sn=-29/3.

3、3、a(n+2)-a(n+ 1)= a(n+ 1)-an

So if an is arithmetic progression, a4-a 1=3d=-6.

Then d=-2, an = a1+(n-1) d = 8-2 (n-1) =-2n+10.

4、4x^2-8x+3=0(2x- 1)(2x-3)= 0

X= 1/2 or x=3/2 {an} is a geometric series whose common ratio q is greater than 1.

So a2004= 1/2, a2005=3/2, and q=a2005/a2004=3.

a2004+a2005=2

a2006+a2007=(a2004+a2005)*q^2=2*3^2= 18

5. Obviously, when an ≥ 0 an+ 1 < 0, Sn is the largest.

-n^2+ 10n+ 1 1≥0 ,n^2- 10n- 1 1≤0

The solution is 0≤n≤ 1 1, where a 1 1=0 and A 12 < 0.

And s11= s10+a11= s10.

So the sum of the first item to the item 10 or the item 1 1 is the largest.