maths problems. AS.GS
plz help me solve the following maths problems as much as u can,thx in advance! 
1. 2 numbers p and q (where p>q) are such that p , 10 , q are the first 3 terms of an A.S. And p,8,q are the first 3 terms of a G.S. Knowing p+q=20 and pq=64,form a quadratic equation in x with roots p and q.Express ur answer in the form ax^2+bx+c=0.
2. x and y are +ve numbers. -6,y,x form an A.S and x,6-,y form a A.S.Find the value of xy and the values of x and y.
3. The first 3 terms of an AS are 1,cos^2θ and sinθ.Form an equation in sinθ.
Update:well i cant be sure if the exercise has got Q.2 wrong~
Comments
(1) Sum of roots = 20
Product of roots = 64
Hence the equation is x2 - 20x + 64 = 0.
(2) From the given, we have:
x - 6 = 2y and xy = 36.
Therefore, sub x = 2y + 6 into x + y = -12:
Solving, we have: y = -6 and x = -6
Finally, xy = 36
(3) From the given, we have:
1 + sin θ = 2 cos2 θ
= 2 (1 - sin2 θ)
= 2 - 2sin2 θ
2sin2 θ + sin θ - 1 = 0
2009-03-18 11:21:00 補充：
For Q2, are u sure that
"-6,y,x form an A.S"
AND
"x,-6,y form a A.S." ?
2009-03-18 11:23:03 補充：
Q1:
For any 2 given roots α and β of a quadratic equation, it can be written as:
(x - α)(x - β) = 0
Expanding:
x^2 - (α + β)x + αβ = 0
So we have:
x^2 - (Sum off roots)x + Product of roots = 0
2009-03-18 11:25:12 補充：
Q3 For any 3 successive terms a, b, c in A.S., we have:
(a + c)/2 = b
So the result will be:
(1 + sin θ)/2 = cos^2 θ
1 + sin θ = 2 cos^2 θ
1 + sin θ = 2 (1 - sin^2 θ)
1 + sin θ = 2 - 2sin^2 θ
2sin^2 θ + sin θ - 1 = 0