急問 數學題

1.
a,b,c成AP, c-b=b-a, b=(a c)/2

a,c,b成GP, c/a=b/c=(a c)/2c=a/2c 1/2

假設 x=c/a,

x=1/2x 1/2
2x^2=1 x
2x^2-x-1=0

x=-1/2 or 1 (neglect as a is not equal to c)


c/a=-1/2

2.
若{an} 是AP,
a3 a8=a2 a9=a1 a10;

由於 a1與a10 是方程 x²-3x-5=0的兩根, 故 a1 a10=3

a3 a8=3

若{an} 是GP，
a5.a6=a4.a7=a3.a8=a2.a9=a1.a10;

由於 a1與a10 是方程 x²-3x-5=0的兩根, 故 a1.a10=-5

a5.a6=-5

3.
如果方程 2x²-5x k=0有一個根為零，

2(0)^2-5(0) k=0, k=0

1.
2b = a + c.............(1)
c/a = b/c, so c^2 = ab..................(2)
Sub. (2) into (1)
2(c^2/a) = a + c
2c^2 = a^2 + ac. Divide all terms by a^2, we get
2(c/a)^2 = 1 + (c/a)
2(c/a)^2 - (c/a) - 1 = 0
[2(c/a) + 1][(c/a) - 1] = 0
so c/a = -1/2 or 1(rej. because a, b and c are not equal).
2.
For AP. a + a + 9d = sum of roots = 3 = 2a + 9d.
So a3 + a8 = a + 2d + a + 7d = 2a + 9d = 3.
For GP. (a)(ar^9) = product of roots = -5 = a^2r^9.
So a5.a6 = (ar^4)(ar^5) = a^2r^9 = -5.
3.
Since one of roots is zero. Put x = 0, we get
k = 0.