f.5maths ,很急

Let the three numbers be a, a + d and a + 2d
a + a + d + a + 2d = -3
3a + 3d = -3
a = -1 - d ... (1)
a(a + d)(a + 2d) = 8 ... (2)
Sub (1) into (2), (-1 - d)(- 1 - d + d)(-1 - d + 2d) = 8
(-1 - d)(-1)(-1 + d) = 8
(d + 1)(d - 1) = 8
d2 - 1 = 8
d2 = 9
d = +/- 3
When d = 3, a = -4 and the 3 numbers are -4, -1 and 2
When d = -3, a = 2 and the 3 numbers are 2, -1 and -4 (same as d = 3)
Let the three numbers be a, a + d and a + 2d
a + a + d + a + 2d = 21
3a + 3d = 21
a = 7 - d
After adding the numbers 2, 2 and 14, they becomes
a + 2, a + d + 2 and a + 2d + 14 or
9 - d, 9 and 21 + d
Since they are in GS,
9 / (9 - d) = (21 + d) / 9
81 = (9 - d)(21 + d)
81 = 189 - 12d - d2
d2 + 12d - 108 = 0
(d - 6)(d + 18) = 0
d = 6 or d = - 18
when d = 6, a =1 and the numbers are 1, 7 and 13
when d = -18, a = 25 and the numbers are 25, 7 and -11
Since a, -2 and b forms a GP,
-2/a = b/-2 =>ab = 4 ... (1)
Since -2, b, a form an AP, the difference of the AP is b - (-2) and (a - b).
So b + 2 = a - b
2b + 2 = a ...(2)
Sub (2) into (1) b(2b + 2) = 4
2b2 + 2b - 4 = 0
b2 + b - 2 = 0
(b + 2)(b - 1) = 0
b = -2 (rejected) or b = 1
b = 1 => a = 4
For the GP 4, -2, 1,... , a = 4, r = -1/2
Sum to infinity = a / (1 - r) = 4 / (1 + 1/2) = 8/3
For the positive terms, they are 4, 1, 1/4,... which is also a GP with a = 4 and r = 1/4.
Sum to infinity = a / (1 - r) = 4 / (1 - 1/4) = 16/3