A.S.& G.S.

1.
The negative terms form a new G.S.: -3, -12, -48, ......
The first term of the new G.S., T(1) = -3
Common ratio, r = -12/(-3) = 4

T(n) = ar^(n - 1) > -6000
(-3)*4^(n - 1) > -6000
4^(n - 1) < 2000
log[4^(n - 1)] < log2000
(n - 1)log4 < log2000
(n - 1) < log2000/log4
(n - 1) < 5.48
n < 6.48

Hence, there are 6 negative terms greater than -6000.


2.
Let T(1) = a

R(1) = T(2) = ar
R(2) = T(4) = ar^3 = R(1)*(r^2)
R(3) = T(6) = ar^5 = R(2)*(r^2)
R(4) = T(8) = ar^7 = R(3)*(r^3)

Obviously, this is a geometric sequence with common ratio r^2.


3(a)
log a, log b, log c are in geometric sequence :
log b / log a = log c / log b
[(3/2)log b] / [(3/2)log a] = [(3/2)log c] / [(3/2)log b]
log b^(3/2) / log a^(3/2) = log c^(3/2) / log b^(3/2)
log √b^3 / log √a^3 = log √c^3 / log √b^3
Hence, log √a^3, log √b^3, log √c^3 are also in geometric sequence.

3(b)
log a, log b, log c are in arithmetic sequence :
log b - log a = log c - log b
log (b / a) = log (c / b)
b / a = c / b
b^(3/2) / a^(3/2) = c^(3/2) / b^(3/2)
√b^3 / √a^3 = √c^3 / √b^3
Hence, √a^3, √b^3, √c^3 are in geometric sequence.


4.
Let the three integers be (a - d), a, (a + d).

(a - d)*a*(a + d) = 11 * [(a - d) + a + (a - d)]
a(a + d)(a - d) = 33a
(a + d)(a - d) = 33
(a + d)(a - d) = 11*3 or (a + d)(a - d) = 33*1

When (a + d)(a - d) = 11*3
a + d = 11 ...... [1]
a - d = 3 ...... [2]
[1] + [2] : 2a = 14, then a = 7

When (a + d)(a - d) = 33*1
a + d = 33 ...... [1]
a - d = 1 ...... [2]
[1] + [2] : 2a = 34, then a = 17

Hence, the three integers are 3, 7, 11 or 1, 17, 33.

1.
Since you consider negative term,
then I just consider -3,-12,...
Common ratio = 4
-3(4)^(k-1) > -6000
4^(k-1) < 2000
4^k < 8000
k < ln8000/ln4
k < 6.48 (corr to 2.d.p.)
Therefore, there are only 6 negative terms greater than -6000.
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2.
Given T(n) : T(n+1) = r, i.e. T(n) : T(n+2) = r^2
R(n) = T(2n) ,i.e. R(n) : R(n+1) = T(2n) : T(2n+2) = r^2 since 2n ~ n as a variable,
Common ratio = r^2
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3(a)
Given log a : log b = r and log b: log c = r,
log a^(3/2) : log b^(3/2) = (3/2) log a : (3/2) log b = log a : log b = r
W.L.O.G., log b^(3/2) : log c^(3/2) = r
Hence, log a^(3/2) , log b^(3/2), log c^(3/2) are in geometric sequence
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3(b)
Given log b - log a = k and log c - log b = k,
i.e. log (b/a) = k => b/a = e^k => a^(3/2) : b^(3/2) = 1 : e^(3k/2)
W.L.O.G., b^(3/2) : c^(3/2) = 1 : e^(3k/2)
Hence, a^(3/2) , b^(3/2) , c^(3/2) are in geometric sequence
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4.
Let the numbers be (x-k), x , (x+k), for x > k > 0,
(x-k)x(x+k) = 11[(x-k)+x+(x+k)]
x(x^2 - k^2) = 11(3x)
(x-k)(x+k) = 33 since x=/=0
Factor of 33 = 1,3,11,33
If x-k = 1, then x+k = 33,
x = [(x-k) + (x+k)]/2 = (1+33)/2 = 17 and k = 33-17 = 16
If x-k = 3, then x+k = 11
x = [(x-k)+(x+k)]/2 = (3+11)/2 = 7 and k = 11-7 = 4
Therefore, the three numbers can be (1,16,33) or (3,7,11)
Since
1*16*33 = 528 = 11(1+16+33) + 2
Therefore, (1,16,33) is rejected.
3*7*11 = 231 = 11(3+7+11) + 0
Therefore, the only possible set of number = (3,7,11)