中五 ---- Arithmetic and Geometric Sequences

1. (a) k / 6 = (3k / 4) / k
k / 6 = 3 / 4
4k = 18
k = 4.5
(b) The common ratio = 4.5 /6 = 3 / 4
Sum to infinity = a / (1 - r)
= 6 / (1 - 3 / 4)
= 6 / (1 / 4)
= 6 * 4
= 24
2. (a) Let a and r be the first term and the common ratio respectively.
ar^3 = 48............(1)
ar^6 = 6..............(2)
(2) / (1), ar^6 / ar^3 = 6 / 48
r^3 = 1 / 8
r = 1 / 2.
Thus a(1 / 2)^3 = 48
(1 / 8)a = 48
T(1) = a = 48 * 8 = 384.
(b) T(3) / T(1) = ar^2 / a = r^2
T(5) / T(3) = ar^4 / ar^2 = r^2
T(7) / T(5) = ar^6 / ar^4 = r^2
...
T(2n + 1) / T(2n - 1) = ar^(2n) / ar^(2n - 2) = r^2
The common ratio is r^2. Thus T(1), T(3), T(5), ... is a geometric sequence.

1a)
Since they are in G.P.
(3k/4)/k = k/6
3/4=k/6
k=9/2

1b)
Commone ratio=(9/2)/6=3/4
Sum= (6x1)/[1-(3/4)]
=24

2a)
T(4)=ar^3 = 48 --------(1) where a is the first term, r is common ratio
T(7)=ar^6 = 6 ---------(2)
(2)/(1): r^3=1/8
r =1/2
So, a = 6
Therefore, T(1) = a = 6 , common ratio = 1/2

2b) T(1) = ar^0
T(3) =ar^2 = T(1) x r^2
T(5) =ar^4 = T(3) x r^2
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T(n+1) / T(n-1) = (ar^n) / [ar^(n-2)]
= r^2
So, T(n+1) = T(n-1) x r^2
Therefore, T(1),T(3),T(5),T(7),...is a geometric sequence.