急 F6 Geometric Sequences 02q10

14.
(a)
Let a and r be the first term and the common ratio of the geometric sequence.

T2 + T3 = 24
ar + ar² = 24 ...... [1]

T6 = 8T3
ar⁵ =8(ar²)
r³ =8
r = 2

Put r = 2 into [1] :
a(2) + a(2)² =24
6a = 24
a = 4
The first term = 4

(b)
Tk = 512
arᵏ⁻¹ =512
4(2)ᵏ⁻¹ =512
(2)ᵏ⁻¹ = 128
(2)ᵏ⁻¹ = (2)⁷
k - 1 = 7
k = 8


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18.
(a)
T1, T2, T3, T4, ......, T8
= T1, T1r, T1r², T­1r³,......, T1r⁷

-kT1, -kT2, -kT3, -kT4, ......, -kT8
= (-kT1), (-kT1)r, (-kT1)r², (-kT­1)r³,......, (-kT1)r⁷

(-kTn+1)/(-kTn) = r
Hence, this is a geometric sequence.
The common ratio = r

(b)
1xT1, 2xT2, 3xT3, 4xT4, ......, 8xT8
= T1, T1(2r), T1­(3r)², T­1(4r)³,......, T1(8r)⁷

(1xT2)/(1xT1) = 2r
(1xT3)/(1xT2) = (3/2)r
Hence, this is NOT a geometric sequence.

(c)
T1², T2², T3², T4², ......,T8²
= (T1)², (T1r)², (T1r²)², (T­1r³)²,......, (T1r⁷)²
= T1², T1²r², T1²r⁴, T­1²r⁶,......, T1²r¹⁴

Tn+1/Tn = r²
Hence, this is a geometric sequence.
The common ratio = r²

14 a) Let a and r be the first term and common ratio of the sequence

T(6) = 8 T(3)
ar^(6-1) = 8 ( ar^(3-1) )
ar^(5) = 8 ar^(2)
r^(3) =8
r =2

T(2) + T(3) = 24
ar^(2-1) + ar^(3-1) = 24
6a = 24
a =4

Therefore, the first term is 4

b) T(k) = 512
4 x 2^(k-1) = 512
2^(k-1) = 128
2^(k-1) = 2^(7)
k-1 = 7
k=8

18 a) Yes,it is. The common ratio is r
b) Yes,it is. The common ratio is 2r
c) Yes, it is. The common ratio is r^(2)