geometric sequences

a.let c be common ratio

c^3T(4)=T(7)
48c^3=6
c=1/2

so (1/2)^3T(1)=T(4)
T(1)/8=48
T(1)=384

b.T(n+2)/T(n)where n is a odd no.=(1/2)^2T(n)/T(n)=1/4

soT(1),T(3),T(5),T(7) is a geometric sequence

設公比為 a
T(4) x a = T(5)
T(5) x a = T(6)
T(6) x a = T(7)

T(4) x a^3 = T(7)
48 x a^3 = 6
a^3 = 1/8
a = 1/2
so the common ratio is 1/2

T(1) = T(4) / a^3
=48 x 8
=384

T(1) x a^2 = T(3)
T(3) x a^2 = T(5)
T(5) x a^2 = T(7)
.
.
T(n-2) x a^2 = T(n)
T(n) x a^2 = T(n+2)

so it is a geometric sequence with common ratio of a^2 = 1/4

a) Let R be the common ratio

T(4)=T(1)x (R to the power 3)=48 [1]
T(7)=T(1)x (R to the power 6)=6 [2]

6/48= (R to the power 6) / (R to the power 3)
1/8= (R to the power 3)
R= 1/2

from [1], T(1)x (1/2 to the power 3)=48
T(1)=48x8
T(1)=384

so T(1)=384, common ratio=1/2

b) T(1)=T(1)
T(3)=T(1)x R to the power 2
T(5)=T(1)x R to the power 4
T(7)=T(1)x R to the power 6

T(3) / T(1)= R to the power 2
T(5) / T(3)= R to the power 2
T(7) / T(5)= R to the power 2
...................
As T(3) / T(1)= T(5) / T(3)=T(7) / T(5).......
T(1) ,T(3),T (5) ,T(7),...is a geometric sequence.