A.P G.P.

1
We have
a+ar+ar^2+...=16
a^2+(ar)^2+(ar)^4+...=153 3/5
So
a/(1-r)=16...(1)
a^2/(1-r^2)=153 3/5...(2)
Sub (1) into (2)
16a/(1+r)=153 3/5
a/(1+r)=48/5
So
16(1-r)=48(1+r)/5
(-128/5)r=-32/5
r=0.25
a=16(0.25-1)=12
The fourth term of the sequence
=ar^3
=12*(1/64)
=3/16
2
T(n)=[3/(2x+1)]^n
a=T(1)=3/(2x+1)
r=3/(2x+1)
The sum to infinity in terms of x
=a/(1-r)
=[3/(2x+1)]/[1-3/(2x+1)]
=[3/(2x+1)]/[(2x-2)/(2x+1)]
=3/(2x-2)
=(3/2)[1/(x-1)]
(b)
Let
(3/2)[1/(x-1)]=22
1/(x-1)=44/3
44x-44=3
x=47/44

2007-12-17 17:44:18 補充：
For question 2absolute value of the common ration is less than 1That is-1<3/(2x+1)<1

1) Let the first term and the common ratio of the geometric sequence be T and r respectively. Then the square of its terms also form a geometric sequence with the first term and the common ration being T^2 and r^2 respectively. We therefore have two equations :
T/(1-r) =16-------------------------(1)
[(T^2)/(1-r^2)] = 768/5---------(2)
Squaring both sides of (1), we have [(T^2)/(1-r)^2]=256--------------(3)
Dividing (2) by (3) to eliminate T, [(1-r)^2)]/(1-r^2) = (768/5)(1/256)=(3/5)
5[(1-2r+r^2)]=3(1-r^2)
8r^2-10r+2=0
2(4r-1)(r-1)=0
r=(1/4) or r=1 (rejected since when r=1, the infinite sum of a geometric sequence does not converge)
Put r=(1/4) into (1), we have T= 12. [Also please check the accuracy of (2)]
The fourth term = T(r)^3=12(1/4)^3= 3/16
(2) (a) the first term = the common ratio = [3/(2x+1)]
sum to infinity = [3/(2x+1)] / [ 1 - (3/2x+1)] = [3/ (2x-2)] provided that the absolute value of the common ration is less than 1, which means that 1<x.
(b) [3/(2x-2)] =22
x = 47/44
Hope the above is of assistance to you