急! F6 Geometric Sequences 02q2

20. Let a be the first term and R be the common ration of the sequence.
a+aR^(3-1)=2(aR^(2-1)+aR^(4-1))
a+aR^2=2aR+2aR^3
2aR^3-aR^2+2aR-a=0
a(2R^3-R^2+2R-1)=0
a=0 (rejected) or 2R^3-R^2+2R-1=0
Let f(R)=2R^3-R^2+2R-1
f(0.5)=2(0.5)^3-(0.5)^2+2(0.5)-1=0
Therefore, by factor theorem, 2R-1 is a factor of f(R).
2R^3-R^2+2R-1=0
(2R-1)(R^2+1)=0
R=0.5 or R^2=-1(rejected)
Hence, the common ratio of the sequence is 0.5.

21. Let a be the first term and R be the common ration of the sequence.
T4=aR^(4-1)=aR^3=3...(1)
T6=aR^(6-1)=aR^5=9...(2)
(2)/(1):R^2=3
R=-√3 or √3
Put R=√3 into (1),
a(√3 )^3=3
3√3 a=3
a=1/√3

Put R=-√3 into (1),
a(-√3)^3=3
-3√3a=3
a=-1/√3

Therefore, (i) the general term Tn=aR^(n-1)
=(1/√3 )(√3 )^(n-1)
=(√3 )^(n-1-1)
=(√3 )^(n-2)
(ii) the general term Tn= aR^(n-1)
=(-1/√3)(-√3)^(n-1)
=(-1)(-1)^(n-1)(√3)(n-1-1)
=(-1)^n (√3 )^(n-2)
23 (a) Let a be the first term and R be the common ratio.
T2=aR=14.........(1)
T4=aR^3=56......(2)
(2)/(1)=R^2=4
R=-2 or R=2
Put R=-2 into (1),
a(-2)=14
a=-7
Put R=2 into (1),
a(2)=14
a=7

Therefore, (i) the general term Tn=(-7)(-2)^(n-1)
=(-1)(-1)^(n-1) 7x2^(n-1)
=(-1)^n 7(2)^(n-1)
(ii) the general term Tn=7x2^(n-1)

For (i), T10=(-1)^10 7x2^9=1x7x512=3584
(ii), T10= 7x2^9=3584

頭先因為超出字數限額，所以我用另一個帳答，繼續：
23(b) Let a be the first term and R be the common ratio.
T3=aR^2=18.........(1)
T5=aR^4=162.......(2)
(2)/(1):R^2=9.........(3)
r=-3 or 3
Put (3) into (1),
a(9)=18
a=2
Therefore, (i) the general term Tn=2(-3)^(n-1)
(ii) the general term Tn=2x3^(n-1)
For (i), T10=2(-3)^9=-39366
For (ii), T10=2x3^9=39366

如果比分，請比Kitty Fung個帳, thx!