急! F6 Geometric Sequences 02q4

4.
Let $Tn be the amount that the nth prize is worth.

T9 = 2T10, T8 = 2T9 ...... and so on
Hence, T2 = (1/2)T1, T3 = (1/2)T2,T4 = (1/2)T3, ......, T3 = (1/2)T2,T2 = (1/2)T1

T1, T2, T3, ......, T10
= T1, (1/2)T1, (1/2)²T1,......, (1/2)⁹T1

T1, T2, T3, ......, T10 is an geometric sequence.
The common ratio, r = 1/2

T10 = 50
(1/2)⁹T1= 50
T1 = 50 x 2⁹
T1 = 25600

The first prize is worth $25600.


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8.
Let Tn km be the distance travelled in the nth minute.

T2 = (3/4)T1
T3 = (3/4)T2 = (3/4)²T1
T4 = (3/4)T3 = (3/4)³T1
......
Tn = (3/4)ⁿ⁻¹T1

Hence, T1­, T2, T3, ......, Tn is an geometric sequence.
The first term, a = T1 = 1.5
The common ratio, r = (3/4)

Sum to infinite term
= a / (1 - r)
= 1.5 / [1 - (3/4)]
= 6

The maximum distance that the car will travel = 6 km