急! F6 Geometric Sequences 02q6

11.
(a)(i)
Let Tn be the number of revolutions in the nth minute.

T1 = 320
T2 = (3/4)T1
T3 = (3/4)T2 = (3/4)²T1
.....
Tn = (3/4)ⁿ⁻¹T1

Hence, T1, T2, T3, ......, Tn is ageometric sequence.
The first term, a = T1 = 320
The common ratio, r = 3/4

T3
= ar²
= 320 x (3/4)²
= 180

The number of revolutions in the third minute = 180

(a)(ii)
T4
= ar³
= 320 x (3/4)³
= 135

The number of revolutions in the third minute = 135

(b)
Sum to infinite terms
= a / [1 - r]
= 320 / [1 - (3/4)]
= 1280

The maximum number of revolutions = 1280


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12.
Let Tn m be the height of the nth rebound.

T1 = 6
T2 = 0.6 x 6 = 0.6T1
T3 = 0.6 x T2 = 0.6²T1
.....
Tn = 0.6 x Tn-1 = 0.6ⁿ⁻¹T1

Hence, T1, T2, T3, ...... Tn is ageometric sequence.
The first term, a = T1 = 6
The common ratio, r = 0.6

S4
= a (1 - rⁿ) /(1 - r)
= 6 (1 - 0.6⁴) /(1 - 0.6)
= 13.056 m

The total vertical distance travelled
= (10 + 2 x S4) m
= (10 + 2 x 13.056) m
= 36.112 m