maths

Assume the three numbers are in the ratio 1:5:11.

Let the three numbers be k, 5k and 11k, where k is not equal to 0.

As they are in geometic sequence,

(5k+3)/(k+3)=(11k+3)/(5k+3)
(5k+3)(5k+3)=(11k+3)(k+3)
25k^2+30K+9=11k^2+36k+9
14k^2-6k=0
2k(7k-3)=0
k=0 (rej.) or k=3/7

So, the three numbers are 3/7, 15/7 and 33/7

ratio = 1:5:11

Let
T1=k+3
T2 = 5k+3
T3 = 11k+3

T2 / T1 = T3 / T2
(5k+3) / (k+3) = (11k+3) / (5k+3)
(5k+3)^2 = (11k+3) (k+3)
25k^2 +30k +9 = 11k^2 +33k +3k +9
25k^2 +30k +9 = 11k^2 +36k +9
14k^2 -6k = 0
2k(7k-3) = 0
2k = 0 or 7k-3 = 0
k = 0(REJ.) or k = 3/7

T1 = 3/7

T2
= 5(3/7)+3
= 15/7 +3
= 15/7+21/7
= 36/7
= 5又7分之1

T3
= 11(3/7)+3
= 33/7 +3
= 54/7
= 7又７分之五

2007-10-27 22:39:20 補充：
Original no. are : 3/7 , 15/7 , 33/7