MATHS~~~

Let the 3 numbers be n , n + d and n + 2d respectively.

As given,

n + n + d + n + 2d = 30

3n + 3d = 30

n + d = 10 --- ( 1 )

n / ( n + 2d ) = 3 / 7

7n = 3n + 6d

4n = 6d

n = 3d / 2 --- ( 2 )

Put ( 2 ) into ( 1 ).

3d / 2 + d = 10

d = 4

n = 3 ( 4 ) / 2 = 6

Then,

n + d = 6 + 4 = 10

n + 2d = 6 + 8 = 14

Hence the 6, 10 and 14 respectively.


2007-09-12 22:38:06 補充：
Reminder, for an arithmetic sequence, T ( n ) = a + ( n - 1 )d Then suppose the 3 consecutive terms is the first three terms, then T ( 1 ) = a , T ( 2 ) = a + d and T ( 3 ) = a + 2d. That's comes the n, n + d and n + 2d in the above.

2007-09-16 01:06:28 補充：
It is better to let the numbers be a, a + d and a + 2d to avoid confusion, sorry for that.

a+(a+R)+(a+2R) = 30 --->1

a/(a+2R)=3/7 --->2

from 1,

3a+3R = 30
a+R = 10
R=10-a ---> 3

Sub 3 into 2

a/[a+2(10-a)] = 3/7
7a = 3a+6(10-a)
7a = 3a+60-6a
10a = 60
a=6

Sub a=6 into 3

R = 10-6
R=4

So, the 3 # are 6, 10 and 14