唔該 !!數學一問 ??

設首行為A,相差值為d,行數為n
第n行有多少件的通式P(n)=A+(n-1)d
玩具n行之和的通式F(n)=n(2A+(n-1)d)/2
a)
P(1)=96
A+(1-1)d=96
A=96
P(6)=61
96+(6-1)d=61
5d=-35
d=-7
每行相差7件玩具

b)
P(n)=96+(n-1)(-7)
當96+(n-1)(-7)=5
(n-1)(-7)=-91
n-1=13
n=14
整個設計共有14行

c)
F(n)=n(2A+(n-1)d)/2
F(14)
=14(2×96+(14-1)(-7))/2
=14(192-91)/2
=14(101)/2
=707
整個設計共擺放玩具707件

d)
F(n)=n(2A+(n-1)d)/2
F(4)
=4(2×96+(4-1)(-7))/2
=4(192-21)/2
=342
共需移342件玩具展品

e)
百分比
=(342/707)×100%
=48.37%
餘下的玩具展品數量較原來的相差了48.37個百分比

Let a be the no. of toys in the first row, d be the common difference between adjacent rows and n be the no. of rows.

The the no, of toys in the nth row R(n) = a - (n-1)d and
the total no. of toys in n rows S(n) = n(R(1) + R(n))/2

a. R(1) = a - (1-1)d = 96 so a = 96
R(6) = a - (6-1)d = 61 so 96 - 5d = 61 gives d = 7
Therefore the common difference is 7
b. let L be the last row, then R(L) = 96 - (L - 1) x 7 = 5 gives L = (96 - 5)/7 -1 = 12
There are 12 rows in the design.
c. S(12) = 12(96 + 96 - (12 - 1) x 7)/2 = 690
There are 690 toys in the whole design.
d. S(4) = 4(96 + 96 - (4 - 1) x 7)/2 = 342
The no. of toys removed is 342.
e. The difference in percentage of the remaining with the original design = % of toys removed = 342/690 = 49.56%

2007-12-15 21:37:45 補充：
Sorry there are errors in the calculation of b. c. and e.b. L = (96-5)/7 1 = 14 There are 14 rows in the design.c. S(14) = 14(96 5)/2 = 707 There are 707 pieces of toys in the whole design.e Percentage of toys removed = 342/707 = 48.37%