MATHS(等差及等比數列)

11. 十名男孩的高度組成一個等差數列。若最高的五名男孩之平均高度為 1.68 m ,
而最矮的三名男孩之平均高度為 1.32 m , 求餘下兩名男孩之平均高度
Sol
a1+a2+a3=396 =>a1+a1+d+a1+2d=a1+3d=396 =>a1+d=132
a1+7d=168
6d=36 =>d=6
Ans=(a4+a5)/2=(2a1+7d)/2=(2a1+2d+5d)/2=(264+30)/2=147cm

12. 若 log x , log y , log z 組成一個等差數列 , 以下哪項是正確的
2logy=logx+logz
y^2=xz
(1) x , y , z 組成一個等差數列=>error
(2) x , y , z 全都是整數=>donott sure
(3) x , y , z 組成一個等比數列=>right
選 (3)

13. 計算 1 + 3 + 3^2 + ... + 3^2003
S=1+3+3^2+....+3^2003
3S= 3+3^2+.....+3^2003+3^2004
2S=3^2004－1
S=(3^2004-1)/2

14. 等差數列 84 , 81 , 78 , ... , －45 , －48 共有多少項
d=81－84=－3
an=a1+(n－1)d
－48=84+(n－1)*(－3)
n－1=44
n=45

11) Let a = height of the shortest boy, d = common difference.
so a + (a + d) + (a + 2d) = 1.32 x 3 = 3.96
3a + 3d = 3.96
a + d = 1.32.................(1)
For tallest, (a + 5d) + (a + 6d) + (a + 7d) + (a + 8d) + (a + 9d) = 1.68 x 5
5a + 35d = 1.68 x 5
a + 7d = 1.68.............(2)
(2) - (1) we get 6d = 0.36, d = 0.06, so a = 1.32 - 0.06 = 1.26.
So height of the remaining 2 boys = (a + 3d) + (a + 4d) = 2a + 7d
= 1.26 x 2 + 0.06 x 7 = 2.52 + 0.42 = 2.94
so average = 2.94/2 = 1.47.
12)
log x + log z = 2 log y
xz = y^2
y/x = z/y
so x, y and z is a G.S. answer is (3)
13)
1 + 3 + 3^2 + 3^3 + .... + 3^2003
= 1 + 3( 1 + 3 + 3^2 + .... + 3^2002)
= 1 + 3(3^2003 - 1)/(3 - 1) = 1 + 3(3^2003 - 1)/2.
14)
a = 84, d = -3
so 84 + (n - 1)(-3) = -48
(n - 1)(-3) = - 132
n - 1 = 132/3 = 44
so n = 44 + 1 = 45.