【纪念当年的帖子（2010）】Add Maths功课讨论区

永遠愛著許瑋倫

怎么我就是不会呢
1)If the number sequence 2,p,q,25/2,r is an arithmetic progression,find hte value of p,q,r
2)In an A.P,the sum of the first three terms is 21 and the sum of the following three terms is 48 . Find the first team and the common difference

乙劍真人

怎么我就是不会呢
If the number sequence 2,p,q,25/2,r is an arithmetic progression,find the value of p,q,r
永遠愛著許瑋倫 发表于 4-2-2010 10:19 PM

不要哭乖..

AP: d 一样, 利用 T₂ - T₁ = T₃ - T₂ 这个条件：

i.e. p - 2 = q - p ---(1)
q - p = 25/2 - q ---(2)
25/2 - q = r - 25/2 ---(3)

From (1), q = 2p - 2 ---(4)
From (2), p = 2q - 25/2 ---(5)
From (3), r - 25 - q ---(6)

Solve (4), (5) & (6) simultaneously, p = 11/2, q = 9, r = 16

乙劍真人

2)In an A.P,the sum of the first three terms is 21 and the sum of the following three terms is 48 .
Find the first term and the commondifference.

S₃ = 21 ---(1)
S₆ - S₃ = 48 ---(2)

Sub (1) into (2), S₆ = 48 + 21
                            = 69 ---(3）

From (1), 3/2 [2a + 2d] = 21
                     2a + 2d = 14
                         a + d = 7 ---(4)

From (3), 6/2 [2a + 5d] = 69
                     2a + 5d = 23 ---(5)

Solve (4) & (5) simultaneously, a = 4, d = 3

永遠愛著許瑋倫

The sum of the first n terms of an Ap.P is given by 1/8(32n-n^2).Calculate the common difference

乙劍真人

The sum of the first n terms of an Ap.P is given by 1/8(32n-n^2).Calculate the common difference
永遠愛著許瑋倫 发表于 4-2-2010 11:23 PM

Sn = 1/8 (32n - n^2) ---(*)

From (*), S₁ = 1/8 [32(1) - 1]
                  = 31/8 ---(a)

From (*), S₂ = 1/8 [32(2) - 4]
                  = 15/2 ---(b)

From (a), S₁ = T₁ = 31/8
From (b), S₂ = T₁ + T₂ = 15/2
                             T₂ = 15/2 - 31/8
                                  = 29/8

d = T₂ - T₁
   = 29/8 - 31/8
   = -1/4

walrein_lim88

The sum of the first n terms of an Ap.P is given by 1/8(32n-n^2).Calculate the common difference
永遠愛著許瑋倫 发表于 4-2-2010 11:23 PM

   

walrein_lim88

Sn = 1/8 (32n - n^2) ---(*)

From (*), S = 1/8 [32(1) - 1]
                  = 31/8 ---(a)

...
乙劍真人 发表于 4-2-2010 11:37 PM

    原来你算了。。哈哈。。PAISEH

乙劍真人

原来你算了。。哈哈。。PAISEH
walrein_lim88 发表于 4-2-2010 11:38 PM

你来了就交给你啦！我要会周公了..有劳..

永遠愛著許瑋倫

Given p,p+4,2p+2 are three consecutive terms of a G.P.
a.Find the value of p such that all the terms are positive.
b.if p is second term,find the seventh term.

a的答案是8
b的我不会..不怎么明白

永遠愛著許瑋倫

Find the smallest term of the G.P   3,5,25/3,... that exceeds 100.

walrein_lim88

Given p,p+4,2p+2 are three consecutive terms of a G.P.
a.Find the value of p such that all the term ...
永遠愛著許瑋倫 发表于 5-2-2010 10:23 PM

walrein_lim88

Find the smallest term of the G.P   3,5,25/3,... that exceeds 100.
永遠愛著許瑋倫 发表于 5-2-2010 10:41 PM

   

superliong

请看看这题 differentiation 的
Find the equation of the tangent to the curve y=(2+x)/(3-2x) at the intersection point of the curve with the straight line y=3

请问intersection parallel perpendicular 的共同点是什么？{:2_80:}这三个好像都一直出现在题目里！

数学神

请看看这题 differentiation 的 Find the equation of the tangent to the curve y=(2+x)/(3-2x) at the intersection point of the curve with the straight line y=3 请问intersection parallel perpendicular 的共同点是什么？{:2_80:}这三个好像都一直出现在题目里！

superliong 发表于 9-2-2010 15:33

y=(2+x)/(3-2x)

Let y=3
3=(2+x)/(3-2x)
3(3-2x)=2+x
9-6x=2+x
7x=7
x=1

so, now we know that the point intersect is (1,3)

y=(2+x)/(3-2x)
Let...
u=2+x
du/dx=2

v=3-2x
dv/dx=-2

dy/dx=[v(du/dx)-u(dv/dx)]/(v^2)
dy/dx=[(3-2x)(2)-(2+x)(-2)]/(3-2x)^2
dy/dx=[(6-4x)-(-4-2x)]/(3-2x)^2
dy/dx=(6-4x+4+2x)/(3-2x)^2
dy/dx=(10-2x)/(3-2x)^2

Let x=1
dy/dx=[10-2(1)]/[3-2(1)]^2
dy/dx=(10-2)/(3-2)^2
dy/dx=8/(1)^2
dy/dx=8 ...... gradient=8

y-y₁=m(x-x₁)
y-3=8(x-1)
y=8x-8+3
y=8x-5

答案对吗？
不好意思，因为现在不方便，所以不能像walrein那样show出来给你看
希望你看得懂

superliong

回复 94# 数学神

    谢谢 ， 但
    答案是 y=7x-4   {:2_71:}

数学神

请问intersection parallel perpendicular 的共同点是什么？{:2_80:}这三个好像都一直出现在题目里！
superliong 发表于 9-2-2010 15:33

intersection 是两条线交叉点
parallel 是两条线平行，永远都不会touch together
perpendicular 是两条线90度连接在一起，就是说两条线连接在一起的角度是90度

又是那句，不好意思我现在不方便
未能画出来show给你看，只能字句解释
希望你看得懂

数学神

回复  数学神

    谢谢 ， 但
    答案是 y=7x-4
superliong 发表于 9-2-2010 17:39

OMG...错掉><
方法应该是酱
你尝试检查看我有没有做错-.-''

superliong

根据你的方法重算了，得到了y = 7x-4
可能是你的减忘了吧～
请看看这题
A curve has the equation y=(2x-3)^2
Find the equation of the normal to the curve that is perpendicular to the straight line 4x-y-4=0.

数学神

根据你的方法重算了，得到了y = 7x-4 可能是你的减忘了吧～ 请看看这题 A curve has the equation y=(2x-3)^2 Find the equation of the normal to the curve that is perpendicular to the straight line 4x-y-4=0.

superliong 发表于 9-2-2010 17:49

-.-''
粗心？><

这题你还是先提供答案吧
做错我不post

superliong

ok
答案是 4y=-x+6

还有这题
A curve has the equation y=(x+1)^3 . Find equations of the two tangents to the curve that is parallel to the straight line 3x-y+3 .
这题答案 ： y=3x+1
                y=3x+5