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Mathematics: Post your doubts here!

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I just completed IGCSE and took "C" in mathematics. I have opted for the same subject in A'levels and I am not confident at all. Can I practically think of achieving an A grade in A'levels. I am really weak.
 
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I just completed IGCSE and took "C" in mathematics. I have opted for the same subject in A'levels and I am not confident at all. Can I practically think of achieving an A grade in A'levels. I am really weak.
grade C in mathematics IGCSE, which is basic stuff means u will hardly achieve a D or an E in mathematics A levels. can u build a building if u didn't build the base properly ? that's the same as ur situation...
 
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(ii)
Sketch an Argand diagram showing the locus of the complex number s such that |s − u| = |u|
.
[3]
show me how the argand diagram looks!!
smzimran plzzzzzzzzz
 
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T3=20
ar2=20
a=20/r2


sum to infinity=3a
a/(1-r)=3a
a=3a(1-r)
20/r2=3(20/r2)(1-r)
20/r2=60/r2(1-r)
60-60r=20
-60r=20-60
-60r=-40
r=2/3

a=20/(2/3)2
=45

T8=3T3
a+7d=3(a+2d)
45+7d=3(45)+6d
7d-6d=135-45
d=90

s8=4s4
s8=8/2[2(45)+7(90)]
=2880

s4=4/2[2(45)+3(90)]
=720
s8=4s4
2880=4(720)
2880=2880
(hence shown)
thanx alot!!But how did u know that the first term of geometric progression is also the first term of an arthimetic?
 
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Someone please help me :
The second term of a geometric progression is 3 and the sum to infinity is 12.
Find the first term of the progression.
[Quoted from June 07 p1]
Second term of G.P = ar = 3 ----- (1)
Sum to infinity = a / (1 - r) = 12
=> a = 12 - 12r

From (1) r = 3/a

Therefore equating the 2 equations
a = 12 - 12(3/a)
a = (12a - 36)/a
a ^2 = (12a - 36)
a^2 - 12a + 36 = 0
(a - 6)(a - 6) = 0
a = 6
 
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Second term of G.P = ar = 3 ----- (1)
Sum to infinity = a / (1 - r) = 12
=> a = 12 - 12r

From (1) r = 3/a

Therefore equating the 2 equations
a = 12 - 12(3/a)
a = (12a - 36)/a
a ^2 = (12a - 36)
a^2 - 12a + 36 = 0
(a - 6)(a - 6) = 0
a = 6

Thank you very much mate :)
 
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