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Re: Maths made easy =) stuck sumwhere?Ask ur doubts here!!
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Mathematics: Post your doubts here!
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Re: Maths made easy =) stuck sumwhere?Ask ur doubts here!!
any question ??
any question ??
Re: Maths made easy =) stuck sumwhere?Ask ur doubts here!!
http://www.xtremepapers.me/CIE/index.ph ... _qp_41.pdf
Qns. 9 (a) (iv) to (vi) and also (b) (iii)... seems like I'm not getting the question :/
http://www.xtremepapers.me/CIE/index.ph ... _qp_41.pdf
Qns. 9 (a) (iv) to (vi) and also (b) (iii)... seems like I'm not getting the question :/
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Re: Maths made easy =) stuck sumwhere?Ask ur doubts here!!
(v) Simply insert n = 150 into the formula given in (ii) = 1/2 x 150 x 151 = 11, 325.
(vi) 11, 325 - 3825 = 7500.
In (b) (iii) again you have to insert n = 20 into the formula given at the start of the part.
Hope this helped.
Q9. (a) (iv) Notice how the terms in this sequence( 3, 6, 9, 12,......150) are mutiples (multiplied by 3)of the sequence 1, 2, 3, 4,......, 50. So multiply your answer of the previous part ( 1275) by 3 to get the answer 3825.Abid357 said:
(v) Simply insert n = 150 into the formula given in (ii) = 1/2 x 150 x 151 = 11, 325.
(vi) 11, 325 - 3825 = 7500.
In (b) (iii) again you have to insert n = 20 into the formula given at the start of the part.
Hope this helped.
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Re: Maths made easy =) stuck sumwhere?Ask ur doubts here!!
assalamoalaikum!!
i was abt to answer it..but anyways...see i said right that sequences are really easy!
assalamoalaikum!!
i was abt to answer it..but anyways...see i said right that sequences are really easy!
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Re: Maths made easy =) stuck sumwhere?Ask ur doubts here!!
Read the question very carefully again! If still stuck, I'll help.Abid357 said:Thanks a lot abcde
Ehhem.. sequences aren't 'that' easy sometimes for me :/. I'm still confused with the part (a) (vi)...
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Re: Maths made easy =) stuck sumwhere?Ask ur doubts here!!
I hope it does. Do lemme know please.
In (v), we found the sum of all the numbers up to 150. Agreed? In part (iv), we found the sum of multiples of 3 up to 150. Right, na? So when they ask us to find the sum of the numbers less than 150 which are not multiples of 3, we simply subtract (iv) from (v). Come on, this makes sense Abid, no?Abid357 said:
I hope it does. Do lemme know please.
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Re: Maths made easy =) stuck sumwhere?Ask ur doubts here!!
There is an easy way to find the formula for the second difference. Perhaps what I am going to tell you isn't mention anywhere in the textbooks so note it down and keep it safe for future generations!
As for finding the formula for the sequence like this :- 2,4,7,11 etc....
It is this :- 0.5(n^2)-0.5n+1
(2nd difference of the sequence/2)x(n^2)-nth formula
First you will take (1/2)x(n^2) and then find the nth formula . For finding the nth formula first do this rough working.
The first term number is 1 and how is that? Since 2,4 has a first difference of 2 and 4,7 has a difference of 3 . The second difference of 2 and 3 is 1. That is why I took (1/2) in the formula for the second difference. If you subtract 1 from 2 , you will get the difference of 1st term and 2nd term which will give you the first term of this sequence which is 1. It is important that the first difference should be one before you start the working because these are second difference sequences. In the first difference type of sequences, they can start with any number. But in these you have to be careful! So whenever you are doing any questions of second difference , always see that any term you have has a first difference of 1!
Having explained that, I will now tell you how to find the nth formula. We know that Nth formula is An= a+(n-1)d
'a' is the first difference of the term. 'd' is the difference of the 2nd and the 1st term.
The working for this is nth term number - (2nd a/2)(nth term)^2 = whatever the answer is.
e.g 1) 1-(0.5)(1)^2 = 0.5
2) 2-(0.5)(2)^2= 0
3) 4-(0.5)(3)^2 = -1/2
d= 2nd term - 1st term or 3rd term - 2nd term
d= (0 - 0.5) = -0.5 , a= 1/2
An = a+(n-1)(d)
An= 0.5+(n-1)(-0.5)
An= 1/2+1/2 - 0.5n
An= -1/2+1
Now you have found the nth formula. Add this into the one that I found earlier.
It will be 0.5(n^2)-0.5n+1
The sequence is 1,2,4,7,11,16
Happy day everyone!
There is an easy way to find the formula for the second difference. Perhaps what I am going to tell you isn't mention anywhere in the textbooks so note it down and keep it safe for future generations!
As for finding the formula for the sequence like this :- 2,4,7,11 etc....
It is this :- 0.5(n^2)-0.5n+1
(2nd difference of the sequence/2)x(n^2)-nth formula
First you will take (1/2)x(n^2) and then find the nth formula . For finding the nth formula first do this rough working.
The first term number is 1 and how is that? Since 2,4 has a first difference of 2 and 4,7 has a difference of 3 . The second difference of 2 and 3 is 1. That is why I took (1/2) in the formula for the second difference. If you subtract 1 from 2 , you will get the difference of 1st term and 2nd term which will give you the first term of this sequence which is 1. It is important that the first difference should be one before you start the working because these are second difference sequences. In the first difference type of sequences, they can start with any number. But in these you have to be careful! So whenever you are doing any questions of second difference , always see that any term you have has a first difference of 1!
Having explained that, I will now tell you how to find the nth formula. We know that Nth formula is An= a+(n-1)d
'a' is the first difference of the term. 'd' is the difference of the 2nd and the 1st term.
The working for this is nth term number - (2nd a/2)(nth term)^2 = whatever the answer is.
e.g 1) 1-(0.5)(1)^2 = 0.5
2) 2-(0.5)(2)^2= 0
3) 4-(0.5)(3)^2 = -1/2
d= 2nd term - 1st term or 3rd term - 2nd term
d= (0 - 0.5) = -0.5 , a= 1/2
An = a+(n-1)(d)
An= 0.5+(n-1)(-0.5)
An= 1/2+1/2 - 0.5n
An= -1/2
+1Now you have found the nth formula. Add this into the one that I found earlier.
It will be 0.5(n^2)-0.5n+1
The sequence is 1,2,4,7,11,16
Happy day everyone!
Re: Maths made easy =) stuck sumwhere?Ask ur doubts here!!
They asked us to find the sum of those numbers in the two parts which are not multiples of 3. Since part (iv) answer is the sum of all the multiples of 3 upto 150, we just subtract that from part (v) answer, taking or cancelling out all the multiples of 3... cool, I finally did it : P
Ooooo I get it now ! After breaking my head for 15 minutes, phew... thanks so much pal, sorry to bother youabcde said:In (v), we found the sum of all the numbers up to 150. Agreed? In part (iv), we found the sum of multiples of 3 up to 150. Right, na? So when they ask us to find the sum of the numbers less than 150 which are not multiples of 3, we simply subtract (iv) from (v). Come on, this makes sense Abid, no?
I hope it does. Do lemme know please.
They asked us to find the sum of those numbers in the two parts which are not multiples of 3. Since part (iv) answer is the sum of all the multiples of 3 upto 150, we just subtract that from part (v) answer, taking or cancelling out all the multiples of 3... cool, I finally did it : P
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Re: Maths made easy =) stuck sumwhere?Ask ur doubts here!!
Yes, you did it! You didn't bother me. I was and will be glad to help.
Yes, you did it!
You didn't bother me. I was and will be glad to help. Re: Maths made easy =) stuck sumwhere?Ask ur doubts here!!
Linear sequences
Eg: 2, 4, 6, 8, 10...
The difference in this case is 2 between every number.
-2--4--6--8--10-
--\_/\_/\_/\_/--
...(2)(2)(2)(2)... --> (differences)
Formula: a + d(n - 1)
where,
a = 1st term
d = difference
Therefore: a = 2 and d = 2 in this case,
a + d(n - 1)
= 2 + 2(n - 1)
= 2 + 2n - 2
= 2n
Check: if n = 3, then...
2(3)
= 6
Quadratic sequences
Eg: 1, 4, 8, 13, 19...
The difference in this case varies between every number.
1--4---8--13--19
_\_/ \_/ \_/ \_/_
..(3).(4).(5).(6).. --> (1st difference)
___\_/\_/\_/___
.....(1)(1)(1)..... --> (2nd difference)
Formula: a + d(n - 1) + ½(n - 1)(n - 2)C
where,
a = 1st term
d = 1st difference between the first two numbers
C = 2nd difference
Therefore: a = 1, d = 3 and C = 1 in this case,
a + d(n - 1) + ½(n - 1)(n - 2)C
= 1 + 3(n - 1) + ½(n - 1)(n - 2) x 1------> (replace letters by their values)
= 1 + 3n - 3 + ½(n^2 - 3n + 2) ----------> (simplify)
= (3n - 2) + ½n^2 - ½ x 3n + ½ x 2 ----> (simplify)
= ½n^2 + (3n - ½ x 3n) + (-2 + 1) ------> (collect like terms together)
= ½n^2 + ½ x 3n - 1 --------------------> (factorize)
= ½n(n + 3) - 1
Check: if n = 3, then...
½ x 3(3 + 3) - 1
= ½ x 3(6) - 1
= ½ x 18 - 1
= 9 - 1
= 8
Many thanks for the information, nightrider95. It's true that this formula isn't usually found in our text books. I would like to explain the same thing in a different waynightrider95 said:There is an easy way to find the formula for the second difference. Perhaps what I am going to tell you isn't mention anywhere in the textbooks so note it down and keep it safe for future generations!
As for finding the formula for the sequence like this :- 2,4,7,11 etc....
It is this :- 0.5(n^2)-0.5n+1
(2nd difference of the sequence/2)x(n^2)-nth formula
First you will take (1/2)x(n^2) and then find the nth formula . For finding the nth formula first do this rough working.
The first term number is 1 and how is that? Since 2,4 has a first difference of 2 and 4,7 has a difference of 3 . The second difference of 2 and 3 is 1. That is why I took (1/2) in the formula for the second difference. If you subtract 1 from 2 , you will get the difference of 1st term and 2nd term which will give you the first term of this sequence which is 1. It is important that the first difference should be one before you start the working because these are second difference sequences. In the first difference type of sequences, they can start with any number. But in these you have to be careful! So whenever you are doing any questions of second difference , always see that any term you have has a first difference of 1!
Having explained that, I will now tell you how to find the nth formula. We know that Nth formula is An= a+(n-1)d
'a' is the first difference of the term. 'd' is the difference of the 2nd and the 1st term.
The working for this is nth term number - (2nd a/2)(nth term)^2 = whatever the answer is.
e.g 1) 1-(0.5)(1)^2 = 0.5
2) 2-(0.5)(2)^2= 0
3) 4-(0.5)(3)^2 = -1/2
d= 2nd term - 1st term or 3rd term - 2nd term
d= (0 - 0.5) = -0.5 , a= 1/2
An = a+(n-1)(d)
An= 0.5+(n-1)(-0.5)
An= 1/2+1/2 - 0.5n
An= -1/2
Now you have found the nth formula. Add this into the one that I found earlier.
It will be 0.5(n^2)-0.5n+1
The sequence is 1,2,4,7,11,16
Happy day everyone!
Linear sequences
Eg: 2, 4, 6, 8, 10...
The difference in this case is 2 between every number.
-2--4--6--8--10-
--\_/\_/\_/\_/--
...(2)(2)(2)(2)... --> (differences)
Formula: a + d(n - 1)
where,
a = 1st term
d = difference
Therefore: a = 2 and d = 2 in this case,
a + d(n - 1)
= 2 + 2(n - 1)
= 2 + 2n - 2
= 2n
Check: if n = 3, then...
2(3)
= 6
Quadratic sequences
Eg: 1, 4, 8, 13, 19...
The difference in this case varies between every number.
1--4---8--13--19
_\_/ \_/ \_/ \_/_
..(3).(4).(5).(6).. --> (1st difference)
___\_/\_/\_/___
.....(1)(1)(1)..... --> (2nd difference)
Formula: a + d(n - 1) + ½(n - 1)(n - 2)C
where,
a = 1st term
d = 1st difference between the first two numbers
C = 2nd difference
Therefore: a = 1, d = 3 and C = 1 in this case,
a + d(n - 1) + ½(n - 1)(n - 2)C
= 1 + 3(n - 1) + ½(n - 1)(n - 2) x 1------> (replace letters by their values)
= 1 + 3n - 3 + ½(n^2 - 3n + 2) ----------> (simplify)
= (3n - 2) + ½n^2 - ½ x 3n + ½ x 2 ----> (simplify)
= ½n^2 + (3n - ½ x 3n) + (-2 + 1) ------> (collect like terms together)
= ½n^2 + ½ x 3n - 1 --------------------> (factorize)
= ½n(n + 3) - 1
Check: if n = 3, then...
½ x 3(3 + 3) - 1
= ½ x 3(6) - 1
= ½ x 18 - 1
= 9 - 1
= 8
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Re: Maths made easy =) stuck sumwhere?Ask ur doubts here!!
Any Hint for 2011 MAy\JUne
What all topics to be revised???????????
Any Hint for 2011 MAy\JUne
What all topics to be revised???????????
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Re: Maths made easy =) stuck sumwhere?Ask ur doubts here!!
http://www.xtremepapers.me/CIE/Cambridg ... 9_qp_4.pdf
mj09 p4
q4 bii
thanks
69 is the asnwer in the marking sheme cao
http://www.xtremepapers.me/CIE/Cambridg ... 9_qp_4.pdf
mj09 p4
q4 bii
thanks
69 is the asnwer in the marking sheme cao
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Re: Maths made easy =) stuck sumwhere?Ask ur doubts here!!
N1AP = (280 - 180) + 23 = 123
N1AB = 360 - 123 - 126 = 111
111 + N2BA = 180 (interior angles)
=> N2BA = 69.
and thus the bearing of A from B is 069*.
Let's call the Norths over A, B and P N1, N2 and N3, respectively.student92 said:
N1AP = (280 - 180) + 23 = 123
N1AB = 360 - 123 - 126 = 111
111 + N2BA = 180 (interior angles)
=> N2BA = 69.
and thus the bearing of A from B is 069*.
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Re: Maths made easy =) stuck sumwhere?Ask ur doubts here!!
Or even simpler: we know BPS3 is 280 - 180 = 100.
By alternate angles, N2BP = 100.
ABP = 180 - 23 - 126 = 31* (angle sum of triangle)
So N2BA = 100 - 31 = 69.
Thus the bearing is 069*.
Hope you understand.
Or even simpler: we know BPS3 is 280 - 180 = 100.
By alternate angles, N2BP = 100.
ABP = 180 - 23 - 126 = 31* (angle sum of triangle)
So N2BA = 100 - 31 = 69.
Thus the bearing is 069*.
Hope you understand.
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Re: Maths made easy =) stuck sumwhere?Ask ur doubts here!!
http://www.xtremepapers.me/CIE/Cambridg ... 9_qp_4.pdf
mj09
q2b
http://www.xtremepapers.me/CIE/Cambridg ... 9_qp_4.pdf
mj09
q2b