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Maths IGCSE exam tomorrow!-PLEASE HELP

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My IGCSE maths exam is tomorrow and I noticed that hard sequences questions now show up at the end of some of the recent past papers (the 2011 ones)
I don't really understand them, especially the parts where you're asked to work out the nth term.
I've attached the papers, the questions are at the very end of both exams.
Please explain how you got the answer
Any help would be appreciated!
 

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  • 0580_w11_qp_21.pdf
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  • 0580_w11_qp_43.pdf
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for 0580_w11_qp_21.pdf For qn 18 a replace n with 23 u get 14.................18 b.....................U1= 5 (T2) - 1 (T1) You get 4...............................U2 = 14(T3) - 5(T2) you get 9..........................
 
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Square root 32 is egual to 32^1/2...........................But 32 is = to 2^5....................Therefore (2^5)^1/2...........law of indices..............therefore..............2^5x1/2.................Therefore............P =5/2

Qn2)

3 square root (1/8) is = to........(1/8)^1/3..................which is equal to (8^-1)^1/3.............therefore (8)^-1/3......................But 8= 2^3................Therefore......................(2^3)^-1/3............law of indices.......................Therefore..........(2)^3 x -1/3................q = -1
 
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Square root 32 is egual to 32^1/2...........................But 32 is = to 2^5....................Therefore (2^5)^1/2...........law of indices..............therefore..............2^5x1/2.................Therefore............P =5/2

Qn2)

3 square root (1/8) is = to........(1/8)^1/3..................which is equal to (8^-1)^1/3.............therefore (8)^-1/3......................But 8= 2^3................Therefore......................(2^3)^-1/3............law of indices.......................Therefore..........(2)^3 x -1/3................q = -1
Bro i cant understand can u upload a pic !
This question part (B) Find Y
 

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My IGCSE maths exam is tomorrow and I noticed that hard sequences questions now show up at the end of some of the recent past papers (the 2011 ones)
I don't really understand them, especially the parts where you're asked to work out the nth term.
I've attached the papers, the questions are at the very end of both exams.
Please explain how you got the answer
Any help would be appreciated!

18
a)

Tn = 1/6 n(n + 1)(2n + 1)

T23 = 1/6 * 23 (23+1)(46+1) = 4324

b)

i)

U1 = T2 – T1 U2 = T3 – T2 U3 = T4 – T3 …….

U1 = 12 + 22 – 12 = 4
U2 = 12 + 22 + 32 – (12 + 22) = 32 = 9
ii) 4 , 9 , …. Then Un = (n+1) 2

C)
V1 = 22 V2 = 22 + 42 V3 = 22 + 42 + 62 V4 = 22 + 42 + 62 + 82.
Recognizing Vn = 4Tn

Vn = 2/3 * (n)(n + 1)(2n+1)
 
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