Guess the next number in line!

• wrexham

  21 Jun 07, 17:12

  408, 571, 332, 377, 577, 989, 664, ___
• RiverFoX

  21 Jun 07, 17:15

  583?
• Darkness_hacker99

  21 Jun 07, 17:23

  Time to reveal the answer liao..
• ^tamago^

  21 Jun 07, 17:24

  oh my goodness, the same question is posted elsewhere too!
• RiverFoX

  21 Jun 07, 17:25

  huh???
  
  1guess nia review ans?
  
  like dat abit cheapo rite the qns?
• wrexham

  21 Jun 07, 17:29

  hahaha. just an intellectual question :O
• Darkness_hacker99

  21 Jun 07, 17:30

  Originally posted by wrexham:

  hahaha. just an intellectual question :O

  
• RiverFoX

  21 Jun 07, 17:34

  
  
  dubyew tee aff?!?
• ^tamago^

  21 Jun 07, 17:38

  843
• ^tamago^

  21 Jun 07, 17:45

  377,408,571,332,377,577,989,664,___
  
  Let T(n) be the nth term in this sequence.
  Let T(2) = 408
  
  ==> T(n+4) = T(n) × sqrt(n)
  
  e.g.
  T(2) = 408, then T(6) = 408 × sqrt(2) = 577
  T(3) = 571, then T(7) = 571 × sqrt(3) = 989
  T(4) = 332, then T(8 ) = 332 × 2 = 664
  T(5) = 377, then T(9) = 377 × sqrt(5) = 843
• Darkness_hacker99

  21 Jun 07, 17:47

  
• RiverFoX

  21 Jun 07, 17:48

  wtf?!?!
  
  in english pls tamago
• ^tamago^

  21 Jun 07, 17:49

  ok, bo bian, see this qn twice i try to crack my brains a bit. so long never do maths.
• dbowie

  21 Jun 07, 17:50

  Originally posted by ^tamago^:

  377,408,571,332,377,577,989,664,___
  
  Let T(n) be the nth term in this sequence.
  Let T(2) = 408
  
  ==> T(n+4) = T(n) × sqrt(n)
  
  e.g.
  T(2) = 408, then T(6) = 408 × sqrt(2) = 577
  T(3) = 571, then T(7) = 571 × sqrt(3) = 989
  T(4) = 332, then T(8 ) = 332 × 2 = 664
  T(5) = 377, then T(9) = 377 × sqrt(5) = 843

  T(n+4) = T(n) × sqrt(n)
  Explain please
• ^tamago^

  21 Jun 07, 17:50

  Originally posted by RiverFoX:

  wtf?!?!
  
  in english pls tamago

  if u understand sequences, u shld understand this
• ^tamago^

  21 Jun 07, 17:50

  Originally posted by dbowie:

  T(n+4) = T(n) × sqrt(n)
  Explain please

  T(6) = T(2) × sqrt(2)
  
  see my examples.....
• dbowie

  21 Jun 07, 17:53

  I understand already.Ty
• dbowie

  21 Jun 07, 17:55

  Originally posted by ^tamago^:

  377,408,571,332,377,577,989,664,___
  
  Let T(n) be the nth term in this sequence.
  Let T(2) = 408
  
  ==> T(n+4) = T(n) × sqrt(n)
  
  e.g.
  T(2) = 408, then T(6) = 408 × sqrt(2) = 577
  T(3) = 571, then T(7) = 571 × sqrt(3) = 989
  T(4) = 332, then T(8 ) = 332 × 2 = 664
  T(5) = 377, then T(9) = 377 × sqrt(5) = 843

  What you mean is
  4 numbers after T(n) = T(n) x square.root (n)
  To find out T(6) we have to find out T(2) number
  T(2) =408
  408(=T2) x squareroot 2= 577
  
  So take 377 x square.root 5 = answer.
  
  
• ^tamago^

  21 Jun 07, 18:11

  Originally posted by dbowie:

  What you mean is
  4 numbers after T(n) = T(n) x square.root (n)
  To find out T(6) we have to find out T(2) number
  T(2) =408
  408(=T2) x squareroot 2= 577
  
  So take 377 x square.root 5 = answer.
  
  

  yeah. i dunno how ur logic flow runs but it sounds like u're about there.
• dbowie

  21 Jun 07, 18:13

  Originally posted by ^tamago^:

  yeah. i dunno how ur logic flow runs but it sounds like u're about there.

  I am not about there,I am there !
• ^tamago^

  21 Jun 07, 18:29

  Originally posted by dbowie:
  
  I am not about there,I am [b]there ! [/b]

  
• Unidentified

  21 Jun 07, 18:32

  Awww.....
  
  i hate algebra
• eviljustice

  21 Jun 07, 18:34

  
• hegu

  21 Jun 07, 18:56

  Originally posted by ^tamago^:

  377,408,571,332,377,577,989,664,___
  
  Let T(n) be the nth term in this sequence.
  Let T(2) = 408
  
  ==> T(n+4) = T(n) × sqrt(n)
  
  e.g.
  T(2) = 408, then T(6) = 408 × sqrt(2) = 577
  T(3) = 571, then T(7) = 571 × sqrt(3) = 989
  T(4) = 332, then T(8 ) = 332 × 2 = 664
  T(5) = 377, then T(9) = 377 × sqrt(5) = 843

  err...looks flawed somewhere
  
  this doesnt justify why 571 comes after 408 and why 332 comes after 571
• dbowie

  21 Jun 07, 18:58

  Because no number given before Term 1 la,