H2 Math Question

• LastRide7

  14 Apr 11, 21:13

  A geometric series has first term 1 and the common ratio r is positive. The sum of the first 5 terms is twice the sum of the terms from the 6th to the 15th exclusive. Prove that r^5 = 0.5(sqrt(5) - 1)
  
  I arrived at the equation r^5 - 1 = 2r^15 - 2r^6
  but I am not sure what to do from here.

• Chemfreak022

  14 Apr 11, 23:01

  Series is 1, r , r^2, r^3...

  First 5 terms is 1 r r^2 r^3 r^4

  6th to 15 th inclusive is r^5, r^6....r^15

  S5 = a(1-r^n)/(1-r)  = (1-r^5)/(1-r)

  S(6 to 15 ) = r^5 + r^6 + ...+ r^15 = r^5 ( 1+ r +... r^9)

  Then Sum(6th to 15th term) = r^5(1-r^10)/(1-r)

  This is becuase its like (Sum of 1 + r+ ...10th term) x r^5

  Then (1-r^5)/(1-r) = r^5[(1-r^10)/(1-r)

  So (1-r^5) = r^5(1-r^10)

  Let y = r^5

  So (1-y)= y(1-y^2)

  (1-y) = y(1+y)(1-y)

  Then y(1+y) = 1...  y^2 + y -1 =0

  Then y = (-1(+/-)sqrt(1+4))/2

  r^5= 0.5(sqrt(5)-1)

• Chemfreak022

  14 Apr 11, 23:03

  Sorry should be until r^14

   

• wee_ws

  15 Apr 11, 09:30

  Dear LastRide7,

    The form to be shown suggests that one may need to use the quadratic formula to find the value of r^5.

    Analysing a question's requirement will help one greatly in problem-solving. Thanks.

  Cheers,
  Wen Shih

• frekiwang

  1 May 11, 18:06

  "Twice the sum", then the answer should not be be 0.5(sqrt(5)-1) anymore. Chemfreak02's working is for if the two sums are equal. sum of the first 5 terms=(1-r^5)/(1-r) sum of the first 15 terms=(1-r^15)/(1-r) S15=1.5*S5 1-r^15=1.5(1-r^5) Let y=r^5 1-y^3=1.5-1.5y 2y^3-3y+1=0 (y-1)(2y^2+2y-1)=0 y=1(rejected) or y=[-2+/-sqrt(12)]/4(reject -) thus r^5=[-1+sqrt(3)]/2