IGCSE/GCSE/O & A Level/IB/University Student Forum

Qualification => Subject Doubts => GCE AS & A2 Level => Math => Topic started by: tmisterr on February 21, 2010, 01:23:04 pm

Title: P1 CIE question geometric sequences
Post by: tmisterr on February 21, 2010, 01:23:04 pm

A person wants to borro $100000 to buy a house. He intends to pay back a fixed sum of $C at the end of each year so that after 25 years he has completely paid off the debt. Assuming a steady interest rate of 4% per year expalin why:


100000=C(1/1.04+1/1.04²+1/1.04³+......1+1.04²⁵

Calculate the value of C

Title: Re: P1 CIE question geometric sequences
Post by: astarmathsandphysics on February 21, 2010, 02:27:58 pm

4 hours

Title: Re: P1 CIE question geometric sequences
Post by: astarmathsandphysics on February 21, 2010, 05:03:14 pm

100000=C(1/1.04+1/1.042+1/1.043+......1+1.04^25)
100000=C(a(1-r^n)/1-r))
100000=C(1(1-1/1.04^25)(1-1/1.04)=C*16.62
C=1662000

Title: Re: P1 CIE question geometric sequences
Post by: tmisterr on February 22, 2010, 02:30:20 pm

can you start from th beginning, the first part, findng C is not the problem, proving the formular is

Title: Re: P1 CIE question geometric sequences
Post by: astarmathsandphysics on February 22, 2010, 02:43:16 pm

Ok. I will do it when Iget home

Title: Re: P1 CIE question geometric sequences
Post by: astarmathsandphysics on February 23, 2010, 08:46:16 am

100000=C(1/1.04+1/1.042+1/1.043+......1+1.0425
multiply both sides by 1/1.04
100000/1.04=C(1/1.042+1/1.043+......1+1.0425+1/1.04^26)
subtract the two sequences
100000-100000/1.04=C(1/1.04-1/1.04^26)
C=(100000-100000/1.04)/(1/1.04-1/1.04^26)