Harder 2U Question (1 Viewer)

[currysauce]

A prize fun is set up with a single investment of $2000 to provide an annual prize of $250. The fund accrues interest at the rate of 10% per annum, compounded yearly. The first prize awarded one year after the investment is initially set up.

i) If $Pn denots the value of the fund at the end of n years (and after the nth prize has been awarded), show that: Pn = 2500 - 500(1.1)^n

ii)Hence find the number of years for which the full amount of the prize can be awarded.

 

[currysauce]

guess not

if anyone can do this it would be appreciated

leave part 2 i worked that out from the previous

 

[Jago]

P = 2000 , r = 1.1 , M = 250

follow your normal steps until you get to:

Pn = 2000 x 1.1ⁿ - 250 [(1.1ⁿ - 1) / 0.1]
Pn = 2000 x 1.1ⁿ - 2500 (1.1ⁿ - 1)
Pn = 2000 x 1.1ⁿ - 2500 x 1.1ⁿ + 2500
Pn = 2500 - 500 x 1.1ⁿ

 

[c_james]

P1 = 2000(1.1) - 250

P2 = [2000(1.1)-250](1.1) - 250
= 2000(1.1)^2 - 250(1.1) - 250

P3 = [2000(1.1)^2 - 250(1.1) - 250](1.1) - 250
= 2000(1.1)^3 - 250(1.1)^2 - 250(1.1) - 250

.: Pn = 2000(1.1)^n - 250[1 + 1.1 ... + 1.1^(n-1)]
= 2000(1.1)^n - 250[1(1.1^n - 1)/1.1-1]
= 2000(1.1)^n - 2500[(1.1)^n -1] (by dividing 250 by 0.1)
= 2000(1.1)^n - 2500(1.1)^n + 2500
= 2500 -500(1.1)^n

 

[currysauce]

thanks alot peoples

i guess i didn't understand the question... i mean whats an award...??????? i ket adding 250 as a award LOL

 

[Jago]

almost all the time payment questions follow the same formula

An = P x R^n - M x [(R^n - 1) / R - 1]

just have to know which number goes where