Question

An Oil and Gas Company offers a position to Amirul with the starting salary of
RM54,000
per year and annual increment of
15%.
(a) Find his salary of fifth years.
(b) Show that his total salary from the first year to the nth years is
RM360,000 (1.15^n-1)
(c) Find the number of years that he has worked if his total salary is exceed
RM3,654,670.

Answer 1

use the compound interest formula :

$a_n = a_0(1+p/100)^n\\ \\ \ \\ a)S_n = 54000(1+0.15)^n\\ S_5 = 54000(1+0.15)^5 = 108613.228\\ b)from \ 1 \ to \ n \ : \\ 54 000 + 54000*1.15 +54000*1.15^2 +... +54000*1.15^n = \\ 54000 \sum_0^n(1.15)^n = 54000*1/3(23*1.15^n -20) = 18000(23*1.15^n -20)\\ c) 18000(23*1.15^n -20) = 3654670\\ 23*1.15^n -20 = 203.04\\ 1.15^n = 9.7\\ n = 16.2571\\ n > 16\\ \text{more than 16 years}$