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

• According to the question, we have that in n years

$T(n) = 54000+54000(1.15)+54000(1.15)^2+....+54000(1.15)^n$

which is clearly a geometric progression, the formula for the nth term of a geometric progression is

$T(n) = ar^{(n-1)}$

therefore $T(5) = 54000*1.15^4$

$T(5) = RM94446.3375$

• The formula for the sum of terms in a geometric progression is $S(n) = a\frac{r^n-1}{r-1}$ where r = 1.15

therefore S(n) = $54000\frac{1.15^n-1}{1.15-1}$ ,hence $S(n)=RM360000(1.15^n-1)$

• Using the formula derived above, we have that

$3654670=360000(1.15^n-1)$ therefore

$1.15^n=\frac{3654670}{360000}+1$ , taking the log of both sides of the equation

$n=\frac{\log(11.15)}{\log(1.15)}$ therefore n = 17.26 years approximately

for Amirui to exceed RM3654670, then he must work for more than 17.26 years