Question

an oil and gas company offerrs a position to amirul with the starting salary of rm54000 per year and annual increment of 15% . find his salary of fifth year. show that his total salary from first year to the nth years is RM360000(1.15'n -1).

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 $S(n)=RM360000(1.15^n-1)$