grade 11 sequences and series question

Question

Lauren's earrings are increasing by rate by 3% every year. in 1998, lauren earned $40000. what did she earn in 2010? determine lauren's total earnings (all the years combined) for the period 1998 to 201

Answer 1

a_n = a₁(r)^n-1

a₁₂ = 40000(1.03)¹¹

 

S₁₂ = 40000(1 - 0.03¹²) / (1 - 0.03)

Answer 2

$40000 in 1998, 3% annual rate.

Now, Lauren would have

at the end of 1998:

$40000+ $4000*(3/100)=$40000(1+0.03)

at the end of 1999:

$40000(1+0.03)+$40000(1+0.03)*(3/100)=$40000(1+0.03)²

at the end of 2000

 ..=$40000(1+0.03)³

 

and so on...after n year, 

$40000(1+0.03)ⁿ

I think you know what n is equal in your question.