Interesting question. Good to know for computer science. Suppose you have a function like an = 3x - 2 Try the first couple a1 = 3(1) - 2 a1 = 3 - 2 a1 = 1
a2 = 3(2) - 2 a2 = 6 - 2 a2 = 4 So each term differs by 3 a2 - a1 = 3
an = a_(n - 1) + 3 a3 = a2 + 3 a3 = 4 + 3 a3 = 7
a4 = a3 + 3 a4 = 7 + 3 a4 = 10
a5 = a4+ 3 a5 = 10 + 3 a5 = 13
I'll do one more and then check it. a6 = a5 + 3 a6 = 13 + 3 a6 = 16
So the general formula is an = a_(n - 1) * k if you were multiplying or an = a_(n - 1) + k if you were adding. The key thing is that you are working with the previous term.
The money has been in the account f<span>or 5 years. This means that the amount of deposit was increased for 5 times. 75*1.04*1.04*1.04*1.04</span>*1.04=75*1.04^5≈91.3 $ In this way i prefer to chose <span>D) $90.00 beacuse it's the nearest number</span>
