The answer is 30.08....................................................................
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The answer is 3√5. You use the distance formula: √((x2-x1)^2+(y2-y1)^2)
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
a6 = 3x -2
a6 = 3*6 - 2
a6 = 18 - 2
a6 = 16 which checks.
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.
