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 2 years ago
3, 12, 21, 30, 39.... What is the arithmetic function of this sequence?
I know we add 9 to it each time but I need to figure out the function. Kind of stuck...
 2 years ago
3, 12, 21, 30, 39.... What is the arithmetic function of this sequence? I know we add 9 to it each time but I need to figure out the function. Kind of stuck...

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AccessDenied
 2 years ago
Best ResponseYou've already chosen the best response.1What are your input values of your function, and what are your output values?

rachelk09
 2 years ago
Best ResponseYou've already chosen the best response.0I don't have any input values. The output values are as I indicated...3, 12, 21, 30, 39...etc.

phi
 2 years ago
Best ResponseYou've already chosen the best response.0add 9 to it each time what do you do to 3 to get each number? add 9, then 2*9 , then 3*9...

rachelk09
 2 years ago
Best ResponseYou've already chosen the best response.0If I do f(n) = 3+9n this would work but only if things start at n=0

AccessDenied
 2 years ago
Best ResponseYou've already chosen the best response.1For the function, if we want it to represent this sequence, it probably should return the first value as 3 ( f(1) = 3 ), the next value as 12 ( f(2) = 12 ), then 21 ( f(3) = 21 ), and so on. So, these are our 'points.' We just have to figure out the general formula. Use the fact that you add 9 a different number of times to 3 to get to all of your function's output values...

rachelk09
 2 years ago
Best ResponseYou've already chosen the best response.0Doesn't the sequence start at n=1

AccessDenied
 2 years ago
Best ResponseYou've already chosen the best response.1Hm.. I think you could technically start it at either n=0 or n=1. It shouldn't matter too much, but I'm not sure if your teacher may have a preference that he/she used?

rachelk09
 2 years ago
Best ResponseYou've already chosen the best response.0She really didn't say...hmm..I guess I'll just try it and indicate that n starts at 0

AccessDenied
 2 years ago
Best ResponseYou've already chosen the best response.1K. If you wanted to move it to n=1 => f(1) = 3, you could just use a function shift: f(n) = 3 + 9n f(n1) = 3 + 9(n1) = 3 + 9n  9 = 6 + 9n < call this the new f(n)

AccessDenied
 2 years ago
Best ResponseYou've already chosen the best response.1I think I would prefer the start at n=0, just because it is easier to see your first term. :)
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