Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

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...

Mathematics
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

What are your input values of your function, and what are your output values?
I don't have any input values. The output values are as I indicated...3, 12, 21, 30, 39...etc.
  • phi
add 9 to it each time what do you do to 3 to get each number? add 9, then 2*9 , then 3*9...

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

If I do f(n) = 3+9n this would work but only if things start at n=0
For 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...
Doesn't the sequence start at n=1
Hm.. 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?
She really didn't say...hmm..I guess I'll just try it and indicate that n starts at 0
K. If you wanted to move it to n=1 => f(1) = 3, you could just use a function shift: f(n) = 3 + 9n f(n-1) = 3 + 9(n-1) = 3 + 9n - 9 = -6 + 9n <-- call this the new f(n)
I think I would prefer the start at n=0, just because it is easier to see your first term. :)

Not the answer you are looking for?

Search for more explanations.

Ask your own question