SOLUTION: I am given a term and the common difference of an arithmetic sequence a21 = -1.4 as the term and the difference is d = 0.6. I have to find the recursive formula and the 3 terms in

Algebra ->  Functions -> SOLUTION: I am given a term and the common difference of an arithmetic sequence a21 = -1.4 as the term and the difference is d = 0.6. I have to find the recursive formula and the 3 terms in       Log On


   

Question 1116678: I am given a term and the common difference of an arithmetic sequence a21 = -1.4 as the term and the difference is d = 0.6. I have to find the recursive formula and the 3 terms in the sequence after the last one given. I have tried +a21+=+1.4+%28.6%29%5E%28n-1%29+ but it doesn’t solve right. Isn’t the recursive formula +an+=+a1%28d%29%5E%28n-1%29?
Found 2 solutions by ikleyn, stanbon:
Answer by ikleyn(52797) About Me  (Show Source):
You can put this solution on YOUR website!
.
Your post is written very inaccurately.


What I got from the post is that you are going to work with arithmetic progression, but mistakenly use the formulas for geometric progression.


For arithmetic progression,  a%5Bn%5D = a%5B1%5D + (n-1)*d.

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For introduction to arithmetic progressions, see the lessons in this site:
    - Arithmetic progressions
    - The proofs of the formulas for arithmetic progressions
    - Problems on arithmetic progressions
    - Word problems on arithmetic progressions

Also,  you have this free of charge online textbook in ALGEBRA-II in this site
    - ALGEBRA-II - YOUR ONLINE TEXTBOOK.

The referred lessons are the part of this online textbook under the topic "Arithmetic progressions".


Save the link to this textbook together with its description

Free of charge online textbook in ALGEBRA-II
https://www.algebra.com/algebra/homework/complex/ALGEBRA-II-YOUR-ONLINE-TEXTBOOK.lesson

into your archive and use when it is needed.



Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
It's aritmetic not geometric.
a(n) = a(1) + (n-1)d
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Cheers,
Stan H.
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