precalculus

posted by Anonymous

how do you find the 12th term of the sequence an= an-1+3 ( the n and n-1 are subscript) n>2 (also equal to)? I understand how to find the first six terms,but how do you use that to find the 12th . Thanks

  1. Reiny

    an = an-1 + 3
    where a1 has to be known, I conclude that a1 = 2
    I don't understand why n>2, I was expecting n>1

    Ok, lets assume our terms are 2, 5, 8 ..... , each term is 3 more than the previous one
    You could just grind it out and find the 12th terms , there are not too many for that.

    However, I can translate this one to
    an = 3n - 1 , that will produce our sequence 2, 5, ...
    so a12 = 3(12) - 1 = 35

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