Algebra

I am so lost on these problems.

Write a geometric sequence that starts with 3 and has a common ratio of 5. What is the 23rd term in the sequence.

Write an arithmetic sequence that has a common difference of 4 and the eighth term is 13. What is the first term? What is the 23rd term in the sequence.

Write an arithmetic sequence where 12 is the first term of 12, and the tenth term is 39. What is the first term? What is the 23rd term in the sequence.

Two consecutive terms of a geometric sequence, in order are 10 and 12. What is the ratio. What is the term before 10? What is the term after 12?

  1. 4
asked by Randy
  1. well, they have said that
    a=3
    r=5
    The 23rd term is thus 3*5^22
    That is, you start with 3 and multiply by 5 22 more times to get to the 23rd term.

    Next, we have
    a + 7*4 = 13
    so, a = -15
    T23 = -15 + 22*4

    Any two consecutive terms of a geometric sequence have a common ratio. In this case, we have

    r = 12/10 = 6/5
    So, to get the term before 10, divide by 6/5
    To get the term after 12, multiply by 6/5

    posted by Steve

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