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Mathematics

Consider the multiple of 7 in between 100 and 500. What are the first term and last term? How many terms and there in the sequence?

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  1. 100/7 = 14 rem 2, so the next multiple of 7 is either down 2 or up 5
    clearly we have to go to 105, sure enough 105/7 = 15

    so we have
    a = 105 and d = 7

    How many terms to get to near 500?
    a + (n-1)d = appr 500
    105 + (n-1)(7) = 500
    n = 57.4 , so it must be 57 terms, since the 58th term > 500
    and the last term is a + 56d = 105 + 56(7) = 497

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