What is the fifth term of an arithmetic sequence whose first term is 40 and whose seventh term is 121?

A.80 

B.64 

C.94 

D.85

ALgebra problem

2 Answers

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  • 1 month ago

    Going from the 1st term to the 7th term is *6* steps. And you go up by 81.

    81/6

    = 27/2

    = 13.5

    So the step size (common difference) is 13.5

    Going backwards 2 steps from 121, we would go back 2*13.5 = 27

    121 - 27

    = 94

    Answer:

    C. 94

  • 1 month ago

    nth term is a + (n - 1)d...where a is the 1st term and d is the common difference.

    so, 7th term is 40 + 6d = 121

    Hence, 6d = 81

    => d = 27/2

    Then, nth term is 40 + (27/2)(n - 1)

    i.e. 53/2 + 27n/2 => (53 + 27n)/2

    so, 5th term is (53 + 135)/2 = 94

    :)>

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