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21 March, 19:58

Find the 3rd term of an arithmetic sequence with t5 = 3 and t7 = 7.

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  1. 21 March, 20:06
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    Tn=t1+d (n-1)

    t5-t7=2d

    7-3=4

    2d=4

    d=2

    tn=t1+2 (n-1)

    t5=3

    3=t1+2 (5-1)

    3=t1+2 (4)

    3=t1+8

    minus 8

    -5=t1

    tn=-5+2 (n-1)

    t3=-5+2 (3-1)

    t3=-5+2 (2)

    t3=-5+4

    t3=-1

    3rd term is - 1
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