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20 July, 05:25

Find t12 term in an arithmetic progression having t3 = 10 and t10 = - 4.

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  1. 20 July, 05:28
    0
    t12 = - 8

    Step-by-step explanation:

    t3

    a+2d=10 (1)

    t10

    a+9d=-4. (2)

    From (1)

    a=10-2d. (3)

    Sub into (3) into equ (2)

    a+9d=-4

    10-2d+9d=-4

    10+7d=-4

    7d=-4-10

    7d=-14

    Divide both sides by 7

    d=-14/7

    d = - 2

    10=a+2d

    10=a+2 (-2)

    10=a-4

    10+4=a

    14=a

    a=14

    t12=a+11d

    = 14+11 (-2)

    =14-22

    = - 8

    t12 = - 8
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