Matematika

Pertanyaan

u1+u2+u3=21, u1.u2.u3=280 berapa u20?

1 Jawaban

  • Jawab

    misalkan Deret Aritmetika
    u1 = (a-b)
    u2 = (a)
    u3 = (a+b)

    u1+u2 +u3 = 21
    a-b + a + a+b = 21
    3a = 21
    a= 7

    u1 = a-b = 7-b
    u2 = a= 7
    u3 = 7+b

    u1 .u2 . u3 = 280
    (7-b)(7)(7+b) = 280
    (7-b)(7+b) = 280/7
    (7-b)(7+b) = 40
    49 - b² = 40
    b² = 9
    b = 3 atau b = - 3

    untuk b = 3
    u1 = 7-b =  7- (3) = 4
    U20 = u1  + 19(b) = 4 + 19(3) = 61

    untk b = -3
    u1 = 7 - b = 7 -(-3)= 10
    u20 = u1 + 19b = 10 + 19(-3) = - 47

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