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The sum of three consecutive terms of an A.P. is 36 and their product is 1140.
Find the terms. (Consider the terms to be in d

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The sum of three consecutive terms of an A.P. is 36 and their product is 1140.
Find the terms. (Consider the terms to be in descending order.)​

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Rose

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2 thoughts on “The sum of three consecutive terms of an A.P. is 36 and their product is 1140.<br />Find the terms. (Consider the terms to be in d”

  1. Answer:

    Let the smallest Number of A.P be x.

    so, the other numbers

    => x+1 and x+2

    Given that,

    sum of these terms = 36

    According to the questions

    x+2+x+1+x = 36

    => 3x+3 = 36

    => 3x = 36–3

    => 3x = 33

    => x = 33/3

    => x = 11

    Therefore the terms are,

    => x+2 = 11+2 = 13

    => x+1 = 11+1 = 12

    => x = 11

    Reply

  2. Answer:

    5,12,19

    Step-by-step explanation:

    let a-d , a , a+d be the 3 terms of A.P such that

    a-d + a + a+d = 36 (given)

    => 3a = 36

    => a = 36/3

    => a= 12

    also given (a-d)* a * (a+d) = 1140

    => (12-d) * (12) * (12+d) = 1140

    => (12-d) (12+d) = 1140/12

    => 12²- d² = 95

    => d² = 144 – 95 = 49

    => d = √49 = 7

    => the 3 terms are , a-d, a, a+d

    12-7, 12,12+7

    5,12,19

    Hope this help you

    Reply

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