Question

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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Answer 1

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

Answer 2

Answer:

$the \: three \: terms \: are \: 5 \: \: 12 \: \: 19$