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

Given :-

The sum of three consecutive terms of an AP = 36 .

The Product of three consecutive terms of an AP = 1140

The terms are in descending order.

Solution :-

Let the three consecutive terms be

x + d, x , x - d

According to the question,

x + d + x + x - d = 36

3x = 36

x = 36/3

x = 12

Now,

Subsitute the value of x in given terms

12 + d , 12 , 12 - d

As , It is also given in the question that the product of three consecutive terms = 1140

Therefore,

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

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

By using identity ,

[( a + b) ( a - b) = a^2 - b^2 ]

12^2 - d^2 = 1140 /12

144 - d^2 = 95

-d^2 = 95 - 144

-d^2 = - 49

-d = -7

d = 7

Thus, The value of d = 7

Therefore,

First terms of an AP = 12 + 7 = 19

Second term of an AP = 12

Third term of an AP = 12 - 7 = 5 $.$