Question

The sum of three numbers in an AP is 27 and their product is 405.find the numbers

Answer 1

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Answer 2

Answer:

Required three numbers in given A.P are (3,9,15) Or (15,9,3)

Explanation:

Let (a-d),a,(a+d) are three consecutive terms in an A.P

According to the problem given,

i) Sum of three numbers = 27

=> a-d+a+a+d=27

=> 3a = 27

Divide each term by 3 , we get

=> a = 9 ----(1)

ii) product of the numbers= 405

=> (a-d)a(a+d)=405

=> (a²-d²)a=405

/* By algebraic identity:

(x-y)(x+y) = x²-y² */

=> (9²-d²)9 = 405

/* substitute a value */

Divide both sides of the equation by 9 , we get

=> 9²-d²=45

=> 81 - d² = 45

=> -d² = 45 - 81

=> -d² = -36

=> d² = 36

=> d =±√6²

=> d = ±6 ---(2)

Therefore,

case1:

If a = 9, d = 6

Required 3 terms are,

(a-d) = 9-6 = 3

a = 9

(a+d) = 9+6 = 15

case 2:

If a = 9, d = -6

Required 3 terms are

(a-d) = 9-(-6) = 9+6 = 15

a = 9

a+d = 9-6 = 3

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