Question

the sum of three numbers in an arthamatic progression is 27 and their products is 405.Find the number​

Answer 1

Answer:

3,9,15

Step-by-step explanation:

Given,

sum=27

product=405

let the numbers in A.P be (a-d),a, (a+d)

sum=a-d+a+a-d

3a=27

a=9

product=(a-d) (a+d) a=(a^2-d^2)a=405

(9^2-d^2)9=405

(81-d^2)=405/9

d^2=81-45=36

d=6

numbers are (9-6),(9),(9+6)

numbers are 3,9,15

Hope this helps you

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Answer 2

Answer:

here is the answer of this.