Question

the sum of three number in a AP is 9 and there product is 24 then write the first three sum of Ap​

Answer 1

Answer:

Let the three numbers be a-d, a and a+d

sum =9

product = 24

a-d+a+a+d=9

3a=9

a=3

(a-d) (a)(a+d) =24

$(a-d)(a+d) = a^{2} - d^{2} \\(a^{2} - d^{2}) a = (9-d^{2}) 3=24\\\\(9-d^{2}) = 24/3\\(9-d^{2}) =8\\-d^{2} = 8 - 9\\\\-d^{2} = -1\\d^{2} = + or - 1$

So a = 3 d =1

AP is a-d = 3-1 =2

            a= 3

        a+d = 3+1 =4

AP is 2, 3, 4...

AP for a=3 d=-1

4, 3,2,...