Question

the sum of three no. in A.P. is -3 and their product is 8. Find the numbers

Answer 1

let the numbers be a-d,a,a+d

sum= a+a+d+a-d= -3

3a= -3

a= -1

now product, (a+d)(a-d)(a)=8

(a^2-d^2)*a=8

(9-d^2*)-3=8

d^2=9-8/3

d= under root19/3

hence the numbers are -1-under root19/3,

-1 and -1+under root19/3