Question

The sum of three no. in an ap is 27 and their product is 405 find the no.s

Answer 1

If the terms are a,a+d,a+2d 
then 
a+a+d+a+2d=27 
3a+3d=27 
3(a+d)=27 
a+d=9 
a=9-d 

(a) x (a+d) x (a+2d)=405 
(9-d)(9-d+d)(9-d+2d)=405 
(9-d)(9)(9+d)=405 
(81-d^2)(9)=405 
729-9d^2=405 
-9d^2=405-729 
-9d^2=-324 
d^2=36 
d=6 

a=9-6 
=3 

check- 
if the terms are a,a+d,a+2d 
3+3+6+3+(2*6)=27 
12+3+12=27 
15+12=27 
27=27 

(a)(a+d)(a+2d)=405 
3*(3+6)(3+2*6)=405 
3*9*15=405 
405=405