Question

The sum of 3 numbers in an ap is15.and their product is 105.Find the numbers.

Answer 1

a-d+a+a+d=15

a=5

(a-d)(a)(a+d)=105

$(a {}^{2} - d {}^{2} )(a) = 105$

$(25 - d { }^{2} ) (5)= 105$

$25 - d {}^{2} = 21$

$d {}^{2} = 4$

d=4

the numbers are

a-d=5-4=1

a=5

a+d=5+4=9

Answer 2

let the 3 numbers be
a-d,a,a+d
sum of 3 no= 15
a-d+a+a+d=15
3a = 15
a= 15/3
a=5
product of 3 no = 105
(a-d)(a)(a+d)=105
(a-d)(5)(a+d)=105
(a-d)(a+d)=105/5
a^2- d^2= 21
put the value of a
5^2 -d^2= 21
25 - d^2=21
25-21 = d^2
4 =d^2
d= +or -2
therefore the no are 3,5,7