Question

find three numbers in ap whose sum is 15 and whose product is 105​

Answer 1

let the no. be,
a-d,a,a+d
a-d+a+a+d=15
3a=15
a=5

a-d(a)(a+d)=105
5-d(5)(5+d)=105
5-d(5+d)=105/5
25-d×d=21
-d×d=-4
d=2

so ap will be
3;5;7

Answer 2

Let three number be :a-d,a,a+d

Given sum = 15

a-d+a+a+d=15

3a=15

a=5-------------(1)

NOW , GIVEN PRODUCT=105

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

substitute a=5 above

(5-d)(5)(5+d)=105

(5-d)(5)(5+d)=105

125+25d-25d-5d=105

-5d^2=105-125

d^2=4

d=2

Hence : (a-d)(a)(a+d)

THEREFORE THE A.P IS

3,5,7