Question

The sum of three numbers in
a.p. is 27, and their product is 504; find them

Answer 1

Let the numbers be a-d,a,a+d
As per the problem,
a-d+a+a+d=27
3a=27
a=9
And, (a-d)(a)(a+d)=504
(a*a-d*d)=56
81-d*d=56
d*d=81-56
d*d=25
d=5
Therefore,the numbers are (9-5),(9),(9+5)
=4,9,14.

Answer 2

sum of three consective term of ap is 27.

their product is 504

Let

The three numbers be a-d , a ,a+d.

a-d+a+a+d=27

3a=27

a=27/3

a=9

(a-d)×a×(a+d)=504

(a^2-d^2)a=504

(9^2-d^2)9=504

81-d^2 = 504/9

81-d^2= 56

d^2 = 81-56

d^2 = 25

d= + -5

If d=+5

Terms of A.p are 4,9,14

If d=-5

Terms of AP are 14,9,4