<h1>math problem in the sequences-puzzle 173-By Panos kalliantasis</h1>

 Three numbers are consecutive terms of arithmetic progression.

Find the numbers if the sum is 30 and the product is 960.

Try to solve this math problem.

a=b-d   and c =b+d

a+b+c=30

b-d +b +b+d=30

3b=30

b=10

a*b*c=960

(b-d)* b * (b+d)= 960

(b-d) * (b+d) *b =960

(b^2-d^2) *b =960

(10^2-d^2) *10 = 960

100- d^2=96

4=d^2

d=2  and d=-2

So

 a=8

b=10

c=12