Find three numbers in an arithmetic sequence whose sum is 33 and whose product is 1320
Been trying to find it all day, could someone shed some light please?
Let the numbers be a, a+d, a+2d (since they are in AP).
Sum of them is a + (a+d) + (a+2d) = 3(a+d) = 33 ===> a + d = 11
The product is 1320:
 (a+2d) = 1320 \\\\ $But $ a+d=11:\\\\ 11a(a+2d) = 1320 \Rightarrow a(a+2d) = 120 \\\\ $But $ a+2d = (a+d)+d \\\\ a(a+d+d) = 120 \\\\ a(11+d) = 120 )
So we have the following:
a+d = 11 ===> a = 11-d
a(11+d) = 120
(11-d)(11+d) = 120
Upon solving, we get d = plus/minus 1.
Sub back into a+d = 11 to get:
a = 10 or a = 12.
So hence our final solutions are:
a x (a+d) x (a+2d) = 1320
or
a x (a+d) x (a+2d) = 1320
Test:
10 x 11 x 12 = 1320
10 + 11 + 12 = 33
If we do the other way around, it's actually the same thing, but swapped the other way around:
12 x 11 x 10 = 1320
12 + 11 + 10 = 33.
