Three. Numbers are in A.P.
First term = a
Second term = a+d
Third term = a+2d
the product of first and last is 216
a(a+2d) = 216
4 times the second together with twice the first is 84
4(a+d) + 2a = 84
So we have this system of two equations in two unknowns:
a(a+2d) = 216
4(a+d)+2a = 84
Simplify the 2nd equation:
4a+4d+2a = 84
6a+4d = 84
Divide through by 2
3a+2d = 42
Solve for 2d
2d = 42-3a
Substitute 42-3a for 2d in
a(a+2d) = 216
a(a+42-3a) = 216
a(-2a+42) = 216
-2aČ+42a = 216
Get 0 on the right
-2aČ+42a-216 = 0
Divide through by -2
aČ-21a+108 = 0
(a-12)(a-9) = 0
a-12=0; a-9=0
a=12; a=9
Substitute a=12 in
2d = 42-3a
2d = 42-3(12)
2d = 42-36
2d = 6
d = 3
That gives the solution
First term = a = 12
Second term = a+d = 12+3 = 15
Third term = a+2d = 12+2(3) = 12+6 = 18
One solution: 12, 15, 18
Substitute a=9 in
2d = 42-3a
2d = 42-3(9)
2d = 42-27
2d = 15
d = 15/2 = 7.5
That gives the solution
First term = a = 9
Second term = a+d = 9+7.5 = 16.5
Third term = a+2d = 9+2(7.5) = 9+15 = 24
Other solution: 9, 16.5, 24
Edwin
