Arithmetic Sequences Question

[Truck]

Three consecutive terms of an arithmetic sequence have a sum of 36 and a product of 1428. Find the three terms.

Just a little bit lost on this one, not quite sure how to apply it within the formula. Cheers for the help .

[luken93]

t1 = a
t2 = a + d
t3 = a + 2d

t1 + t2 + t3 = 36
3a + 3d = 36
a + d = 12

Also;
t1 x t2 x t3 = 1428

[Truck]

K I worked it out by doing:
a + a + d + a + 2d = 36
3a + 3d = 36
a + d = 12
a = 12 - d

a x (a + d) x (a + 2d) = 1428
12 - d x (12 - d + d) x ( 12 -d + 2d) = 1428
12(12-d)(12+d) = 1428
12(144-d^2)=1428
1728 - 12d^2 = 1428
300 = 12d^2
d^2 = 25
d = +- root25... therefore d=5

3a + 3d = 36
3a + 3(5) = 36
a = 7

Therefore 7, 12, 17 = your numbers.
So that's right, but are there any other ways to solve such a Q?

EDIT: lol luken beat me to it, thanks man .

[Truck]

Just another question:

IF the sum of n terms of an AP is nx(n+1) and the nth term is 2nx, find the AP.

[brightsky]

n/2(a + 2nx) = 1/2(na + 2n^2x) = na/2  + n^2x = n^2x + nx
Hence nx = na/2
a = 2x
since 2nx = a + (n-1)d
2nx = 2x + (n-1)d
2nx - 2x = (n-1)d
d = (2nx - 2x)/(n-1) = 2x(n-1)/(n-1) = 2x

So clearly the arithmetic sequence is 2x, 4x, 6x, ..., 2nx, where x can be anything.

[Truck]

n/2(a + 2nx) = 1/2(na + 2n^2x) = na/2  + n^2x = n^2x + nx
Hence nx = na/2
a = 2x
since 2nx = a + (n-1)d
2nx = 2x + (n-1)d
2nx - 2x = (n-1)d
d = (2nx - 2x)/(n-1) = 2x(n-1)/(n-1) = 2x

So clearly the arithmetic sequence is 2x, 4x, 6x, ..., 2nx, where x can be anything.

Makes sense, thanks.