find the first four terms of the geometric progression if the first term is 12 and sum to infinity is 24?

24 = 12 / [ 1 - r ]
24 - 24 r = 12
12 = 24 r
r = 1 / 2
Term1__Term 2___Term 3__Term 4
12________6_______3______1.5

First term, a = 12
Common difference = r

Sum to infinity = 24
a/(1 - r) = 24
12/(1 - r) = 24
1 - r = 1/2
r = 1/2

T(1) = 12
T(2) = 12 × (1/2) = 6
T(3) = 6 × (1/2) = 3
T(4) = 3 × (1/2) = 3/2

The first four terms are 12, 6, 3 and 3/2.

The formula for the infinite sum of a geometric sequence (where |r| < 1) is:
S = a/(1 - r)

S : infinite sum (24)
a : first term (12)
r : common ratio

Plug in your values for the infinite sum and the first term:
24 = 12/(1 - r)

Now solve for r:
24(1 - r) = 12
1 - r = 12/24
1 - r = 1/2
r = 1 - 1/2
r = 1/2

So the common ratio is 1/2. You already know the first term is 12; now just find each of the subsequent terms by multiplying by 1/2.

Answer:
12, 6, 3, 1½, ...

24 is 2 times 12.
Therefore, if you "factor out" the 12, you are looking for a series that is a geometrical progression (equal factor between each term) that will begin with 1, and add up to 2.

Such an infinite series is already known:

Sum[n = 0 to infinity] of (1/2)^n

Anything to the power 0 is equal to 1,
then the next term is 1/2, then 1/4, then 1/8...
and there exists a way to show that this series tend to 2 (as n goes to infinity)

(1 + 1/2 + 1/4 + 1/8 + ... forever) = 2

If you multiply both sides by 12, you get
12(1 + 1/2 + 14 + 1/8 + ... forever) = 12 * 2 = 24

The 12 is a factor of the whole series, therefore it gets to multiply every term, when you "distribute" it into the series.