Find four numbers that form a geometric progression such that the third term is greater than the first by 12 and the fourth is greater than the second by 36.

5, 4.5, 13.5 and 40.5

Step-by-step explanation:

Since the numbers are in geometric progression, their form is essentially:

a, ar, ar^2 and ar^3

Where a and r are first term and common ratio respectively.

From the information given in the catalog:

Third term is greater than the first by 12 while fourth is greater than second by 36.

Let's now translate this to mathematics.

ar^2 - a = 12

ar^3 - ar = 36

From 1, a (r^2 - 1) = 12 and 2:

ar (r^2 - 1) = 36

From 2 again r[a (r^2 - 1] = 36

The expression inside square bracket looks exactly like equation 1 and equals 12.

Hence, 12r = 36 and r = 3

Substituting this in equation 1,

a (9 - 1) = 12

8a = 12

a = 12/8 = 1.5

Thus, the numbers are 1.5, (1.5 * 3), (1.5 * 9), (1.5 * 27) = 1.5, 4.5, 13.5 and 40.5