The first three of four given numbers are in G.P and the last three are in A.P. with common difference 6. If the first and fourth numbers are equal, find the numbers.

The first and last numbers are given to be equal. Let's assume them to be a.

The last three numbers are said to be in A.P. and the common difference given is 6. So, the second and third numbers will be (a - 12) and (a - 6) respectively.

Now, since it is given that the first three numbers are in G.P. (2nd no/1st no) should be equal to (3rd no/2nd no).

=> (a - 12)/a = (a - 6)/(a - 12)
=> (a - 12)^2 = a(a - 6)
=> a^2 - 24a + 144 = a^2 - 6a
=> - 24a + 6a = -144
=> -18a = -144
=> a = 8

The rest of the numbers can be easily determined now. The four numbers thus come out to be 8, -4, 2, 8.