To solve the problem, let's consider the setup described:
The three numbers in an Arithmetic Progression (AP) are denoted as
a,a+6, and
a+12, where the common difference is 6 .
The sequence of four numbers has the first and last numbers equal.
Thus, the four numbers are
(a+12),a,a+6, and
(a+12).
The problem states that the first three numbers are in a Geometric Progression (GP). Therefore, the following relationship should hold:
a2=(a+12)(a+6)Expanding and solving the equation:
a2=a2+18a+72Simplifying:
0=18a+72Solving for
a :
18a=−72a=−‌=−4Given this value of
a, the four numbers are calculated as:
‌a+12=8‌a=−4‌a+6=2‌a+12=8Therefore, the sequence of four numbers is
8,−4,2,8.
The two intermediate numbers among the four (excluding the repeated elements at the start and end) are -4 and 2 . Hence, the two other numbers in this sequence are
−4,2.