BITSAT 2024 Shift 1 Paper

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+72
Simplifying:
0=18a+72
Solving for a :
18a=−72
a=−‌

72

18

=−4
Given this value of a, the four numbers are calculated as:
‌a+12=8
‌a=−4
‌a+6=2
‌a+12=8
Therefore, 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.