In a set of four numbers, the first three are in a geometric sequence and the last three are in an arithmetic sequence with a common difference of 6. If the first number is the same as the fourth, find the four numbers.
+21 Let our 4 numbers be \(a_1, a_2, a_3, a_4\). Since we know that \(a_1 = a_4\), we replace it, getting us:
\(a_1, a_2, a_3, a_1\)
Therefore, we know that the last three numbers form an arithmetic sequence. We can represent each of the other terms in terms of \(a_2\). We know that the common difference is 6, therefore, we can write our 4 numbers as:
\(a_1, a_2, a_2 + 6, a_2 + 12\)
Since the first number is the same as the last number, we can also replace the first number with the last number, which gets us:
\(a_2 + 12, a_2, a_2 + 6, a_2 + 12\)
Let r be the common ratio between our numbers. Then, we know that:
\(r(a_2 + 12) = a_2,\\ ra_2 = a_2 + 6,\\ \)
Solving this systems of equations by subsitution, we find out that \(a_2 = -4, r = -\frac{1}{2}\). Therefore, our 4 numbers are:
\(8, -4, 2, 8\)
