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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.

Jul 27, 2021

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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:

$$8, -4, 2, 8$$

Jul 28, 2021