The problem states that we have an increasing arithmetic sequence ${a_n}$. The first term $a_1 = 2$. Also, $a_1$, $a_2$, and $a_3$ form a geometric sequence. We need to find the common difference of the arithmetic sequence ${a_n}$.

1. Problem Description

The problem states that we have an increasing arithmetic sequence ${a_n}$. The first term $a_1 = 2$. Also, $a_1$, $a_2$, and $a_3$ form a geometric sequence. We need to find the common difference of the arithmetic sequence ${a_n}$.

2. Solution Steps

Since ${a_n}$ is an arithmetic sequence, we can express $a_2$ and $a_3$ in terms of $a_1$ and the common difference $d$.

$a_2 = a_1 + d = 2 + d$

$a_3 = a_1 + 2d = 2 + 2d$

Since $a_1, a_2, a_3$ form a geometric sequence, we have:

$\frac{a_2}{a_1} = \frac{a_3}{a_2}$

$a_2^2 = a_1 a_3$

$(2+d)^2 = 2(2+2d)$

$4 + 4d + d^2 = 4 + 4d$

$d^2 = 0$

$d = 0$

However, it is stated that the sequence is increasing, so $d>0$. This indicates that the given terms $a_1, a_2, a_3$ form a geometric sequence only approximately. Let us re-examine the condition that $a_1, a_2, a_3$ form a geometric sequence. Then:

$a_2^2 = a_1 a_3$

$(a_1+d)^2 = a_1(a_1+2d)$

$(2+d)^2 = 2(2+2d)$

$4 + 4d + d^2 = 4 + 4d$

$d^2 = 0$

$d = 0$

Since the sequence is increasing, $d$ must be positive. Let us assume there is some error in the text, the ratio instead may be nearly equal, meaning that $a_1, a_2, a_3$ form a near-geometric sequence. Let's reconsider the equation $(2+d)^2 = 2(2+2d)$, which simplifies to $d^2=0$. The prompt says the sequence is strictly increasing, meaning that the common difference $d$ must be greater than zero. However, the equation leads to $d=0$.

Let's assume there was a typo in the problem. Perhaps the three terms were not perfectly geometric, and we just want the closest value. Since $d^2=0$, it implies $d=0$. But since ${a_n}$ is a increasing arithmetic sequence, $d>0$. If $d$ is slightly greater than 0, $a_1, a_2, a_3$ almost form a geometric sequence. Because $d$ can only be 0, and the problem asks for the common difference, the most sensible answer is $d=0$.

However, this conflicts with the statement that the sequence is strictly increasing. Thus, there is no solution with the strict interpretation of all given constraints.

Let's assume the question is slightly modified to: given an arithmetic sequence, $a_1=2$, and $a_1, a_2, a_3$ form a geometric progression, what is the common difference. This results in $d=0$, implying all terms are the same.

Let us try another approach. If we are *interpreting* $a_1, a_2, a_3$ as a geometric sequence, we can write $a_n = a_1 + (n-1)d$. We have $a_1 = 2$, $a_2 = 2 + d$, and $a_3 = 2 + 2d$.

For these to form a geometric sequence, we require $a_2/a_1 = a_3/a_2$.

$(2+d)/2 = (2+2d)/(2+d)$.

$(2+d)^2 = 2(2+2d)$.

$4 + 4d + d^2 = 4 + 4d$.

$d^2 = 0$.

$d = 0$.

Since the sequence is increasing, $d > 0$, but we found $d = 0$. There must be an error in the problem statement or the question intended something different. I suspect there may be a typo in the problem, or it is set up to be a contradiction, and hence has no solution. With the given information, the only value for the common difference we can find is 0, though this implies that it is not an increasing sequence.

3. Final Answer

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