We are given an arithmetic sequence $\{a_n\}$ with the first term $a_1 = 2$ and common difference $d = 2$. We are also given that $a_1, a_2, a_3$ form a geometric sequence. We need to find the value of $k$. It's likely the problem has a typo and it asks to find k such that a1, a2, a3 are in geometric progression. However since no k has been given we have to figure out what this means. Let's assume that the problem is asking for a constant $k$ so that $a_1, a_2, a_3$ form a geometric progression where $a_1=2$ and the common difference is 2. The variable $k$ is not really clearly used, but probably the problem asks for $a_3$.

1. Problem Description

We are given an arithmetic sequence $\{a_n\}$ with the first term $a_1 = 2$ and common difference $d = 2$. We are also given that $a_1, a_2, a_3$ form a geometric sequence. We need to find the value of $k$. It's likely the problem has a typo and it asks to find k such that a1, a2, a3 are in geometric progression. However since no k has been given we have to figure out what this means. Let's assume that the problem is asking for a constant $k$ so that $a_1, a_2, a_3$ form a geometric progression where $a_1=2$ and the common difference is

2. The variable $k$ is not really clearly used, but probably the problem asks for $a_3$.

2. Solution Steps

Since $\{a_n\}$ is an arithmetic sequence with $a_1 = 2$ and $d = 2$, we can find $a_2$ and $a_3$ using the formula:

$a_n = a_1 + (n-1)d$

$a_2 = a_1 + (2-1)d = 2 + 1(2) = 2 + 2 = 4$

$a_3 = a_1 + (3-1)d = 2 + 2(2) = 2 + 4 = 6$

For $a_1, a_2, a_3$ to form a geometric sequence, the ratio between consecutive terms must be constant. That is,

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

$\frac{4}{2} = \frac{6}{4}$

$2 = \frac{3}{2}$

Since this is false, $a_1, a_2, a_3$ do not form a geometric sequence.

Assuming that question means to find $a_3$ :

$a_3 = 6$

We can suppose that the problem asks what value to add to $a_3$ so that $a_1, a_2, a_3+k$ is a geometric progression. Thus

$a_3+k=x$

Then $2, 4, x$ should be in geometric progression.

Then $\frac{4}{2} = \frac{x}{4}$

$2 = \frac{x}{4}$

$x=8$

$a_3+k = 8$

$6+k=8$

$k=2$

3. Final Answer

The value of $a_3 = 6$.

If the problem meant to ask to find k such that a1, a2, a3+k are in GP, then k=

2.