the first three terms of an arithmetic sequence are (k+2) , (2k+3) and (4k-2) respectively. Find the value of the constant k

To solve for the constant \( k \), we start by using the property of an arithmetic sequence: the difference between consecutive terms is constant. 

Given the first three terms of the arithmetic sequence are:
1. \( a_1 = k + 2 \)
2. \( a_2 = 2k + 3 \)
3. \( a_3 = 4k - 2 \)

The common difference \( d \) of an arithmetic sequence can be found as:

\[
d = a_2 - a_1 = a_3 - a_2
\]

Let's compute each difference:

1. Difference between the second and first terms:
\[
a_2 - a_1 = (2k + 3) - (k + 2) = 2k + 3 - k - 2 = k + 1
\]

2. Difference between the third and second terms:
\[
a_3 - a_2 = (4k - 2) - (2k + 3) = 4k - 2 - 2k - 3 = 2k - 5
\]

Since these differences are equal, we set them equal to each other:

\[
k + 1 = 2k - 5
\]

Now, solve for \( k \):

\[
k + 1 = 2k - 5
\]

Subtract \( k \) from both sides:

\[
1 = k - 5
\]

Add 5 to both sides:

\[
k = 6
\]

### Conclusion

The value of the constant \( k \) is **6**.

Would you like more details or have any questions?

Here are 5 related questions for further exploration:

1. What are the first five terms of this arithmetic sequence with \( k = 6 \)?
2. How do you find the nth term of an arithmetic sequence?
3. Can you derive a formula for the sum of the first n terms of an arithmetic sequence?
4. What changes in the arithmetic sequence if \( k \) were a different value?
5. How can we find the common difference if we only know two non-consecutive terms of the sequence?

**Tip:** Remember, in arithmetic sequences, knowing any two consecutive terms allows you to find the common difference!

The first three terms of an arithmetic sequence are (k+2), (2k+3), and (4k-2) respectively. Find the value of the constant k.

Learn how to solve for the constant k in an arithmetic sequence where the first three terms are (k+2), (2k+3), and (4k-2). This problem involves algebra and arithmetic sequences, perfect for students in Grades 9-10. Step-by-step solution provided.

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