Find the value of x, if x, x + 4, and .x + 6 are consecutive terms of geometric sequence.

Given that \( x \), \( x + 4 \), and \( x + 6 \) are consecutive terms of a geometric sequence, we know that the ratio between consecutive terms is constant. Therefore, the ratio between the second and first terms should be equal to the ratio between the third and second terms.

\[
\text{Let the common ratio be } r.
\]

We can set up the equation based on the common ratio:

\[
\frac{x + 4}{x} = \frac{x + 6}{x + 4}
\]

To solve for \( x \), we'll cross-multiply:

\[
(x + 4)^2 = x(x + 6)
\]

Expanding both sides:

\[
x^2 + 8x + 16 = x^2 + 6x
\]

Now, subtract \( x^2 \) from both sides:

\[
8x + 16 = 6x
\]

Subtract \( 6x \) from both sides:

\[
2x + 16 = 0
\]

Subtract 16 from both sides:

\[
2x = -16
\]

Divide by 2:

\[
x = -8
\]

So, the value of \( x \) is \( \boxed{-8} \).

Would you like any further details or have any questions? Here are some related questions:

1. What is the value of the common ratio \( r \) in this geometric sequence?
2. How do you verify that \( x = -8 \) satisfies the condition of being a geometric sequence?
3. Can you solve a similar problem with different initial values?
4. What if the sequence was arithmetic instead of geometric? How would the solution change?
5. Can you generalize the process for solving problems involving geometric sequences?

**Tip:** Always verify your solutions by substituting the value back into the original equation to check if the conditions are met.

Learn how to find the value of x in a geometric sequence problem where x, x + 4, and x + 6 are consecutive terms. This problem involves using the concept of geometric sequences and solving equations. Suitable for advanced high school students.

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