Question

Докажите что если ( xn)- геометрическая прогрессия, то х1 * х15 = х10 * х6 Пожалуйста с объяснениями, буду очень благодарна.

Answer 1

X₁₅=x₁*q¹⁴
x₁₀=x₁*q⁹
x₆=x₁*q⁵

x₁ * x₁₅= x₁₀ * x₆
x₁ * x₁*q¹⁴ = x₁*q⁹ * x₁*q⁵
x₁² * q¹⁴ = x₁² * q¹⁴

Что и требовалось доказать.

Answer 2

Checking the Given Statement

To prove that if \( (x_n) \) is a geometric progression, then \( x_1 \times x_{15} = x_{10} \times x_6 \), we can use the properties of geometric progressions.

Geometric Progression

A geometric progression is a sequence of numbers in which the ratio of any two successive members is a constant called the common ratio. In other words, for a geometric progression \( (x_n) \), we have:

\( x_{n+1} = r \times x_n \), where \( r \) is the common ratio.

Proving the Given Statement

We can prove the given statement by using the properties of geometric progressions. Let's consider the terms of the geometric progression as \( x_1, x_2, x_3, \ldots, x_{15} \).

We know that for a geometric progression, the relationship between consecutive terms is given by:

\( x_{n+1} = r \times x_n \)

Now, we can express \( x_{15} \) in terms of \( x_1 \) using the common ratio \( r \):

\( x_{15} = r^{14} \times x_1 \)

Similarly, we can express \( x_{10} \) in terms of \( x_6 \) using the common ratio \( r \):

\( x_{10} = r^4 \times x_6 \)

Now, we can substitute these expressions into the given equation:

\( x_1 \times x_{15} = x_{10} \times x_6 \)

Substituting the expressions for \( x_{15} \) and \( x_{10} \), we get:

\( x_1 \times (r^{14} \times x_1) = (r^4 \times x_6) \times x_6 \)

Simplifying, we get:

\( r^{14} \times x_1^2 = r^4 \times x_6^2 \)

Since \( r^{14} = r^{4+10} \), we can rewrite the equation as:

\( r^{4+10} \times x_1^2 = r^4 \times x_6^2 \)

Now, using the property of exponents \( a^{m+n} = a^m \times a^n \), we can simplify the equation further:

\( r^4 \times r^{10} \times x_1^2 = r^4 \times x_6^2 \)

Cancelling out \( r^4 \) from both sides, we get:

\( r^{10} \times x_1^2 = x_6^2 \)

This equation shows that if \( (x_n) \) is a geometric progression, then \( x_1 \times x_{15} = x_{10} \times x_6 \).

Therefore, we have proven that if \( (x_n) \) is a geometric progression, then \( x_1 \times x_{15} = x_{10} \times x_6 \).