If a, b and c are in Geometric Progression and $a^\frac{1}{x} = b^\frac{1}{y} = c^\frac{1}{z}$ then, x, y, z are in ________.
The problem asks us to find the relationship between the exponents $x$, $y$, and $z$ given that $a$, $b$, and $c$ are in Geometric Progression (GP) and satisfy the condition $a^\frac{1}{x} = b^\frac{1}{y} = c^\frac{1}{z}$.
If three terms $a$, $b$, and $c$ are in Geometric Progression, it means the ratio between consecutive terms is constant. This implies that the square of the middle term is equal to the product of the other two terms:
$b^2 = ac$
We are given the relation:
$a^\frac{1}{x} = b^\frac{1}{y} = c^\frac{1}{z}$
Let's introduce a constant, $k$, such that:
$a^\frac{1}{x} = b^\frac{1}{y} = c^\frac{1}{z} = k$
From this, we can express $a$, $b$, and $c$ in terms of $k$:
Now, we substitute these expressions for $a$, $b$, and $c$ into the geometric progression condition $b^2 = ac$:
$(k^y)^2 = (k^x)(k^z)$
Using the rules of exponents ($(p^m)^n = p^{mn}$ and $p^m \cdot p^n = p^{m+n}$), we get:
$k^{2y} = k^{x+z}$
Since the bases are the same (and assuming $k$ is positive and not equal to 1), the exponents must be equal:
$2y = x + z$
The equation $2y = x + z$ is the defining condition for an Arithmetic Progression (AP). This means that $y$ is the arithmetic mean of $x$ and $z$. Therefore, the sequence $x$, $y$, $z$ forms an Arithmetic Progression.
This matches Option 1.
Which of the following statement is true about the geometric series
$ 1 + r +r^2 + r^3 + ...............; (r > 0) $?
$6240$ रुपये की राशि $30$ किस्तों में इस प्रकार चुकाई जाती है कि प्रत्येक किस्त पिछली किस्त से $10$ रुपये अधिक है । पहली किस्त की मूल्य ____________है।