If x, y, z are in Arithmetic Progression (AP) | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given that $x, y, z$ are in $AP$,we have $2y = x + z$. This implies $z = 2y - x$. Given that $x, y, t$ are in $GP$,we have $y^2 = xt$. This implie

Vedclass · Accepted Answer

Geometric Progression

Vedclass · Answer

(A) Given that $x, y, z$ are in $AP$,we have $2y = x + z$. This implies $z = 2y - x$. Given that $x, y, t$ are in $GP$,we have $y^2 = xt$. This implies $t = \frac{y^2}{x}$. Now,consider the sequence $x, x - y, t - z$. Substitute $t$ and $z$: $t - z = \frac{y^2}{x} - (2y - x) = \frac{y^2 - 2xy + x^2}{x} = \frac{(x - y)^2}{x}$. For the sequence $x, x - y, t - z$ to be in $GP$,the square of the middle term must equal the product of the first and third terms: $(x - y)^2 = x 	imes (t - z)$. Substituting the expression for $(t - z)$: $x 	imes \frac{(x - y)^2}{x} = (x - y)^2$. Since the condition holds,$x, x - y, t - z$ are in $GP$.

If $x, y, z$ are in Arithmetic Progression $(AP)$ and $x, y, t$ are in Geometric Progression $(GP)$,then in which progression are $x, x - y, t - z$?

Similar Questions

Given a sequence of $4$ numbers,the first three of which are in $G.P.$ and the last three are in $A.P.$ with a common difference of $6$. If the first and last terms in this sequence are equal,then the last term is:

If $a, b, c$ are three unequal numbers such that $a, b, c$ are in $A.P.$ and $b - a, c - b, a$ are in $G.P.$,then $a : b : c$ is

Let $\frac{1}{16}, a$ and $b$ be in $G.P.$ and $\frac{1}{a}, \frac{1}{b}, 6$ be in $A.P.,$ where $a, b > 0$. Then $72(a + b)$ is equal to ...... .

The harmonic mean between two numbers is $14\frac{2}{5}$ and the geometric mean is $24$. The greater number among them is

If $a, b, c$ are in $A.P.$ and $a^2, b^2, c^2$ are in $G.P.$ such that $a < b < c$ and $a+b+c = \frac{3}{4}$,then the value of $a$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module