If x, G_1, G_2, y are the consecutive terms o | vedclass

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

Question

(C) Given that $x, G_1, G_2, y$ are in a Geometric Progression $(G.P.)$.Let the common ratio of the $G.P.$ be $r$.Then,$G_1 = xr$,$G_2 = xr^2$,and $y

Vedclass · Accepted Answer

$xy$

Vedclass · Answer

(C) Given that $x, G_1, G_2, y$ are in a Geometric Progression $(G.P.)$.Let the common ratio of the $G.P.$ be $r$.Then,$G_1 = xr$,$G_2 = xr^2$,and $y = xr^3$.From the last term,we have $r^3 = y/x$,so $r = (y/x)^{1/3}$.Now,the product $G_1 G_2 = (xr)(xr^2) = x^2 r^3$.Substituting $r^3 = y/x$ into the expression:$G_1 G_2 = x^2 (y/x) = xy$.Alternatively,in any $G.P.$,the product of terms equidistant from the beginning and the end is constant and equal to the product of the first and last terms. Thus,$G_1 G_2 = x \cdot y$.

If $x, G_1, G_2, y$ are the consecutive terms of a $G.P.$,then the value of $G_1 G_2$ will be

Similar Questions

If $\log _{10} 2, \log _{10} (2^x - 1), \log _{10} (2^x + 3)$ are in $A.P.,$ then :-

The common difference of the $A.P.$ $b_{1}, b_{2}, \ldots, b_{m}$ is $2$ more than the common difference of $A.P.$ $a_{1}, a_{2}, \ldots, a_{n}.$ If $a_{40} = -159,$ $a_{100} = -399$ and $b_{100} = a_{70},$ then $b_{1}$ is equal to

The sum of an infinite number of terms in a $G.P.$ is $20$ and the sum of their squares is $100$. The common ratio of the $G.P.$ is:

Let $a, b, c, d \in R^+$ such that $256 abcd \geq (a+b+c+d)^4$ and $3a + b + 2c + 5d = 11$. Then,the value of $a^3 + b + c^2 + 5d$ is equal to:

If the sum of an infinite $G.P.$ and the sum of the squares of its terms is $3$,then the common ratio of the first series is

Vedclass Test Series

Exam Paper Generator

Online Exam Module