If x = sumlimits_{n = 0}^infty {{a^n}} ,;y = | vedclass

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

Question

(B) Given $x = \sum\limits_{n = 0}^\infty {{a^n}} = \frac{1}{{1 - a}}$.$\Rightarrow 1 - a = \frac{1}{x}$ $\Rightarrow a = 1 - \frac{1}{x} = \frac{x -

Vedclass · Accepted Answer

$xz + yz = xy + z$

Vedclass · Answer

(B) Given $x = \sum\limits_{n = 0}^\infty {{a^n}} = \frac{1}{{1 - a}}$.$\Rightarrow 1 - a = \frac{1}{x}$ $\Rightarrow a = 1 - \frac{1}{x} = \frac{x - 1}{x}$.Similarly,$y = \sum\limits_{n = 0}^\infty {{b^n}} = \frac{1}{{1 - b}} \Rightarrow b = \frac{y - 1}{y}$.And $z = \sum\limits_{n = 0}^\infty {{{(ab)}^n}} = \frac{1}{{1 - ab}}$ $\Rightarrow 1 - ab = \frac{1}{z}$ $\Rightarrow ab = 1 - \frac{1}{z} = \frac{z - 1}{z}$.Substituting the values of $a$ and $b$ in the expression for $ab$:$\left( \frac{x - 1}{x} \right) \left( \frac{y - 1}{y} \right) = \frac{z - 1}{z}$.$\frac{xy - x - y + 1}{xy} = \frac{z - 1}{z}$.$z(xy - x - y + 1) = xy(z - 1)$.$xyz - xz - yz + z = xyz - xy$.$-xz - yz + z = -xy$.$xy + z = xz + yz$.

If $x = \sum\limits_{n = 0}^\infty {{a^n}} ,\;y = \sum\limits_{n = 0}^\infty {{b^n},\;z = \sum\limits_{n = 0}^\infty {{{(ab)}^n}} } $,where $a, b < 1$,then

Similar Questions

If $\frac{a + bx}{a - bx} = \frac{b + cx}{b - cx} = \frac{c + dx}{c - dx}$ and $x \neq 0$,then $a, b, c$ and $d$ are in:

The number which should be added to the numbers $2, 14, 62$ so that the resulting numbers may be in $G.P.$ is

If three numbers are in $G.P.$,then their logarithms will be in

If the roots of the equation $x^3-13x^2+Kx-27=0$ are in geometric progression,then $K=$

Let $a$ and $b$ be two distinct positive real numbers. Let the $11^{\text{th}}$ term of a $GP$,whose first term is $a$ and third term is $b$,be equal to the $p^{\text{th}}$ term of another $GP$,whose first term is $a$ and fifth term is $b$. Then $p$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module