If x = sum_{n=0}^{infty} a^n,y = sum_{n=0}^{i | vedclass

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

Question

(A) Given that $x = \sum_{n=0}^{\infty} a^n = \frac{1}{1-a}$,$y = \sum_{n=0}^{\infty} b^n = \frac{1}{1-b}$,and $z = \sum_{n=0}^{\infty} c^n = \frac{1}

Vedclass · Accepted Answer

Harmonic Progression

Vedclass · Answer

(A) Given that $x = \sum_{n=0}^{\infty} a^n = \frac{1}{1-a}$,$y = \sum_{n=0}^{\infty} b^n = \frac{1}{1-b}$,and $z = \sum_{n=0}^{\infty} c^n = \frac{1}{1-c}$.Since $a, b, c$ are in arithmetic progression,we have $2b = a + c$.We need to check if $x, y, z$ are in harmonic progression,which requires $\frac{1}{x}, \frac{1}{y}, \frac{1}{z}$ to be in arithmetic progression.Note that $\frac{1}{x} = 1-a$,$\frac{1}{y} = 1-b$,and $\frac{1}{z} = 1-c$.Let $A = 1-a$,$B = 1-b$,and $C = 1-c$.Then $2B = 2(1-b) = 2 - 2b = 2 - (a+c) = (1-a) + (1-c) = A + C$.Since $A, B, C$ are in arithmetic progression,their reciprocals $\frac{1}{A}, \frac{1}{B}, \frac{1}{C}$ are in harmonic progression.Thus,$x, y, z$ are in harmonic progression.

If $x = \sum_{n=0}^{\infty} a^n$,$y = \sum_{n=0}^{\infty} b^n$,and $z = \sum_{n=0}^{\infty} c^n$,where $a, b, c$ are in arithmetic progression and $|a| < 1, |b| < 1, |c| < 1$,then $x, y, z$ are in which progression?

Similar Questions

If $a, b, c, d$ are in $H.P.$,then

If $a, b, c$ are in $H.P.$,then the value of $\left( \frac{1}{b} + \frac{1}{c} - \frac{1}{a} \right) \left( \frac{1}{c} + \frac{1}{a} - \frac{1}{b} \right)$ is

Given that the roots of $x^3+3px^2+3qx+r=0$ are in harmonic progression,then:

If $\ln(a+c), \ln(c-a), \ln(a-2b+c)$ are in $A.P.$,then

The harmonic mean of the numbers $2, 3, 4$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module