The value of x that satisfies the relation x | vedclass

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

Question

(C) The given expression is an infinite geometric series: $x = 1 - x + x^2 - x^3 + x^4 - x^5 + \dots \infty$.This is a geometric series with the first

Vedclass · Accepted Answer

$2 \sin 18^\circ$

Vedclass · Answer

(C) The given expression is an infinite geometric series: $x = 1 - x + x^2 - x^3 + x^4 - x^5 + \dots \infty$.This is a geometric series with the first term $a = 1$ and common ratio $r = -x$.The sum of an infinite geometric series is given by $S = \frac{a}{1 - r}$,provided $|r| Substituting the values,we get $x = \frac{1}{1 - (-x)} = \frac{1}{1 + x}$.This simplifies to $x(1 + x) = 1$,which is $x^2 + x - 1 = 0$.Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we get $x = \frac{-1 \pm \sqrt{1^2 - 4(1)(-1)}}{2(1)} = \frac{-1 \pm \sqrt{5}}{2}$.Since the series must converge,we require $|-x| The value $\frac{-1 - \sqrt{5}}{2} \approx -1.618$ is rejected as it is less than $-1$.Thus,$x = \frac{\sqrt{5} - 1}{2}$.We know that $\sin 18^\circ = \frac{\sqrt{5} - 1}{4}$.Therefore,$x = 2 \sin 18^\circ$.

The value of $x$ that satisfies the relation $x = 1 - x + x^2 - x^3 + x^4 - x^5 + \dots \infty$.

Similar Questions

If $\tan \theta + \sin \theta = m$ and $\tan \theta - \sin \theta = n,$ then

If $\sin \theta = \frac{12}{13}$ where $0 < \theta < \frac{\pi}{2}$ and $\cos \phi = -\frac{3}{5}$ where $\pi < \phi < \frac{3\pi}{2}$,then $\sin(\theta + \phi)$ is:

What is the value of $(1+\cot A)^{2}+(1-\cot A)^{2}$?

Let ${f_k}(x) = \frac{1}{k}(\sin^k x + \cos^k x)$ for $k = 1, 2, 3, ...$. Then for all $x \in R$,the value of $f_4(x) - f_6(x)$ is equal to

If $1+\cos^{2} \theta = 3 \sin \theta \cos \theta$,then the integral value of $\cot \theta$ is $(0 < \theta < \frac{\pi}{2})$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module