The value of x that satisfies the relation x | vedclass

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(C) The given series is an infinite geometric progression with first term $a = 1$ and common ratio $r = -x$. For the series to converge,$|r| < 1$,i.e.

Vedclass · Accepted Answer

$2 \sin 18^\circ$

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(C) The given series is an infinite geometric progression with first term $a = 1$ and common ratio $r = -x$. For the series to converge,$|r| The sum of the infinite geometric series is given by $S = \frac{a}{1 - r}$. 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 $x = \frac{-1 \pm \sqrt{1 - 4(1)(-1)}}{2} = \frac{-1 \pm \sqrt{5}}{2}$. Since $|x| Thus,$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 $x = 1 - x + x^2 - x^3 + x^4 - x^5 + \dots \infty$ is:

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