If x = frac{4}{3} - frac{4x}{9} + frac{4x^2}{ | vedclass

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(A) The given series is an infinite geometric progression with the first term $a = \frac{4}{3}$ and common ratio $r = -\frac{x}{3}$.The sum of an infi

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only $1$

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(A) The given series is an infinite geometric progression with the first term $a = \frac{4}{3}$ and common ratio $r = -\frac{x}{3}$.The sum of an infinite geometric series is given by $S = \frac{a}{1-r}$,provided $|r| Substituting the values,we get $x = \frac{4/3}{1 - (-x/3)} = \frac{4/3}{1 + x/3} = \frac{4}{3+x}$.Cross-multiplying gives $x(3+x) = 4$,which simplifies to $x^2 + 3x - 4 = 0$.Factoring the quadratic equation,we get $(x+4)(x-1) = 0$,so $x = 1$ or $x = -4$.For the infinite series to converge,we must satisfy $|r| Since $|-4| = 4 > 3$,$x = -4$ is rejected. Thus,$x = 1$ is the only valid solution.

If $x = \frac{4}{3} - \frac{4x}{9} + \frac{4x^2}{27} - \dots \infty$,then $x$ is equal to

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