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8. Sequences and Series
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if $x = \,\frac{4}{3}\, - \,\frac{{4x}}{9}\, + \,\,\frac{{4{x^2}}}{{27}}\, - \,\,.....\,\infty $ , then $x$ is equal to
A
only $1$
B
$1$ or $-4$
C
only $-4$
D
$-1$ or $4$
Solution
$\Rightarrow x=\frac{4}{3}\left(\frac{1}{1+\frac{x}{3}}\right)=\frac{4}{3+x}$
$\Rightarrow 3 x+x^{2}=4$
$\Rightarrow x^{2}+3 x-4 \Rightarrow(x+4)(x-1)=0$
$\Rightarrow x=1,-4$
$\Rightarrow x=1$ only as $\left|-\frac{4}{3}\right|>1$
Standard 11
Mathematics
