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If $x+1,4 x+1$, and $8 x+1$ are in geometric progression, then what is the non-trivial value of $\mathrm{x}$ ?
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Verified Answer
The correct answer is:
$\frac{1}{8}$
$\operatorname{If}(\mathrm{x}+1),(4 \mathrm{x}+1)$ and $(8 \mathrm{x}+1)$ are in GP.
then $(4 x+1)^{2}=(x+1)(8 x+1)$
$\Rightarrow 16 x^{2}+8 x+1=8 x^{2}+x+8 x+1$
$\Rightarrow 8 x^{2}-x=0 \Rightarrow x(8 x-1)=0$
$\Rightarrow \mathrm{x}=0, \frac{1}{8},\left[\frac{1}{8}\right.$ is non-trivial value]
then $(4 x+1)^{2}=(x+1)(8 x+1)$
$\Rightarrow 16 x^{2}+8 x+1=8 x^{2}+x+8 x+1$
$\Rightarrow 8 x^{2}-x=0 \Rightarrow x(8 x-1)=0$
$\Rightarrow \mathrm{x}=0, \frac{1}{8},\left[\frac{1}{8}\right.$ is non-trivial value]
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