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The sum of an infinite $G P$ is $x$ and the common ratio $r$ is such that $|r| < 1$. If the first term of the $G P$ is 2, then which one of the following is correct?
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The correct answer is:
$\quad 1 < x < \infty$
$\quad \mathrm{GP}=\mathrm{x}$
$\qquad \frac{\mathrm{a}}{1-\mathrm{r}}=\mathrm{x} \quad$ (where, $\mathrm{a}=$ lst term and $\mathrm{r}=$ common ratio $)$
$\Rightarrow \frac{2}{1-r}=x \quad$...(1) $(\because$ Given $a=2$ and $|r| < 1)$
$\Rightarrow-1 < \mathrm{r} < 1 \Rightarrow 1>-\mathrm{r}>-1$
$\Rightarrow 1+1>1-r>1-1$
$\Rightarrow 0 < 1-\mathrm{r} < 2$
$\Rightarrow \frac{1}{1-r}>\frac{1}{2}, \frac{2}{1-r}>1$
from equation (i) $x>1$
Hence, $1 < \mathrm{x} < \infty$
$\qquad \frac{\mathrm{a}}{1-\mathrm{r}}=\mathrm{x} \quad$ (where, $\mathrm{a}=$ lst term and $\mathrm{r}=$ common ratio $)$
$\Rightarrow \frac{2}{1-r}=x \quad$...(1) $(\because$ Given $a=2$ and $|r| < 1)$
$\Rightarrow-1 < \mathrm{r} < 1 \Rightarrow 1>-\mathrm{r}>-1$
$\Rightarrow 1+1>1-r>1-1$
$\Rightarrow 0 < 1-\mathrm{r} < 2$
$\Rightarrow \frac{1}{1-r}>\frac{1}{2}, \frac{2}{1-r}>1$
from equation (i) $x>1$
Hence, $1 < \mathrm{x} < \infty$
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