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If $1, x, y, z, 16$ are in geometric progression, then what is the value of $x+y+z$ ?
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The correct answer is:
14
As given $1, \mathrm{x}, \mathrm{y}, \mathrm{z} 16$ are in geometric progression. Let common ratio be $\mathrm{r}$, $\mathrm{x}=1 . \mathrm{r}=\mathrm{r}$
$\mathrm{y}=1 \cdot \mathrm{r}^{2}=\mathrm{r}^{2}$
$\mathrm{z}=1 \cdot \mathrm{r}^{3}=\mathrm{r}^{3}$
and $16=1 \cdot r^{4} \Rightarrow 16=r^{4}$
$\Rightarrow \mathrm{r}=2$
$\therefore \quad \mathrm{x}=1 . \mathrm{r}=2, \mathrm{y}=1 \cdot \mathrm{r}^{2}=4$
$\mathrm{z}=1 \cdot \mathrm{r}^{3}=8$
$\therefore \quad x+y+z=2+4+8=14$
$\mathrm{y}=1 \cdot \mathrm{r}^{2}=\mathrm{r}^{2}$
$\mathrm{z}=1 \cdot \mathrm{r}^{3}=\mathrm{r}^{3}$
and $16=1 \cdot r^{4} \Rightarrow 16=r^{4}$
$\Rightarrow \mathrm{r}=2$
$\therefore \quad \mathrm{x}=1 . \mathrm{r}=2, \mathrm{y}=1 \cdot \mathrm{r}^{2}=4$
$\mathrm{z}=1 \cdot \mathrm{r}^{3}=8$
$\therefore \quad x+y+z=2+4+8=14$
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