Question

If $1,\mathrm{x},\mathrm{y},\mathrm{z},2$ are in geometric progression, then $\mathrm{xyz}=?$

• A. $2\sqrt{2}$
• B. $4$
• C. $8$
• D. None of these

Answer 1

The correct option is A

$2\sqrt{2}$

Explanation for the correct answer:

Let $\mathrm{a}$ be the first term of the geometric progression and $\mathrm{r}$ be the common ratio

Then, $\mathrm{a}=1$

$\mathrm{a}\times \mathrm{r}=\mathrm{x}$

$\mathrm{a}\times {\mathrm{r}}^2=\mathrm{y}$

$\mathrm{a}\times {\mathrm{r}}^3=\mathrm{z}$

$\mathrm{a}\times {\mathrm{r}}^4=2$

${\mathrm{r}}^4=2$ $...\left(\because \mathrm{a}=1\right)$

$\Rightarrow$ ${\mathrm{r}}^2=\sqrt{2}$

$\mathrm{xyz}=ar\times a{r}^2\times a{r}^3$

$\Rightarrow$ $\mathrm{xyz}=a^3{r}^6$

$\Rightarrow$ $\mathrm{xyz}=a^{3^{}}\times {\left(r^2\right)}^3$

Substitute the values of $a=1,r^2=\sqrt{2}$ we get,

$\Rightarrow$ $\mathrm{xyz}=2\sqrt{2}$

Hence, option $\left(A\right)$ is the correct answer.