Question

The product of three geometric means between $4$ and $\frac{1}{4}$ will be

• A. $4$
• B. $2$
• C. $-1$
• D. $1$

Answer 1

The correct option is D

$1$

Explanation for correct option:

Step 1: Find the common ratio.

Let the geometric means between $4$ and $\frac{1}{4}$ be $4,g_1,g_2,g_3,\frac{1}{4}$

Let the first term be $a$ and common ratio be $r$

$$\begin{gathered} \Rightarrow a{r}^{n-1}=a_n \\ \Rightarrow \frac{1}{4}=4\times r^{5-1} \\ \Rightarrow \frac{1}{16}=r^4 \\ \Rightarrow r=\pm \frac{1}{2} \end{gathered}$$

Step 2: Find the three geometric numbers between the given numbers.

Since, $g_1=ar,g_2=a{r}^2,g_3=a{r}^3$.

Compute $g_1$.

$$\begin{gathered} g_1=ar \\ \Rightarrow g_1=4\times \pm \frac{1}{2} \\ \Rightarrow g_1=\pm 2 \end{gathered}$$

Compute $g_2$.

$$\begin{gathered} g_2=a{r}^2 \\ \Rightarrow g_2=4\times \frac{1}{4} \\ \Rightarrow g_2=1 \end{gathered}$$

Compute $g_3$.

$$\begin{gathered} g_3=a{r}^3 \\ \Rightarrow g_3=4\times \pm \frac{1}{8} \\ \Rightarrow g_3=\pm \frac{1}{2} \end{gathered}$$

Step 3: Evaluate the product.

Compute the required product.

$$\begin{gathered} g_1·g_2·g_3=\pm 2\times 1\times \pm \frac{1}{2} \\ \Rightarrow g_1·g_2·g_3=1 \end{gathered}$$

Therefore, the required product is $1$.

Hence, option D is the correct anwer.