Question

Let $a$ be a positive number such that the arithmetic mean of $a$ and $2$ exceeds their geometric mean by $1$.Then, the value of $a$ is

• A. $3$
• B. $5$
• C. $9$
• D. $8$

Answer 1

The correct option is D

$8$

Determine the value of $a$

The Arithmetic mean (A.M) of a series containing $n$ observations are the sum of $n$ terms divided by the number of terms.

$AM=\frac{a_1+a_2...+a_n}{n}$
The Geometric Mean (G.M) of a series containing $n$ observations is the $n^{th}$ root of the product of the values.

$GM={\sqrt[n]{a_1\times a_2...\times a}}_n$

Arithmetic mean of $a$ and $2$ is $\frac{a+2}{2}$

Geometric mean of $a$ and $2$ is $\sqrt{2a}$

Given, AM exceeds GM by $1$.

$$\begin{gathered} \Rightarrow \text{AM-GM}=1 \\ \Rightarrow \frac{a+2}{2}-\sqrt{2a}=1 \\ \Rightarrow a+2-2\sqrt{2a}=2 \\ \Rightarrow a-2\sqrt{2a}=2-2 \\ \Rightarrow \sqrt{a}\left(\sqrt{a}-2\sqrt{2}\right)=0 \\ \Rightarrow \sqrt{a}=2\sqrt{2} \end{gathered}$$

Squaring on both sides

$\Rightarrow$ ${\left(\sqrt{a}\right)}^2={\left(2\sqrt{2}\right)}^2$

$\Rightarrow$ $a=8$

Hence, option (D) is the correct answer