Question

If we insert two numbers between $\sqrt{2}$ and $4$ so that the resulting sequence is in G.P., then the inserted numbers in the order are

• A. $4, \sqrt{2}$
• B. $2, 2\sqrt{2}$
• C. $\sqrt{8}, 2$
• D. $2\sqrt{2}, 4$

Hint:

A useful trick for simplifying expressions with radicals is to express integers as powers of roots. For example, recognizing that $4 = (\sqrt{2})^4$ allows you to immediately simplify $\frac{4}{\sqrt{2}}$ to $\frac{(\sqrt{2})^4}{\sqrt{2}} = (\sqrt{2})^3$, making the cube root calculation trivial.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Inserting $n$ numbers between two given numbers $a$ and $b$ to form a Geometric Progression (G.P.) implies creating a sequence of $n+2$ terms where the first term is $a$, the last term is $b$, and the sequence has a constant common ratio. The inserted numbers are called geometric means.

Step 2: Key Formula or Approach:
Let the newly formed G.P. sequence be $t_1, t_2, t_3, t_4$. We are given $t_1 = a = \sqrt{2}$ and $t_4 = b = 4$. The general formula for the $n$-th term of a G.P. is $t_n = a \cdot r^{n-1}$. We can use the 4th term to solve for the common ratio $r$. Once $r$ is found, the inserted numbers are $t_2 = a \cdot r$ and $t_3 = a \cdot r^2$.

Step 3: Detailed Explanation:
Let the two numbers inserted be $x$ and $y$. The resulting sequence is: $\sqrt{2}, x, y, 4$. This sequence is a Geometric Progression. The first term is $a = \sqrt{2}$. Since there are 4 terms in total, the fourth term is $t_4 = 4$. Let $r$ be the common ratio of this G.P. Using the formula for the $n$-th term, $t_n = a \cdot r^{n-1}$, we have: \[ t_4 = a \cdot r^3 \] Substitute the known values into the equation: \[ 4 = \sqrt{2} \cdot r^3 \] Now, solve for $r^3$: \[ r^3 = \frac{4}{\sqrt{2}} \] To simplify, multiply the numerator and denominator by $\sqrt{2}$ to rationalize the denominator: \[ r^3 = \frac{4\sqrt{2}}{\sqrt{2} \cdot \sqrt{2}} = \frac{4\sqrt{2}}{2} = 2\sqrt{2} \] We can express $2\sqrt{2}$ as a cube to easily find $r$: \[ 2\sqrt{2} = (\sqrt{2})^2 \cdot (\sqrt{2})^1 = (\sqrt{2})^3 \] So, we have: \[ r^3 = (\sqrt{2})^3 \] Taking the real cube root of both sides gives the common ratio: \[ r = \sqrt{2} \] Now we calculate the inserted numbers $x$ and $y$, which are the second and third terms of the G.P.: First inserted number, $x = t_2 = a \cdot r = \sqrt{2} \cdot \sqrt{2} = 2$. Second inserted number, $y = t_3 = a \cdot r^2 = \sqrt{2} \cdot (\sqrt{2})^2 = \sqrt{2} \cdot 2 = 2\sqrt{2}$. Therefore, the inserted numbers in order are $2$ and $2\sqrt{2}$.

Step 4: Final Answer:
The inserted numbers are $2, 2\sqrt{2}$.