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

To solve the problem of inserting two numbers between \( \sqrt{2} \) and \( 4 \) such that the sequence forms a Geometric Progression (G.P.), we start by considering the structure of a G.P.

In a G.P, the ratio between consecutive terms remains constant. Let the sequence be \( \sqrt{2}, a, b, 4 \). The common ratio is denoted by \( r \).

Let's find the common ratio, \( r \), starting with:

\(a = \sqrt{2} \cdot r\)

\(b = a \cdot r = \sqrt{2} \cdot r^2\)

\(4 = b \cdot r = \sqrt{2} \cdot r^3\)

Solving for \( r \) from the last equation:

\(4 = \sqrt{2} \cdot r^3\)

\(r^3 = \frac{4}{\sqrt{2}}\)

\(r^3 = \frac{4\sqrt{2}}{2} = 2\sqrt{2}\)

Solving for \( r \):

\(r = (2\sqrt{2})^{1/3}\)

Or, to express this more conveniently, recognize that \( r = 2^{1/3} \cdot (\sqrt{2})^{1/3} \), a simplification leads to:

\(r = 2^{2/3}\)

Now, compute the inserted numbers:

• \(a = \sqrt{2} \cdot r = \sqrt{2} \cdot 2^{1/3} = 2\)
• \(b = \sqrt{2} \cdot r^2 = \sqrt{2} \cdot 2^{2/3} = 2\sqrt{2}\)

Hence, the inserted numbers, in order, are \(2\) and \(2\sqrt{2}\).

Therefore, the correct answer is:

$2, 2\sqrt{2}$