Question

Which term of the geometric progression \(2,2\sqrt{2},4,…..\)is \(128?\)

• A. 11th
• B. 12th
• C. 13th
• D. 14th

Answer 1

The Correct Option is C

To find which term of the geometric progression \(2, 2\sqrt{2}, 4, \ldots\) is \(128\), we start by identifying the first term (\(a\)) and the common ratio (\(r\)).

Given:

• First term (\(a\)) = \(2\)
• Second term = \(2\sqrt{2}\)

To find the common ratio \(r\), divide the second term by the first term:

\(r = \frac{2\sqrt{2}}{2} = \sqrt{2}\)

The formula for the \(n\)-th term of a geometric progression is:

\(a_n = a \cdot r^{n-1}\)

We are given \(a_n = 128\), so:

\(128 = 2 \cdot (\sqrt{2})^{n-1}\)

Simplify:

\(64 = (\sqrt{2})^{n-1}\)

Express \(64\) as a power of \(2\):

\(64 = 2^6\)

Since \((\sqrt{2})^{n-1} = 2^{\frac{n-1}{2}}\), equate the exponents:

\(\frac{n-1}{2} = 6\)

Solve for \(n\):

\(n-1 = 12\)

\(n = 13\)

This means the 13th term of the geometric progression is 128.

Option	Result
11th	Incorrect
12th	Incorrect
13th	Correct

Answer 2

In a geometric progression (G.P.), the general term is given by: 

\( a_n = a r^{(n-1)} \)

From the given sequence: \( 2, 2\sqrt{2}, 4, \dots \)

The first term is: \( a = 2 \)

The common ratio is:

\( r = \frac{2\sqrt{2}}{2} = \sqrt{2} \)

We need to find \( n \) such that:

\( 2 (\sqrt{2})^{(n-1)} = 128 \)

Dividing by 2:

\( (\sqrt{2})^{(n-1)} = 64 \)

Since \( 64 = (\sqrt{2})^{12} \), we equate:

\( n-1 = 12 \)

\( n = 13 \)

Thus, the required term is: 13th.