Question

Which term of G.P. \( 2, 2\sqrt{2}, 4, \dots \) is 128?

• A. \( 7 \)
• B. \( 8 \)
• C. \( 13 \)
• D. \( 10 \)

Hint:

For geometric progressions, use the formula \( T_n = ar^{n-1} \) and simplify logarithmic equations when necessary.

Answer 1

The Correct Option is C

Step 1: Formula for the nth term of a geometric progression.
The nth term of a geometric progression (G.P.) is given by: \[ T_n = ar^{n-1} \] Here, \( a = 2 \) and \( r = \sqrt{2} \). We need to find \( n \) such that: \[ 2 (\sqrt{2})^{n-1} = 128 \] Step 2: Simplify the equation.
First, divide both sides by 2: \[ (\sqrt{2})^{n-1} = 64 \] Since \( 64 = 2^6 \), we have: \[ (\sqrt{2})^{n-1} = 2^6 \] Step 3: Solve the equation.
Since \( \sqrt{2} = 2^{1/2} \), we can write: \[ (2^{1/2})^{n-1} = 2^6 \quad \Rightarrow \quad 2^{(n-1)/2} = 2^6 \] Now, equate the exponents: \[ \frac{n-1}{2} = 6 \quad \Rightarrow \quad n-1 = 12 \quad \Rightarrow \quad n = 13 \] Thus, the 13th term of the G.P. is 128.