Question

Which term of GP. \(\sqrt3\),\(3,3\sqrt3...\) is 729 ?

• A. 10
• B. 12
• C. 14
• D. 16

Answer 1

The Correct Option is B

Step 1: Identify the first term and common ratio of the G.P.

First term, \( a = \sqrt{3} \)

Second term, \( a_2 = 3 \)

So, common ratio \( r = \frac{a_2}{a_1} = \frac{3}{\sqrt{3}} = \sqrt{3} \)

Step 2: Use the formula for the \( n^{\text{th}} \) term of a G.P.:

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

We are given \( a_n = 729 \), \( a = \sqrt{3} \), \( r = \sqrt{3} \)

Step 3: Set up the equation:

\( \sqrt{3} \cdot (\sqrt{3})^{n-1} = 729 \)

\( (\sqrt{3})^n = 729 \)

Now, express 729 as power of 3:

\( 729 = 3^6 \), so:

\( (\sqrt{3})^n = 3^6 \)

\( (3^{1/2})^n = 3^6 \Rightarrow 3^{n/2} = 3^6 \)

Equating powers of 3:

\( \frac{n}{2} = 6 \Rightarrow n = 12 \)
 

The correct option is (B): \(12\)