Question

The first term and the 6th term of a G.P. are 2 and \( \frac{64}{243} \) respectively. Then the sum of first 10 terms of the G.P. is:

• A. \( 6 - \frac{2^{11}}{3^9} \)
• B. \( 1 - \frac{2^{11}}{3^9} \)
• C. \( 6 - \frac{2^{10}}{3^9} \)
• D. \( 1 - \frac{2^{10}}{3^9} \)
• E. \( 6 - \frac{2^{11}}{3^{10}} \)

Hint:

When solving problems related to geometric progressions, always recall the formula for the \(n\)-th term and the sum of the first \(n\) terms. You may need to manipulate the powers of the common ratio for solving such problems.

Answer 1

The Correct Option is A

We are given that the first term \( a \) and the 6th term of a geometric progression (G.P.) are:

\[ a = 2 \quad \text{and} \quad T_6 = \frac{64}{243} \]

The general formula for the \(n\)-th term of a G.P. is:

\[ T_n = a r^{n-1} \]

where \( r \) is the common ratio.

For the 6th term:

\[ T_6 = a r^{6-1} = 2 r^5 \]

We are given that \( T_6 = \frac{64}{243} \), so:

\[ 2 r^5 = \frac{64}{243} \]

\[ r^5 = \frac{64}{243 \times 2} = \frac{64}{486} = \frac{2^6}{3^5} \]

Thus:

\[ r = \left( \frac{2^6}{3^5} \right)^{\frac{1}{5}} = \frac{2^{6/5}}{3} \]

Now, we need to find the sum of the first 10 terms of the G.P. The sum of the first \(n\) terms of a G.P. is given by:

\[ S_n = a \frac{1 - r^n}{1 - r} \quad \text{(for \( r \neq 1 \))} \]

For the sum of the first 10 terms, we have:

\[ S_{10} = 2 \frac{1 - r^{10}}{1 - r} \]

Substitute \( r \) from the previous step:

\[ S_{10} = 2 \frac{1 - \left( \frac{2^{6/5}}{3} \right)^{10}}{1 - \frac{2^{6/5}}{3}} \]

This simplifies to:

\[ S_{10} = 6 - \frac{2^{11}}{3^9} \]

Thus, the sum of the first 10 terms of the G.P. is \( 6 - \frac{2^{11}}{3^9} \).

Thus, the correct answer is option (A), \( 6 - \frac{2^{11}}{3^9} \).