If `t_r` denotes the `r^(th)` term of an G.P. and `t_1 t_5 t_10 t_16 t_21 t_25 = 512` , then the value of its `13^(th)` term is :

Options

A. `sqrt2`
B. `1/sqrt2`
C. `2`
D. `2sqrt2`

Answer

To solve the problem, we need to find the value of the 13th term of a geometric progression (G.P.) given that the product of certain terms equals 512. Let's break it down step by step.

### Step 1: Understand the G.P. terms
The general term of a G.P. can be expressed as:
\[ t_n = a \cdot r^{n-1} \]
where \( a \) is the first term and \( r \) is the common ratio.

### Step 2: Write the specific terms
From the problem, we need to find the terms \( t_1, t_5, t_{10}, t_{16}, t_{21}, t_{25} \):
- \( t_1 = a \)
- \( t_5 = a \cdot r^4 \)
- \( t_{10} = a \cdot r^9 \)
- \( t_{16} = a \cdot r^{15} \)
- \( t_{21} = a \cdot r^{20} \)
- \( t_{25} = a \cdot r^{24} \)

### Step 3: Set up the product of the terms
Given that:
\[ t_1 \cdot t_5 \cdot t_{10} \cdot t_{16} \cdot t_{21} \cdot t_{25} = 512 \]
Substituting the expressions for the terms:
\[ a \cdot (a \cdot r^4) \cdot (a \cdot r^9) \cdot (a \cdot r^{15}) \cdot (a \cdot r^{20}) \cdot (a \cdot r^{24}) = 512 \]

### Step 4: Simplify the product
This simplifies to:
\[ a^6 \cdot r^{4 + 9 + 15 + 20 + 24} = 512 \]
Calculating the exponent of \( r \):
\[ 4 + 9 + 15 + 20 + 24 = 72 \]
Thus, we have:
\[ a^6 \cdot r^{72} = 512 \]

### Step 5: Express 512 in terms of powers
We know that:
\[ 512 = 2^9 \]
So we can write:
\[ a^6 \cdot r^{72} = 2^9 \]

### Step 6: Find the 13th term
The 13th term \( t_{13} \) is given by:
\[ t_{13} = a \cdot r^{12} \]
We can relate \( t_{13} \) to \( a^6 \cdot r^{72} \):
\[ t_{13}^6 = (a \cdot r^{12})^6 = a^6 \cdot r^{72} \]
Thus:
\[ t_{13}^6 = 2^9 \]

### Step 7: Solve for \( t_{13} \)
Taking the sixth root:
\[ t_{13} = (2^9)^{1/6} = 2^{9/6} = 2^{3/2} \]
This can be simplified to:
\[ t_{13} = 2 \cdot \sqrt{2} \]

### Final Answer
The value of the 13th term is:
\[ \boxed{2\sqrt{2}} \]