Let `T_(r)` be the `r^(th)` term of a sequence, for r=1,2,3,......... If `3T_(r+1)=T_(r)` and `T_(7)=(1)/(243)` then the value of `sum_(r=1)^(oo)(T_(r).T_(r+1))` is :

Options

A. `(9)/(2)`
B. `(27)/(8)`
C. `(81)/(8)`
D. `(81)/(4)`

Explanation

To solve the problem, we need to analyze the given sequence defined by the recurrence relation \(3T_{r+1} = T_r\) and the value of \(T_7 = \frac{1}{243}\).

### Step 1: Find the general term of the sequence

From the relation \(3T_{r+1} = T_r\), we can express \(T_{r+1}\) in terms of \(T_r\):
\[ T_{r+1} = \frac{T_r}{3} \]
This indicates that the sequence is a geometric progression (GP) with a common ratio of \(\frac{1}{3}\).

### Step 2: Express \(T_r\) in terms of \(T_1\)

Using the recurrence relation, we can express the terms of the sequence:
- \(T_2 = \frac{T_1}{3}\)
- \(T_3 = \frac{T_2}{3} = \frac{T_1}{3^2}\)
- \(T_4 = \frac{T_3}{3} = \frac{T_1}{3^3}\)
- Continuing this pattern, we find:
\[ T_r = \frac{T_1}{3^{r-1}} \]

### Step 3: Use the value of \(T_7\) to find \(T_1\)

We know that \(T_7 = \frac{1}{243}\). Using our expression for \(T_r\):
\[ T_7 = \frac{T_1}{3^{7-1}} = \frac{T_1}{3^6} \]
Setting this equal to \(\frac{1}{243}\):
\[ \frac{T_1}{729} = \frac{1}{243} \]
Since \(3^6 = 729\) and \(243 = 3^5\), we can rewrite the equation:
\[ T_1 = \frac{729}{243} = 3 \]

### Step 4: Write the general term explicitly

Now that we have \(T_1\):
\[ T_r = \frac{3}{3^{r-1}} = \frac{3}{3^{r-1}} = 3^{2-r} \]

### Step 5: Calculate \(T_r \cdot T_{r+1}\)

Next, we need to find the product \(T_r \cdot T_{r+1}\):
\[ T_{r+1} = \frac{T_r}{3} = 3^{2-(r+1)} = 3^{1-r} \]
Thus,
\[ T_r \cdot T_{r+1} = 3^{2-r} \cdot 3^{1-r} = 3^{(2-r) + (1-r)} = 3^{3-2r} \]

### Step 6: Find the sum \( \sum_{r=1}^{\infty} T_r \cdot T_{r+1} \)

We need to find:
\[ \sum_{r=1}^{\infty} T_r \cdot T_{r+1} = \sum_{r=1}^{\infty} 3^{3-2r} \]
This is a geometric series with the first term \(3^{3-2 \cdot 1} = 3^{1} = 3\) and a common ratio \(r = \frac{1}{9}\) (since \(3^{-2} = \frac{1}{9}\)):
\[ \text{Sum} = \frac{a}{1 - r} = \frac{3}{1 - \frac{1}{9}} = \frac{3}{\frac{8}{9}} = \frac{3 \cdot 9}{8} = \frac{27}{8} \]

### Final Answer

Thus, the value of \(\sum_{r=1}^{\infty} T_r \cdot T_{r+1}\) is:
\[ \boxed{\frac{27}{8}} \]