Sequence `{t_n}` of positive terms is a G.P If `t_6 2, 5, t_14` form another G.P in that order then the product `t_1 t_2 t_3........t_18 t_19` is equal to

Options

A. `10^(9)`
B. `10 ^(10)`
C. `10^(17//2)`
D. `10 ^(19//2)`

Answer

To solve the problem step by step, we will follow the reasoning outlined in the video transcript.

### Step 1: Define the terms of the G.P.
Let the first term of the G.P. be \( a \) and the common ratio be \( r \). Therefore, we can express the terms as follows:
- \( t_1 = a \)
- \( t_2 = ar \)
- \( t_3 = ar^2 \)
- \( t_4 = ar^3 \)
- \( t_5 = ar^4 \)
- \( t_6 = ar^5 \)
- \( t_{14} = ar^{13} \)

### Step 2: Set up the equations based on the given conditions.
According to the problem, \( t_6, t_2, t_5, t_{14} \) form another G.P. This means that the ratio of consecutive terms is constant. Therefore, we can write:

1. From \( t_6, t_2, t_5 \):
\[ \left( \frac{t_2}{t_6} \right) = \left( \frac{t_5}{t_2} \right) \]
This gives us:
\[ \frac{ar}{ar^5} = \frac{ar^4}{ar} \]
Simplifying this, we get:
\[ \frac{1}{r^4} = r^3 \implies 1 = r^7 \implies r = 1 \]

2. From \( t_2, t_5, t_{14} \):
\[ \left( \frac{t_5}{t_2} \right) = \left( \frac{t_{14}}{t_5} \right) \]
This gives us:
\[ \frac{ar^4}{ar} = \frac{ar^{13}}{ar^4} \]
Simplifying this, we get:
\[ r^3 = r^9 \implies 1 = r^6 \implies r = 1 \]

### Step 3: Calculate the product \( t_1 t_2 t_3 \ldots t_{19} \).
Since we have established that \( r = 1 \), all terms are equal to \( a \):
- \( t_1 = a \)
- \( t_2 = a \)
- \( t_3 = a \)
- ...
- \( t_{19} = a \)

Thus, the product can be calculated as:
\[ t_1 t_2 t_3 \ldots t_{19} = a^{19} \]

### Step 4: Find the value of \( a \).
From the earlier equations, we can find \( a \) using the conditions given in the problem:
1. From \( t_6 \) and \( t_2 \):
\[ 2^2 = 5 \cdot t_6 \implies 4 = 5 \cdot ar^5 \implies ar^5 = \frac{4}{5} \]

2. From \( t_2 \) and \( t_{14} \):
\[ 5^2 = 2 \cdot t_{14} \implies 25 = 2 \cdot ar^{13} \implies ar^{13} = \frac{25}{2} \]

### Step 5: Solve for \( a \).
Using the equations:
1. \( ar^5 = \frac{4}{5} \)
2. \( ar^{13} = \frac{25}{2} \)

Since \( r = 1 \):
- \( a = \frac{4}{5} \)

### Step 6: Substitute back to find the product.
Now substituting \( a \) back into the product:
\[ t_1 t_2 t_3 \ldots t_{19} = a^{19} = \left(\frac{4}{5}\right)^{19} \]

### Final Answer
The product \( t_1 t_2 t_3 \ldots t_{19} \) is equal to \( \left(\frac{4}{5}\right)^{19} \).