Let `a_1,a_2,a_3` ….and `b_1 , b_2 , b_3 …` be two geometric progressions with `a_1= 2 sqrt(3)` and `b_1= 52/9 sqrt(3)` If `3a_99b_99=104` then find the value of `a_1 b_1+ a_2 b_2+…+a_nb_n`

Explanation

To solve the problem step by step, we will follow the given information and derive the required result.

### Step 1: Understand the given information
We have two geometric progressions (GPs):
- The first GP: \( a_1, a_2, a_3, \ldots \) with \( a_1 = 2\sqrt{3} \)
- The second GP: \( b_1, b_2, b_3, \ldots \) with \( b_1 = \frac{52}{9}\sqrt{3} \)

We also know that:
\[ 3a_{99}b_{99} = 104 \]

### Step 2: Express \( a_{99} \) and \( b_{99} \)
In a geometric progression, the \( n \)-th term can be expressed as:
- \( a_n = a_1 \cdot r_1^{n-1} \)
- \( b_n = b_1 \cdot r_2^{n-1} \)

Thus, we can express \( a_{99} \) and \( b_{99} \) as:
\[ a_{99} = a_1 \cdot r_1^{98} \]
\[ b_{99} = b_1 \cdot r_2^{98} \]

### Step 3: Substitute into the equation
Substituting \( a_{99} \) and \( b_{99} \) into the equation:
\[ 3a_{99}b_{99} = 3(a_1 \cdot r_1^{98})(b_1 \cdot r_2^{98}) = 104 \]

This simplifies to:
\[ 3(a_1 b_1) (r_1 r_2)^{98} = 104 \]

### Step 4: Substitute the values of \( a_1 \) and \( b_1 \)
Now substituting \( a_1 = 2\sqrt{3} \) and \( b_1 = \frac{52}{9}\sqrt{3} \):
\[ 3(2\sqrt{3} \cdot \frac{52}{9}\sqrt{3})(r_1 r_2)^{98} = 104 \]

Calculating \( a_1 b_1 \):
\[ a_1 b_1 = 2\sqrt{3} \cdot \frac{52}{9}\sqrt{3} = \frac{104}{9} \cdot 3 = \frac{312}{9} = \frac{104}{3} \]

### Step 5: Substitute \( a_1 b_1 \) back into the equation
Now substituting \( a_1 b_1 \) into the equation:
\[ 3 \cdot \frac{104}{3} (r_1 r_2)^{98} = 104 \]

This simplifies to:
\[ 104 (r_1 r_2)^{98} = 104 \]

### Step 6: Solve for \( r_1 r_2 \)
Dividing both sides by 104:
\[ (r_1 r_2)^{98} = 1 \]
Thus:
\[ r_1 r_2 = 1 \]

### Step 7: Find the sum \( a_1b_1 + a_2b_2 + \ldots + a_nb_n \)
The sum of the products of the corresponding terms in the two GPs can be expressed as:
\[ S_n = a_1b_1 + a_2b_2 + \ldots + a_nb_n \]

Using the formula for the sum of products in GPs:
\[ S_n = a_1b_1 \cdot \frac{1 - (r_1 r_2)^n}{1 - r_1 r_2} \]

Since \( r_1 r_2 = 1 \):
\[ S_n = a_1b_1 \cdot n \]

### Step 8: Substitute \( n = 99 \) and \( a_1 b_1 = \frac{104}{3} \)
Thus:
\[ S_{99} = \frac{104}{3} \cdot 99 \]

Calculating this gives:
\[ S_{99} = \frac{104 \cdot 99}{3} = \frac{10296}{3} = 3432 \]

### Final Answer
The value of \( a_1b_1 + a_2b_2 + \ldots + a_nb_n \) is:
\[ \boxed{3432} \]