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, we need to find the value of \( S_n = a_1b_1 + a_2b_2 + a_3b_3 + \ldots + a_nb_n \) given that \( a_1 = 2\sqrt{3} \), \( b_1 = \frac{52}{9}\sqrt{3} \), and \( 3a_{99}b_{99} = 104 \).

### Step 1: Understand the Geometric Progressions
We know that both \( a_n \) and \( b_n \) are terms of geometric progressions (GP). The \( n \)-th term of a GP can be expressed as:
\[ a_n = a_1 r_1^{n-1} \]
\[ b_n = b_1 r_2^{n-1} \]
where \( r_1 \) and \( r_2 \) are the common ratios of the GPs for \( a_n \) and \( b_n \), respectively.

### Step 2: Find \( a_{99} \) and \( b_{99} \)
Using the formulas for the \( n \)-th terms:
\[ a_{99} = a_1 r_1^{98} = 2\sqrt{3} \cdot r_1^{98} \]
\[ b_{99} = b_1 r_2^{98} = \frac{52}{9}\sqrt{3} \cdot r_2^{98} \]

### Step 3: Use the Given Condition
We know from the problem statement:
\[ 3a_{99}b_{99} = 104 \]
Substituting the expressions for \( a_{99} \) and \( b_{99} \):
\[ 3 \left( 2\sqrt{3} \cdot r_1^{98} \right) \left( \frac{52}{9}\sqrt{3} \cdot r_2^{98} \right) = 104 \]

### Step 4: Simplify the Equation
This simplifies to:
\[ 3 \cdot 2 \cdot \frac{52}{9} \cdot 3 \cdot r_1^{98} \cdot r_2^{98} = 104 \]
\[ \frac{312}{9} \cdot r_1^{98} \cdot r_2^{98} = 104 \]
Multiplying both sides by 9:
\[ 312 \cdot r_1^{98} \cdot r_2^{98} = 936 \]
Dividing both sides by 312:
\[ r_1^{98} \cdot r_2^{98} = 3 \]

### Step 5: Find \( r_1 r_2 \)
Taking the 98th root:
\[ r_1 r_2 = 3^{1/98} \]

### Step 6: Calculate \( S_n \)
Now, we can find \( S_n \):
\[ S_n = a_1b_1 + a_2b_2 + a_3b_3 + \ldots + a_nb_n \]
\[ = \sum_{i=1}^{n} a_i b_i = \sum_{i=1}^{n} (a_1 r_1^{i-1})(b_1 r_2^{i-1}) = a_1b_1 \sum_{i=1}^{n} (r_1 r_2)^{i-1} \]
Using \( r_1 r_2 = 3^{1/98} \):
\[ = a_1b_1 \sum_{i=0}^{n-1} (3^{1/98})^i \]
This is a geometric series with first term 1 and common ratio \( 3^{1/98} \):
\[ = a_1b_1 \cdot \frac{1 - (3^{1/98})^n}{1 - 3^{1/98}} \]

### Step 7: Substitute Values
Now substituting \( a_1 \) and \( b_1 \):
\[ = 2\sqrt{3} \cdot \frac{52}{9}\sqrt{3} \cdot \frac{1 - (3^{1/98})^n}{1 - 3^{1/98}} \]
Calculating \( a_1b_1 \):
\[ = \frac{104 \cdot 3}{9} = \frac{312}{9} \]

### Final Step: Conclude
Thus, the value of \( S_n \) becomes:
\[ S_n = \frac{104n}{3} \]