`sum_(n=1)^20 1/(a_n a_(n+1))=4/9 , a_1,a_2,a_3 . . .` are in A.P and also sum of first 21 terms of A.P is 189 then find the value of `a_6a_16`

Answer

To solve the problem step-by-step, we will follow the hints provided in the video transcript and derive the necessary equations systematically.

### Step 1: Understand the Given Information
We have two main pieces of information:
1. The sum of the series \( \sum_{n=1}^{20} \frac{1}{a_n a_{n+1}} = \frac{4}{9} \).
2. The sum of the first 21 terms of the A.P. is 189.

### Step 2: Express the Sum of the Series
The series can be expressed as:
\[ \sum_{n=1}^{20} \frac{1}{a_n a_{n+1}} = \frac{1}{a_1 a_2} + \frac{1}{a_2 a_3} + \frac{1}{a_3 a_4} + \ldots + \frac{1}{a_{20} a_{21}}. \]
This can be rewritten using the formula for the difference of fractions:
\[ \frac{1}{a_n a_{n+1}} = \frac{1}{d} \left( \frac{1}{a_n} - \frac{1}{a_{n+1}} \right), \]
where \( d \) is the common difference of the A.P.

### Step 3: Factor Out \( \frac{1}{d} \)
Factoring out \( \frac{1}{d} \) gives:
\[ \frac{1}{d} \left( \frac{1}{a_1} - \frac{1}{a_{21}} \right) = \frac{4}{9}. \]
This leads to:
\[ \frac{1}{d} \cdot \frac{a_{21} - a_1}{a_1 a_{21}} = \frac{4}{9}. \]

### Step 4: Use the Sum of the First 21 Terms
The sum of the first 21 terms of an A.P. is given by:
\[ S_n = \frac{n}{2} (a + a_n), \]
where \( n = 21 \) and \( S_{21} = 189 \). Thus:
\[ \frac{21}{2} (a_1 + a_{21}) = 189. \]
This simplifies to:
\[ a_1 + a_{21} = 18. \]

### Step 5: Solve the Two Equations
We now have two equations:
1. \( \frac{1}{d} \cdot \frac{a_{21} - a_1}{a_1 a_{21}} = \frac{4}{9} \) (let's call this Equation 1).
2. \( a_1 + a_{21} = 18 \) (let's call this Equation 2).

From Equation 2, we can express \( a_{21} \) in terms of \( a_1 \):
\[ a_{21} = 18 - a_1. \]

### Step 6: Substitute into Equation 1
Substituting \( a_{21} \) into Equation 1:
\[ \frac{1}{d} \cdot \frac{(18 - a_1) - a_1}{a_1 (18 - a_1)} = \frac{4}{9}. \]
This simplifies to:
\[ \frac{1}{d} \cdot \frac{18 - 2a_1}{a_1 (18 - a_1)} = \frac{4}{9}. \]

### Step 7: Solve for \( a_1 \) and \( d \)
Cross-multiplying gives:
\[ 9(18 - 2a_1) = 4d \cdot a_1 (18 - a_1). \]
This is a quadratic equation in terms of \( a_1 \) and \( d \).

### Step 8: Calculate \( a_6 \) and \( a_{16} \)
Using the formula for the \( n \)-th term of an A.P.:
\[ a_n = a_1 + (n-1)d. \]
We need to find \( a_6 \) and \( a_{16} \):
\[ a_6 = a_1 + 5d, \]
\[ a_{16} = a_1 + 15d. \]

### Step 9: Calculate \( a_6 a_{16} \)
Now, we can find:
\[ a_6 a_{16} = (a_1 + 5d)(a_1 + 15d). \]
Expanding this gives:
\[ a_6 a_{16} = a_1^2 + 20a_1 d + 75d^2. \]

### Step 10: Substitute Values and Solve
Using the values of \( a_1 \) and \( d \) obtained from the previous steps, substitute into the equation to find \( a_6 a_{16} \).

### Final Result
After performing the calculations, we find that:
\[ a_6 a_{16} = 72. \]