Key Concepts and Formulas

• Arithmetic Progression (AP): A sequence where the difference between consecutive terms is constant (common difference, $d$). The $n$-th term is $a_n = a_1 + (n-1)d$. The sum of the first $N$ terms is $S_N = \frac{N}{2}(a_1 + a_N)$.
• Telescoping Sums: A series where intermediate terms cancel out, leaving only the first and last terms. The identity $\frac{1}{a_n a_{n+1}} = \frac{1}{d} \left( \frac{1}{a_n} - \frac{1}{a_{n+1}} \right)$ is key for sums of reciprocals of products in an AP.
• Symmetry of AP: For an AP with an odd number of terms, the middle term is the average of the first and last terms. Also, $a_k + a_{N-k+1} = a_1 + a_N$. For terms equidistant from the middle, their product can be expressed as $a_{mid}^2 - (kd)^2$.

Step-by-Step Solution

Step 1: Simplify the given summation using the telescoping sum technique. We are given the sum $\sum\limits_{n = 1}^{20} {{1 \over {{a_n}{a_{n + 1}}}} = {4 \over 9}}$. For an AP, $a_{n+1} - a_n = d$. Consider the term $\frac{1}{a_n a_{n+1}}$. We can rewrite it as: $\frac{1}{a_n a_{n+1}} = \frac{1}{d} \left( \frac{a_{n+1} - a_n}{a_n a_{n+1}} \right) = \frac{1}{d} \left( \frac{1}{a_n} - \frac{1}{a_{n+1}} \right)$ Now, apply this to the sum: $\sum\limits_{n = 1}^{20} {{1 \over {{a_n}{a_{n + 1}}}}} = \sum\limits_{n = 1}^{20} {\frac{1}{d} \left( \frac{1}{a_n} - \frac{1}{a_{n+1}} \right)}$ Factor out the constant $\frac{1}{d}$: $= \frac{1}{d} \sum\limits_{n = 1}^{20} {\left( \frac{1}{a_n} - \frac{1}{a_{n+1}} \right)}$ This is a telescoping sum: $= \frac{1}{d} \left[ \left(\frac{1}{a_1} - \frac{1}{a_2}\right) + \left(\frac{1}{a_2} - \frac{1}{a_3}\right) + \dots + \left(\frac{1}{a_{20}} - \frac{1}{a_{21}}\right) \right]$ The intermediate terms cancel out, leaving: $= \frac{1}{d} \left( \frac{1}{a_1} - \frac{1}{a_{21}} \right)$ Combine the terms inside the parenthesis: $= \frac{1}{d} \left( \frac{a_{21} - a_1}{a_1 a_{21}} \right)$ Since $a_{21} = a_1 + 20d$, we have $a_{21} - a_1 = 20d$. Substituting this: $= \frac{1}{d} \left( \frac{20d}{a_1 a_{21}} \right) = \frac{20}{a_1 a_{21}}$ We are given that this sum is $\frac{4}{9}$: $\frac{20}{a_1 a_{21}} = \frac{4}{9}$ Solving for $a_1 a_{21}$: $a_1 a_{21} = \frac{20 \times 9}{4} = 5 \times 9 = 45$

Step 2: Use the sum of the AP to find the sum of the first and last terms. The AP has 21 terms ($a_1, a_2, \dots, a_{21}$). The sum is given as $S_{21} = 189$. The sum formula for an AP is $S_N = \frac{N}{2}(a_1 + a_N)$. For $N=21$: $S_{21} = \frac{21}{2}(a_1 + a_{21})$ Substitute the given sum: $189 = \frac{21}{2}(a_1 + a_{21})$ Solving for $a_1 + a_{21}$: $a_1 + a_{21} = \frac{189 \times 2}{21} = 9 \times 2 = 18$

Step 3: Determine the properties of the AP using $a_1 a_{21}$ and $a_1 + a_{21}$. We have two equations:

1. $a_1 a_{21} = 45$
2. $a_1 + a_{21} = 18$ Consider the middle term of the AP, which is $a_{11}$ (since there are 21 terms). For an AP with an odd number of terms, the middle term is the average of the first and last terms: $a_{11} = \frac{a_1 + a_{21}}{2} = \frac{18}{2} = 9$ Now, express $a_1$ and $a_{21}$ in terms of $a_{11}$ and the common difference $d$: $a_1 = a_{11} - (11-1)d = a_{11} - 10d$ $a_{21} = a_{11} + (21-11)d = a_{11} + 10d$ Their product is: $a_1 a_{21} = (a_{11} - 10d)(a_{11} + 10d) = a_{11}^2 - (10d)^2$ Substitute the known values $a_1 a_{21} = 45$ and $a_{11} = 9$: $45 = 9^2 - (10d)^2$ $45 = 81 - 100d^2$ Solve for $d^2$: $100d^2 = 81 - 45 = 36$ $d^2 = \frac{36}{100} = \frac{9}{25}$

Step 4: Calculate the product $a_6 a_{16}$. We need to find $a_6 a_{16}$. Express $a_6$ and $a_{16}$ in terms of $a_{11}$ and $d$: $a_6 = a_{11} + (6-11)d = a_{11} - 5d$ $a_{16} = a_{11} + (16-11)d = a_{11} + 5d$ Now, calculate their product: $a_6 a_{16} = (a_{11} - 5d)(a_{11} + 5d) = a_{11}^2 - (5d)^2$ Substitute the values $a_{11}=9$ and $d^2 = \frac{9}{25}$: $a_6 a_{16} = 9^2 - 25d^2$ $a_6 a_{16} = 81 - 25 \left( \frac{9}{25} \right)$ $a_6 a_{16} = 81 - 9$ $a_6 a_{16} = 72$

Common Mistakes & Tips

• Assuming $a_6 a_{16} = a_1 a_{21}$: While $6+16 = 1+21$, this property applies to the sum of terms, not their product. Always calculate the product explicitly using the common difference or middle term.
• Forgetting the $\frac{1}{d}$ factor: The telescoping sum identity for reciprocals of products in an AP inherently involves the common difference $d$. Ensure it's included and handled correctly.
• Division by zero: Verify that $d \ne 0$ and no $a_n = 0$. If $d=0$, all terms are equal, say $a$. $S_{21}=21a=189 \implies a=9$. Then $\sum \frac{1}{a^2} = \frac{20}{81} \ne \frac{4}{9}$, so $d \ne 0$. Since $a_1 a_{21} = 45 \ne 0$, neither $a_1$ nor $a_{21}$ is zero.

Summary

This problem requires a combination of understanding telescoping sums and the properties of arithmetic progressions. By first simplifying the given summation to relate $a_1$ and $a_{21}$, and then using the sum of the AP to find $a_1 + a_{21}$, we can determine the middle term $a_{11}$ and the square of the common difference $d^2$. Finally, expressing the required product $a_6 a_{16}$ in terms of $a_{11}$ and $d$ allows for its calculation.

The final answer is $\boxed{72}$.