Key Concepts and Formulas

• Arithmetico-Geometric Progression (AGP): A series where each term is the product of a term from an Arithmetic Progression (AP) and a term from a Geometric Progression (GP). The general form is $a, (a+d)r, (a+2d)r^2, \dots$.
• Sum of a finite AGP: The sum $S_n$ of a finite AGP $a + (a+d)r + \dots + (a+(n-1)d)r^{n-1}$ can be found by multiplying the series by $r$ and subtracting the result from the original series. This leads to a geometric series which can be summed using the formula $G_n = \frac{a_G(1-r_G^n)}{1-r_G}$ for a GP with first term $a_G$, common ratio $r_G$, and $n$ terms.

Step-by-Step Solution

1. Identify the AGP structure and its components: The given series is $S=109+\frac{108}{5}+\frac{107}{5^{2}}+\ldots .+\frac{2}{5^{107}}+\frac{1}{5^{108}}$. This series can be viewed as an Arithmetico-Geometric Progression (AGP). The numerators form an Arithmetic Progression (AP): $109, 108, 107, \ldots, 2, 1$.

   • The first term of the AP is $a = 109$.
   • The common difference of the AP is $d = 108 - 109 = -1$. The denominators involve powers of $5$, which form a Geometric Progression (GP) when considered as $1, \frac{1}{5}, \frac{1}{5^2}, \ldots, \frac{1}{5^{108}}$.
   • The common ratio of the GP is $r = \frac{1}{5}$.

   To find the number of terms ($n$), let's consider the AP: The $k$-th term of the AP is $a_k = a + (k-1)d$. For the last term, $1 = 109 + (n-1)(-1) \implies 1 = 109 - n + 1 \implies n = 109$. The GP terms are $(1/5)^0, (1/5)^1, \ldots, (1/5)^{108}$. The number of terms is $108 - 0 + 1 = 109$. Thus, the AGP has $n=109$ terms.

2. Apply the standard method to sum the AGP: Write the series $S$: $S = 109 + \frac{108}{5} + \frac{107}{5^2} + \ldots + \frac{2}{5^{107}} + \frac{1}{5^{108}} \quad \ldots(1)$ Multiply $S$ by the common ratio of the GP, $r = \frac{1}{5}$: $\frac{1}{5}S = \frac{109}{5} + \frac{108}{5^2} + \frac{107}{5^3} + \ldots + \frac{2}{5^{108}} + \frac{1}{5^{109}} \quad \ldots(2)$ Subtract equation (2) from equation (1) to eliminate most terms: $S - \frac{1}{5}S = \left(109 + \frac{108}{5} + \ldots + \frac{1}{5^{108}}\right) - \left(\frac{109}{5} + \frac{108}{5^2} + \ldots + \frac{1}{5^{109}}\right)$ Align terms with the same power of $5$: $\frac{4}{5}S = 109 + \left(\frac{108}{5} - \frac{109}{5}\right) + \left(\frac{107}{5^2} - \frac{108}{5^2}\right) + \ldots + \left(\frac{1}{5^{108}} - \frac{2}{5^{108}}\right) - \frac{1}{5^{109}}$ $\frac{4}{5}S = 109 - \frac{1}{5} - \frac{1}{5^2} - \frac{1}{5^3} - \ldots - \frac{1}{5^{108}} - \frac{1}{5^{109}}$ $\frac{4}{5}S = 109 - \left(\frac{1}{5} + \frac{1}{5^2} + \frac{1}{5^3} + \ldots + \frac{1}{5^{109}}\right)$

3. Sum the resulting Geometric Progression: The terms inside the parenthesis form a finite Geometric Progression: $\frac{1}{5}, \frac{1}{5^2}, \ldots, \frac{1}{5^{109}}$.

   • First term $a_G = \frac{1}{5}$.
   • Common ratio $r_G = \frac{1}{5}$.
   • Number of terms $n_G = 109$. The sum of this GP is: $G_{109} = \frac{a_G(1-r_G^{n_G})}{1-r_G} = \frac{\frac{1}{5}\left(1 - \left(\frac{1}{5}\right)^{109}\right)}{1 - \frac{1}{5}} = \frac{\frac{1}{5}\left(1 - \frac{1}{5^{109}}\right)}{\frac{4}{5}}$ $G_{109} = \frac{1}{4}\left(1 - \frac{1}{5^{109}}\right) = \frac{1}{4} - \frac{1}{4 \cdot 5^{109}}$ Substitute this sum back into the expression for $\frac{4}{5}S$: $\frac{4}{5}S = 109 - \left(\frac{1}{4} - \frac{1}{4 \cdot 5^{109}}\right)$ $\frac{4}{5}S = 109 - \frac{1}{4} + \frac{1}{4 \cdot 5^{109}}$ Combine the constant terms: $\frac{4}{5}S = \frac{436 - 1}{4} + \frac{1}{4 \cdot 5^{109}}$ $\frac{4}{5}S = \frac{435}{4} + \frac{1}{4 \cdot 5^{109}}$

4. Solve for S: Multiply both sides by $\frac{5}{4}$ to isolate $S$: $S = \frac{5}{4} \left(\frac{435}{4} + \frac{1}{4 \cdot 5^{109}}\right)$ $S = \frac{5 \cdot 435}{16} + \frac{5}{16 \cdot 5^{109}}$ $S = \frac{2175}{16} + \frac{1}{16 \cdot 5^{108}}$

5. Calculate the required expression: We need to find the value of $\left(16 S-(25)^{-54}\right)$. First, calculate $16S$: $16S = 16 \left(\frac{2175}{16} + \frac{1}{16 \cdot 5^{108}}\right)$ $16S = 2175 + \frac{1}{5^{108}}$ Next, simplify $(25)^{-54}$: $(25)^{-54} = (5^2)^{-54} = 5^{-108} = \frac{1}{5^{108}}$ Now, substitute these values into the expression: $16 S-(25)^{-54} = \left(2175 + \frac{1}{5^{108}}\right) - \frac{1}{5^{108}}$ $16 S-(25)^{-54} = 2175$

Common Mistakes & Tips

• Number of terms: Ensure the number of terms in both the AP and the resulting GP is correctly identified. Miscounting can lead to errors, especially with series starting from index 0 or 1.
• Subtraction of series: Be very careful when subtracting the two series ($S$ and $rS$). Align terms correctly and handle the first and last terms appropriately.
• Algebraic manipulation: Pay close attention to fractions and exponents during the simplification process. A small error can propagate and lead to a wrong final answer.

Summary

The given series is an Arithmetico-Geometric Progression. The sum $S$ was calculated by writing the series, multiplying by the common ratio ($1/5$), subtracting the second series from the first, and then summing the resulting geometric progression. After finding the expression for $S$, we calculated $16S$ and $(25)^{-54}$, and then found their difference, which resulted in a constant integer value.

The final answer is $\boxed{2175}$.