1. Key Concepts and Formulas

• Arithmetic Progression (A.P.): A sequence where the difference between consecutive terms is constant.
  • $n^{th}$ term: $a_n = a_1 + (n-1)d$, where $a_1$ is the first term and $d$ is the common difference.
  • Sum of the first $n$ terms: $S_n = \frac{n}{2}(2a_1 + (n-1)d)$.
• Ratio of Sums: The problem involves a ratio of sums of terms in an A.P., which will be used to establish a relationship between $a_1$ and $d$.

2. Step-by-Step Solution

Step 1: Express the given ratio using the sum formula. We are given the ratio: $\frac{a_1 + a_2 + \dots + a_{10}}{a_1 + a_2 + \dots + a_p} = \frac{100}{p^2}$ Using the sum formula $S_n = \frac{n}{2}(2a_1 + (n-1)d)$, we can write $S_{10}$ and $S_p$: $S_{10} = \frac{10}{2}(2a_1 + (10-1)d) = 5(2a_1 + 9d)$ $S_p = \frac{p}{2}(2a_1 + (p-1)d)$ Substituting these into the given ratio: $\frac{5(2a_1 + 9d)}{\frac{p}{2}(2a_1 + (p-1)d)} = \frac{100}{p^2}$

Step 2: Simplify the ratio and derive a relationship between $a_1$ and $d$. $\frac{10(2a_1 + 9d)}{p(2a_1 + (p-1)d)} = \frac{100}{p^2}$ Cross-multiplying and simplifying: $10(2a_1 + 9d) \cdot p^2 = 100 \cdot p(2a_1 + (p-1)d)$ Since $p \ne 10$ and we can assume $p \ne 0$ (as it represents the number of terms), we can divide both sides by $10p$: $p(2a_1 + 9d) = 10(2a_1 + (p-1)d)$ Expand both sides: $2pa_1 + 9pd = 20a_1 + 10pd - 10d$ Group terms with $a_1$ on one side and terms with $d$ on the other: $2pa_1 - 20a_1 = 10pd - 9pd - 10d$ $a_1(2p - 20) = d(p - 10)$ $2a_1(p - 10) = d(p - 10)$ Given that $p \ne 10$, $(p-10) \ne 0$. Therefore, we can divide both sides by $(p-10)$: $2a_1 = d$ This implies $\frac{a_1}{d} = \frac{1}{2}$ (assuming $d \ne 0$. If $d=0$, then $a_1=0$, making all terms zero, which leads to an undefined ratio in the problem statement, so $d \ne 0$ is implied).

Step 3: Calculate the desired ratio $\frac{a_{11}}{a_{10}}$. Using the formula for the $n^{th}$ term, $a_n = a_1 + (n-1)d$: $a_{11} = a_1 + (11-1)d = a_1 + 10d$ $a_{10} = a_1 + (10-1)d = a_1 + 9d$ The ratio we need to find is: $\frac{a_{11}}{a_{10}} = \frac{a_1 + 10d}{a_1 + 9d}$ Substitute the relationship $d = 2a_1$ into this expression: $\frac{a_{11}}{a_{10}} = \frac{a_1 + 10(2a_1)}{a_1 + 9(2a_1)}$ $\frac{a_{11}}{a_{10}} = \frac{a_1 + 20a_1}{a_1 + 18a_1}$ $\frac{a_{11}}{a_{10}} = \frac{21a_1}{19a_1}$ Since $a_1 \ne 0$ (as $d=2a_1$ and $d \ne 0$), we can cancel $a_1$: $\frac{a_{11}}{a_{10}} = \frac{21}{19}$

3. Common Mistakes & Tips

• Algebraic Errors: Be extremely careful when expanding and rearranging terms in the equation involving $a_1$ and $d$. A small mistake can lead to an incorrect relationship.
• Division by Zero: The condition $p \ne 10$ is critical. It allows us to divide by $(p-10)$ to uniquely determine the relationship between $a_1$ and $d$. If $p=10$, the original equation would be $S_{10}/S_{10} = 100/10^2$, which simplifies to $1=1$, providing no information about $a_1$ and $d$.
• Expressing Terms: When calculating the final ratio, substituting $d=2a_1$ is one way. Alternatively, dividing the numerator and denominator by $d$ and using $a_1/d = 1/2$ can also be efficient.

4. Summary

The problem requires us to find the ratio of two consecutive terms in an Arithmetic Progression given a ratio of sums. We first used the formula for the sum of an A.P. to express the given ratio in terms of $a_1$ and $d$. Through careful algebraic manipulation, and utilizing the condition $p \ne 10$, we established a fundamental relationship between the first term and the common difference: $d = 2a_1$. Finally, we substituted this relationship into the expression for the desired ratio $\frac{a_{11}}{a_{10}}$, which simplified to $\frac{21}{19}$.

The final answer is \boxed{\frac{21}{19}} which corresponds to option (A).