Key Concepts and Formulas

• Arithmetic Progression (AP) Formulas:
  • The $n$-th term: $a_n = a_1 + (n-1)d$, where $a_1$ is the first term and $d$ is the common difference.
  • The sum of the first $n$ terms: $S_n = \frac{n}{2}[2a_1 + (n-1)d]$.
• Natural Numbers: The set of natural numbers is $\{1, 2, 3, \ldots\}$. This constraint implies that $a_1$ and $d$ must be such that all terms $a_i$ are positive integers.

Step-by-Step Solution

Step 1: Use the given ratio of sums to establish a relationship between $a_1$ and $d$. We are given that the ratio of the sum of the first five terms ($S_5$) to the sum of the first nine terms ($S_9$) is $5:17$. $\frac{S_5}{S_9} = \frac{5}{17}$ Using the formula $S_n = \frac{n}{2}[2a_1 + (n-1)d]$: $S_5 = \frac{5}{2}[2a_1 + (5-1)d] = \frac{5}{2}[2a_1 + 4d]$ $S_9 = \frac{9}{2}[2a_1 + (9-1)d] = \frac{9}{2}[2a_1 + 8d]$ Substituting these into the ratio equation: $\frac{\frac{5}{2}(2a_1 + 4d)}{\frac{9}{2}(2a_1 + 8d)} = \frac{5}{17}$ Cancel out the $\frac{1}{2}$ from the numerator and denominator, and the factor of $5$ from the left side with the $5$ on the right side: $\frac{2a_1 + 4d}{9(2a_1 + 8d)} = \frac{1}{17}$ Cross-multiply: $17(2a_1 + 4d) = 9(2a_1 + 8d)$ $34a_1 + 68d = 18a_1 + 72d$ Rearrange the terms to group $a_1$ and $d$: $34a_1 - 18a_1 = 72d - 68d$ $16a_1 = 4d$ Divide by $4$ to get the relationship between $a_1$ and $d$: $d = 4a_1$

Step 2: Use the given range of the 15th term to find the value of $a_1$. We are given that $110 < a_{15} < 120$. The formula for the $n$-th term is $a_n = a_1 + (n-1)d$. For $n=15$: $a_{15} = a_1 + (15-1)d = a_1 + 14d$ Substitute the relationship $d = 4a_1$ into the expression for $a_{15}$: $a_{15} = a_1 + 14(4a_1)$ $a_{15} = a_1 + 56a_1$ $a_{15} = 57a_1$ Now, substitute this into the given inequality: $110 < 57a_1 < 120$ Divide all parts of the inequality by $57$: $\frac{110}{57} < a_1 < \frac{120}{57}$ Calculating the approximate values: $1.929... < a_1 < 2.105...$ Since $a_1$ must be a natural number, the only possible integer value for $a_1$ in this range is $2$. $a_1 = 2$

Step 3: Calculate the common difference $d$. Using the relationship $d = 4a_1$ and the value $a_1 = 2$: $d = 4(2)$ $d = 8$ The first term is $a_1=2$ and the common difference is $d=8$. Since $a_1$ is a natural number and $d$ is a positive integer, all terms of the AP will be natural numbers ($a_n = 2 + (n-1)8$).

Step 4: Calculate the sum of the first ten terms ($S_{10}$). We need to find $S_{10}$ using $a_1 = 2$, $d = 8$, and $n = 10$: $S_{10} = \frac{10}{2}[2a_1 + (10-1)d]$ $S_{10} = 5[2(2) + 9(8)]$ $S_{10} = 5[4 + 72]$ $S_{10} = 5[76]$ $S_{10} = 380$

Common Mistakes & Tips

• Natural Number Constraint: Always remember that terms are natural numbers. This is critical for selecting the correct integer value for $a_1$.
• Algebraic Simplification: Simplify fractions and common factors in the ratio equation early to avoid complex calculations.
• Formula Accuracy: Ensure correct application of the $n$-th term and sum formulas, particularly the $(n-1)$ factor for the common difference.

Summary

We utilized the given ratio of sums of an arithmetic progression to derive a relationship between the first term ($a_1$) and the common difference ($d$), finding $d=4a_1$. Subsequently, we used the information about the 15th term's range and the derived relationship to determine the precise value of $a_1$. Given that $a_1$ must be a natural number, we found $a_1=2$, which in turn yielded $d=8$. Finally, we calculated the sum of the first ten terms ($S_{10}$) using these values, resulting in $S_{10}=380$.

The final answer is $\boxed{380}$.