Key Concepts and Formulas

• Arithmetic Progression (AP): The general term is $a_n = a_1 + (n-1)d$, where $a_1$ is the first term and $d$ is the common difference. The sum of an AP can be found using specific formulas, but here we will express terms individually.
• Sum of Squares of First n Natural Numbers: $\sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6}$.
• Algebraic Manipulation: Solving systems of linear equations and simplifying expressions.

Step-by-Step Solution

Step 1: Utilize the given condition $a_9 + a_{43} = 66$ to establish a relationship between $a_1$ and $d$. We express $a_9$ and $a_{43}$ using the general formula for an AP: $a_9 = a_1 + (9-1)d = a_1 + 8d$ $a_{43} = a_1 + (43-1)d = a_1 + 42d$

Substituting these into the given equation: $(a_1 + 8d) + (a_1 + 42d) = 66$ $2a_1 + 50d = 66$ Dividing by 2, we get our first equation: $a_1 + 25d = 33 \quad (1)$

Step 2: Utilize the given condition $\sum_{k=0}^{12} a_{4k+1} = 416$ to establish another relationship between $a_1$ and $d$. The summation $\sum_{k=0}^{12} a_{4k+1}$ expands to: $a_{4(0)+1} + a_{4(1)+1} + a_{4(2)+1} + \dots + a_{4(12)+1}$ $= a_1 + a_5 + a_9 + \dots + a_{49}$ There are $12 - 0 + 1 = 13$ terms in this sum. We can express each term as: $a_1 = a_1 + 0d$ $a_5 = a_1 + 4d$ $a_9 = a_1 + 8d$ ... $a_{49} = a_1 + 48d$

The sum is: $(a_1) + (a_1 + 4d) + (a_1 + 8d) + \dots + (a_1 + 48d) = 416$ This can be rewritten as the sum of 13 $a_1$'s and the sum of the common differences: $13a_1 + (0d + 4d + 8d + \dots + 48d) = 416$ $13a_1 + d(0 + 4 + 8 + \dots + 48) = 416$ The series $0, 4, 8, \dots, 48$ is an AP with first term 0, common difference 4, and 13 terms. The sum of this series is $\frac{13}{2}(0 + 48) = \frac{13}{2}(48) = 13 \times 24 = 312$. So, the equation becomes: $13a_1 + 312d = 416$ Dividing by 13, we get our second equation: $a_1 + 24d = 32 \quad (2)$

Step 3: Solve the system of linear equations (1) and (2) for $a_1$ and $d$. We have:

1. $a_1 + 25d = 33$
2. $a_1 + 24d = 32$

Subtract equation (2) from equation (1): $(a_1 + 25d) - (a_1 + 24d) = 33 - 32$ $d = 1$

Substitute $d=1$ into equation (2): $a_1 + 24(1) = 32$ $a_1 + 24 = 32$ $a_1 = 32 - 24$ $a_1 = 8$

So, the first term $a_1 = 8$ and the common difference $d = 1$.

Step 4: Calculate the sum $a_1^2 + a_2^2 + \dots + a_{17}^2$. The terms are $a_n = a_1 + (n-1)d = 8 + (n-1)(1) = 8 + n - 1 = n+7$. So, the terms are: $a_1 = 1+7 = 8$ $a_2 = 2+7 = 9$ ... $a_{17} = 17+7 = 24$

We need to calculate $8^2 + 9^2 + 10^2 + \dots + 24^2$. This sum can be expressed as the sum of squares from 1 to 24 minus the sum of squares from 1 to 7: $\sum_{n=1}^{17} a_n^2 = \sum_{n=1}^{17} (n+7)^2 = 8^2 + 9^2 + \dots + 24^2$ $= (1^2 + 2^2 + \dots + 24^2) - (1^2 + 2^2 + \dots + 7^2)$

Using the formula $\sum_{i=1}^{N} i^2 = \frac{N(N+1)(2N+1)}{6}$: Sum of squares from 1 to 24: $\frac{24(24+1)(2 \times 24 + 1)}{6} = \frac{24(25)(49)}{6} = 4 \times 25 \times 49 = 100 \times 49 = 4900$

Sum of squares from 1 to 7: $\frac{7(7+1)(2 \times 7 + 1)}{6} = \frac{7(8)(15)}{6} = \frac{840}{6} = 140$

Therefore, $a_1^2 + a_2^2 + \dots + a_{17}^2 = 4900 - 140 = 4760$.

Step 5: Determine the value of $m$. We are given that $a_1^2 + a_2^2 + \dots + a_{17}^2 = 140m$. We found that $a_1^2 + a_2^2 + \dots + a_{17}^2 = 4760$. So, $4760 = 140m$. Dividing both sides by 140: $m = \frac{4760}{140} = \frac{476}{14} = 34$.

Common Mistakes & Tips

• Index Errors: Be careful when calculating the number of terms in a summation or the subscripts of terms. For example, $\sum_{k=0}^{12}$ has 13 terms.
• Formula Application: Ensure you are using the correct formula for the sum of squares and applying it to the correct range of numbers.
• Algebraic Errors: Double-check all arithmetic and algebraic manipulations, especially when solving systems of equations or simplifying fractions.

Summary The problem involves an arithmetic progression. We first used the given conditions to form a system of linear equations in terms of the first term ($a_1$) and the common difference ($d$). Solving this system yielded $a_1=8$ and $d=1$. Subsequently, we calculated the sum of squares of the first 17 terms by expressing them in terms of $n$ and using the formula for the sum of squares of natural numbers. Finally, we equated this sum to $140m$ to find the value of $m$.

The final answer is \boxed{34}.