Key Concepts and Formulas

• Arithmetic Progression (AP): A sequence where the difference between consecutive terms is constant. The $n$-th term is given by $a_n = a_1 + (n-1)d$, where $a_1$ is the first term and $d$ is the common difference.
• Difference of Squares: The algebraic identity $x^2 - y^2 = (x-y)(x+y)$ is fundamental for simplifying the expression for $A_k$.
• Sum of an AP: The sum of the first $n$ terms of an AP is $S_n = \frac{n}{2}(a_1 + a_n) = \frac{n}{2}(2a_1 + (n-1)d)$.

Step-by-Step Solution

Step 1: Simplify the expression for $A_k$ The given expression for $A_k$ is $A_k=a_1^2-a_2^2+a_3^2-a_4^2+\ldots+a_{2 k-1}^2-a_{2 k}^2$. We can group terms in pairs and apply the difference of squares formula: $A_k = (a_1^2 - a_2^2) + (a_3^2 - a_4^2) + \ldots + (a_{2k-1}^2 - a_{2k}^2)$ $A_k = (a_1 - a_2)(a_1 + a_2) + (a_3 - a_4)(a_3 + a_4) + \ldots + (a_{2k-1} - a_{2k})(a_{2k-1} + a_{2k})$ In an AP, the difference between consecutive terms is the common difference $d$, so $a_n - a_{n+1} = -d$. Substituting this into the expression for $A_k$: $A_k = (-d)(a_1 + a_2) + (-d)(a_3 + a_4) + \ldots + (-d)(a_{2k-1} + a_{2k})$ $A_k = -d[(a_1 + a_2) + (a_3 + a_4) + \ldots + (a_{2k-1} + a_{2k})]$ The sum inside the bracket is the sum of the first $2k$ terms of the AP. Let $a_1 = a$. The sum of the first $2k$ terms is $S_{2k} = \frac{2k}{2}(2a + (2k-1)d) = k(2a + (2k-1)d)$. Therefore, $A_k = -d \cdot k(2a + (2k-1)d) = -dk(2a + (2k-1)d) \quad (*)$

Step 2: Use the given values of $A_3$ and $A_5$ to form equations We are given $A_3 = -153$ and $A_5 = -435$. Using the formula $(*)$: For $k=3$: $A_3 = -3d(2a + (2 \cdot 3 - 1)d) = -3d(2a + 5d)$ $-3d(2a + 5d) = -153$ $d(2a + 5d) = 51 \quad (1)$ For $k=5$: $A_5 = -5d(2a + (2 \cdot 5 - 1)d) = -5d(2a + 9d)$ $-5d(2a + 9d) = -435$ $d(2a + 9d) = 87 \quad (2)$

Step 3: Solve the system of equations for $a$ and $d$ We have two equations:

1. $2ad + 5d^2 = 51$
2. $2ad + 9d^2 = 87$ Subtract equation (1) from equation (2): $(2ad + 9d^2) - (2ad + 5d^2) = 87 - 51$ $4d^2 = 36$ $d^2 = 9$ Since the terms of the AP are positive, $a_n > 0$ for all $n$. If $d$ were negative, eventually terms would become negative. For instance, if $a=1$ and $d=-3$, then $a_1=1, a_2=-2$. Thus, we must have $d > 0$ to maintain positive terms, especially if $a_1$ is small. So, $d = 3$. Substitute $d=3$ into equation (1): $3(2a + 5(3)) = 51$ $3(2a + 15) = 51$ $2a + 15 = 17$ $2a = 2$ $a = 1$ So, $a_1 = 1$ and $d = 3$.

Step 4: Verify the condition $a_1^2 + a_2^2 + a_3^2 = 66$ With $a_1 = 1$ and $d = 3$: $a_1 = 1$ $a_2 = a_1 + d = 1 + 3 = 4$ $a_3 = a_1 + 2d = 1 + 2(3) = 7$ $a_1^2 + a_2^2 + a_3^2 = 1^2 + 4^2 + 7^2 = 1 + 16 + 49 = 66$ The condition is satisfied.

Step 5: Calculate $a_{17}$ and $A_7$ The 17th term of the AP is: $a_{17} = a_1 + (17-1)d = 1 + 16(3) = 1 + 48 = 49$ Now, calculate $A_7$ using the formula $(*)$ with $k=7$, $a=1$, and $d=3$: $A_7 = -7d(2a + (2 \cdot 7 - 1)d)$ $A_7 = -7(3)(2(1) + 13(3))$ $A_7 = -21(2 + 39)$ $A_7 = -21(41)$ $A_7 = -861$

Step 6: Calculate $a_{17} - A_7$ $a_{17} - A_7 = 49 - (-861)$ $a_{17} - A_7 = 49 + 861$ $a_{17} - A_7 = 910$

Common Mistakes & Tips

• Sign Errors: Be very careful with signs, especially when simplifying $a_n - a_{n+1}$ and when substituting negative values (though in this problem $d$ turns out to be positive).
• Algebraic Manipulation: Ensure accuracy when solving the system of equations. Errors in expansion or subtraction can lead to incorrect values for $a$ and $d$.
• Formula Application: Double-check the correct application of the AP sum formula and the derived formula for $A_k$.

Summary

The problem requires simplifying the expression for $A_k$ using the difference of squares, which leads to a formula involving the first term ($a$) and the common difference ($d$). By using the given values of $A_3$ and $A_5$, we form a system of linear equations in terms of $a$ and $d$, which we solve to find their values. The condition on the sum of squares of the first three terms is used to verify these values. Finally, we calculate $a_{17}$ and $A_7$ using the determined values of $a$ and $d$, and then compute their difference.

The final answer is $\boxed{910}$.