Question

Let \( a_1, a_2, a_3, \dots \) be in an arithmetic progression of positive terms.
Let \( A_k = a_1^2 - a_2^2 + a_3^2 - a_4^2 + \dots + a_{2k-1}^2 - a_{2k}^2 \).  
If \( A_3 = -153 \), \( A_5 = -435 \), and \( a_1^2 + a_2^2 + a_3^2 = 66 \), then \( a_{17} - A_7 \) is equal to _________.

Answer 1

Correct Answer: 910

Given:

\[ d \rightarrow \text{common difference.} \]

The general term:

\[ A_k = kd \left[ 2a + (2k - 1)d \right] \]

Given:

\[ A_3 = -153 \] \[ \Rightarrow 153 = 13d \left[ 2a + 5d \right] \]

Simplifying:

\[ 51 = d \left[ 2a + 5d \right] \quad \dots (1) \]

Also, given:

\[ A_5 = -435 \] \[ 435 = 5d \left[ 2a + 9d \right] \]

Simplifying:

\[ 87 = d \left[ 2a + 9d \right] \quad \dots (2) \]

Subtracting equation (1) from equation (2):

\[ 36 = 4d^2 \] \[ d = 3, \quad a = 1 \]

Finally:

\[ a_{17} - A_7 = 49 - \left[ -7.3 \left[ 2 + 39 \right] \right] = 910 \]

Answer 2

Step 1: Represent the arithmetic progression
Let the arithmetic progression be \( a_1, a_2, a_3, \dots \) with common difference \( d \).
Thus \( a_n = a_1 + (n - 1)d \).

Step 2: Express \( A_k \) in terms of \( a_1 \) and \( d \)
By definition: \[ A_k = a_1^2 - a_2^2 + a_3^2 - a_4^2 + \dots + a_{2k-1}^2 - a_{2k}^2. \] We can group terms as pairs: \[ A_k = \sum_{r=1}^{k} (a_{2r-1}^2 - a_{2r}^2) = \sum_{r=1}^{k} (a_{2r-1} - a_{2r})(a_{2r-1} + a_{2r}). \] Now \( a_{2r-1} - a_{2r} = -d \). Also, \[ a_{2r-1} + a_{2r} = [a_1 + (2r-2)d] + [a_1 + (2r-1)d] = 2a_1 + (4r - 3)d. \] Hence, \[ A_k = \sum_{r=1}^{k} [-d(2a_1 + (4r - 3)d)] = -2a_1 d k - d^2 \sum_{r=1}^{k}(4r - 3). \]

Step 3: Simplify the summation
\[ \sum_{r=1}^{k}(4r - 3) = 4\sum_{r=1}^{k}r - 3k = 2k(k+1) - 3k = 2k^2 - k. \] So, \[ A_k = -2a_1 d k - d^2(2k^2 - k) = -2a_1 d k - 2d^2k^2 + d^2k. \]

Step 4: Use given values \( A_3 = -153 \) and \( A_5 = -435 \)
For \( k = 3 \): \[ A_3 = -6a_1 d - 18d^2 + 3d^2 = -6a_1 d - 15d^2 = -153. \] For \( k = 5 \): \[ A_5 = -10a_1 d - 50d^2 + 5d^2 = -10a_1 d - 45d^2 = -435. \] Simplify: \[ -6a_1 d - 15d^2 = -153 \Rightarrow 6a_1 d + 15d^2 = 153. \quad (1) \] \[ -10a_1 d - 45d^2 = -435 \Rightarrow 10a_1 d + 45d^2 = 435. \quad (2) \]

Step 5: Solve equations (1) and (2)
From (1): \( 2a_1 d + 5d^2 = 51. \) From (2): \( 2a_1 d + 9d^2 = 87. \)
Subtract (1) from (2): \[ (2a_1 d + 9d^2) - (2a_1 d + 5d^2) = 87 - 51 \Rightarrow 4d^2 = 36 \Rightarrow d^2 = 9 \Rightarrow d = 3 \text{ (since terms are positive).} \] Substitute \( d = 3 \) into (1): \[ 2a_1(3) + 5(9) = 51 \Rightarrow 6a_1 + 45 = 51 \Rightarrow a_1 = 1. \]

Step 6: Verify with the given sum \( a_1^2 + a_2^2 + a_3^2 = 66 \)
\[ a_1 = 1,\; a_2 = 4,\; a_3 = 7. \] Then \( a_1^2 + a_2^2 + a_3^2 = 1 + 16 + 49 = 66, \) verified.

Step 7: Find \( a_{17} - A_7 \)
Compute \( a_{17} = a_1 + 16d = 1 + 48 = 49. \)
Now find \( A_7 \): \[ A_7 = -2a_1 d (7) - 2d^2(7^2) + d^2(7) = -14a_1 d - 2d^2(49) + 7d^2 = -14(1)(3) - 98(9) + 63 = -42 - 882 + 63 = -861. \] Hence, \[ a_{17} - A_7 = 49 - (-861) = 910. \]

Final answer

910