Let $$a_1, a_2, a_3, \ldots$$ be in an arithmetic progression of positive terms. Let $$A_k = a_1^2 - a_2^2 + a_3^2 - a_4^2 + \ldots + a_{2k-1}^2 - a_{2k}^2$$. 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 ______

AP with first term $$a_1 = a$$, common difference $$d$$.

$$A_k = \sum_{i=1}^{k}(a_{2i-1}^2 - a_{2i}^2) = \sum_{i=1}^{k}(a_{2i-1} - a_{2i})(a_{2i-1} + a_{2i})$$.

$$a_{2i-1} - a_{2i} = -d$$ and $$a_{2i-1} + a_{2i} = 2a + (4i-3)d$$.

$$ A_k = -d \sum_{i=1}^{k}(2a + (4i-3)d) = -d[2ak + d\sum_{i=1}^{k}(4i-3)] $$

$$ = -d[2ak + d(4 \cdot \frac{k(k+1)}{2} - 3k)] = -d[2ak + d(2k^2 + 2k - 3k)] = -d[2ak + d(2k^2 - k)] $$

$$ A_k = -d[2ak + dk(2k-1)] = -dk[2a + d(2k-1)] $$

$$A_3 = -3d[2a + 5d] = -153$$ ... (1)

$$A_5 = -5d[2a + 9d] = -435$$ ... (2)

From (2)/(1): $$\frac{5(2a+9d)}{3(2a+5d)} = \frac{435}{153} = \frac{145}{51} = \frac{145}{51}$$.

Simplify: $$\frac{145}{51} = \frac{5 \times 29}{3 \times 17}$$. So $$\frac{5(2a+9d)}{3(2a+5d)} = \frac{5 \times 29}{3 \times 17}$$.

$$\frac{2a+9d}{2a+5d} = \frac{29}{17}$$.

$$17(2a+9d) = 29(2a+5d)$$

$$34a + 153d = 58a + 145d$$

$$8d = 24a \Rightarrow d = 3a$$.

From (1): $$-3(3a)[2a + 15a] = -153 \Rightarrow -9a \cdot 17a = -153 \Rightarrow 153a^2 = 153 \Rightarrow a = 1$$.

So $$d = 3$$.

Check: $$a_1^2 + a_2^2 + a_3^2 = 1 + 16 + 49 = 66$$ ✓ ($$a_1=1, a_2=4, a_3=7$$).

$$a_{17} = a + 16d = 1 + 48 = 49$$.

$$A_7 = -7d[2a + 13d] = -7(3)[2 + 39] = -21 \times 41 = -861$$.

$$a_{17} - A_7 = 49 - (-861) = 49 + 861 = 910$$.

The answer is 910.