Let a_1, a_2, a_3, ldots be in an arithmetic | vedclass

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(C) Let $d$ be the common difference and $a$ be the first term.$A_k = (a_1^2 - a_2^2) + (a_3^2 - a_4^2) + \ldots + (a_{2k-1}^2 - a_{2k}^2)$.Using $x^2

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$910$

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(C) Let $d$ be the common difference and $a$ be the first term.$A_k = (a_1^2 - a_2^2) + (a_3^2 - a_4^2) + \ldots + (a_{2k-1}^2 - a_{2k}^2)$.Using $x^2 - y^2 = (x-y)(x+y)$,we have $a_{2n-1}^2 - a_{2n}^2 = (a_{2n-1} - a_{2n})(a_{2n-1} + a_{2n}) = (-d)(2a + (4n-3)d)$.Summing this,$A_k = -d \sum_{n=1}^k (2a + (4n-3)d) = -d [2ak + d(4 \frac{k(k+1)}{2} - 3k)] = -kd(2a + (2k-1)d)$.Given $A_3 = -3d(2a + 5d) = -153 \Rightarrow d(2a + 5d) = 51$ $(1)$.Given $A_5 = -5d(2a + 9d) = -435 \Rightarrow d(2a + 9d) = 87$ $(2)$.Subtracting $(1)$ from $(2)$: $d(2a + 9d - 2a - 5d) = 87 - 51$ $\Rightarrow 4d^2 = 36$ $\Rightarrow d = 3$.Substituting $d=3$ into $(1)$: $3(2a + 15) = 51$ $\Rightarrow 2a + 15 = 17$ $\Rightarrow a = 1$.Check: $a_1^2 + a_2^2 + a_3^2 = 1^2 + 4^2 + 7^2 = 1 + 16 + 49 = 66$. This matches.$a_{17} = a + 16d = 1 + 16(3) = 49$.$A_7 = -7d(2a + 13d) = -7(3)(2(1) + 13(3)) = -21(2 + 39) = -21(41) = -861$.$a_{17} - A_7 = 49 - (-861) = 49 + 861 = 910$.

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:

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