Let [a_1, a_2, a_3, ] be in an arithmetic progression of positive terms.

Question

Let $$a_1, a_2, a_3, \ldots$$ be in an arithmetic progression of positive terms.
Let $$A_k=a_1^2-a_2^2+a_3^2-a_4^2+\ldots+a_{2 k-1}^2-a_{2 k}^2$$.
If $$\mathrm{A}_3=-153, \mathrm{~A}_5=-435$$

Accepted Answer

Let ${a}_{n}=a+(n-1)d\foralln\in N$

$\begin{gathered} {A}_{k} =({a}_{2}^{1}-{a}_{2}^{2})+({a}_{2}^{3}-{a}_{2}^{4})+\ldots {a}_{2}^{2k-1}-{a}_{2}^{2k} \ =(-d)({a}_{1}+{a}_{2}+\ldots +{a}_{2k}) \ {A}_{k}=(-dk)(2a+(2k-1)d) \ \Rightarrow {A}_{3}=(-3d)(2a+5d)=-153 \ \Rightarrow d(2a+5d)=51\, \ {A}_{5}=(-5d)(2a+9d)=-435 $

$ \Rightarrow d(2a+9d)=87 \ \Rightarrow 4{d}^{2}=36\Rightarrow d=\pm 3(d=3) \ \Rightarrow 3(2a+27)=87 \ \Rightarrow 2a=29-27 \ \Rightarrow a=1 \ {a}_{17}-{A}_{7}=(a+16d)-(-7d)(2+13d) \ =49+7	imes 3(2+39) \ =49+21	imes 41=910 \end{gathered}$

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