JEE MAIN - Mathematics (2024 - 5th April Morning Shift - No. 27)
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 $$\mathrm{A}_3=-153, \mathrm{~A}_5=-435$$ and $$\mathrm{a}_1^2+\mathrm{a}_2^2+\mathrm{a}_3^2=66$$, then $$\mathrm{a}_{17}-\mathrm{A}_7$$ is equal to ________.
Explanation
Let $$a_n=a+(n-1) d \forall n \in N$$
$$\begin{aligned} A_k & =\left(a_1^2-a_2^2\right)+\left(a_3^2-a_4^2\right)+\ldots a_{2 k-1}^2-a_{2 k}^2 \\ & =(-d)\left(a_1+a_2+\ldots+a_{2 k}\right) \\ & A_k=(-d k)(2 a+(2 k-1) d) \\ \Rightarrow & A_3=(-3 d)(2 a+5 d)=-153 \\ \Rightarrow & d(2 a+5 d)=51 \quad \text{... (i)}\\ & A_5=(-5 d)(2 a+9 d)=-435 \end{aligned}$$
$$\begin{aligned} \Rightarrow & d(2 a+9 d)=87 \\ \Rightarrow & 4 d^2=36 \Rightarrow d= \pm 3(d=3 \text { positive terms }) \\ \Rightarrow & 3(2 a+27)=87 \\ \Rightarrow & 2 a=29-27 \\ \Rightarrow & a=1 \\ & a_{17}-A_7=(a+16 d)-(-7 d)(2+13 d) \\ & =49+7 \times 3(2+39) \\ & =49+21 \times 41=910 \end{aligned}$$
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