Let a_1, a_2, a_3, … be in an arithmetic progression of positive terms. Let A_k=a_1^2-a_2^2+a_3^2-a_4^2+…+a_2 k-1^2-a_2 k^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______

Correct answer: 910. d arrow common diff. A_k=-kd[2 a+(2 k-1) d] A_3=-153 ⇒ 153=13 ~d[2 a+5 ~d] 51=d[2 a+5 ~d]...(1) A_5=-435 435=5 ~d[2 a+9 ~d] 87=d[2 a+9 ~d] (2)-(1) 36=4 ~d^2 ~d=3, a=1 a_17-A_7=49-[-7.3[2+39]]=910

Q57 / 90 · JEE Main 2024 (05 Apr Shift 1) · Mathematics

Let $a_1, a_2, a_3, \ldots$ be in an arithmetic progression of positive terms. Let $\mathrm{A}_{\mathrm{k}}=\mathrm{a}_1^2-\mathrm{a}_2^2+\mathrm{a}_3^2-\mathrm{a}_4^2+\ldots+\mathrm{a}_{2 \mathrm{k}-1}^2-\mathrm{a}_{2 \mathrm{k}}^2$. 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______

Correct answer: 910

Solution

$\begin{aligned} & \mathrm{d} \rightarrow \text { common diff. } \\ & \mathrm{A}_{\mathrm{k}}=-\mathrm{kd}[2 \mathrm{a}+(2 \mathrm{k}-1) \mathrm{d}] \\ & \mathrm{A}_3=-153 \\ & \Rightarrow 153=13 \mathrm{~d}[2 \mathrm{a}+5 \mathrm{~d}] \\ & 51=\mathrm{d}[2 \mathrm{a}+5 \mathrm{~d}]...(1) \\ & \mathrm{A}_5=-435 \\ & 435=5 \mathrm{~d}[2 \mathrm{a}+9 \mathrm{~d}] \\ & 87=\mathrm{d}[2 \mathrm{a}+9 \mathrm{~d}] \\ & (2)-(1) \\ & 36=4 \mathrm{~d}^2 \\ & \mathrm{~d}=3, \mathrm{a}=1 \\ & \mathrm{a}_{17}-\mathrm{A}_7=49-[-7.3[2+39]]=910\end{aligned}$

Hence, the correct answer is 910.

Solution

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