Let a1,a2,a3,... be in A.P. with a1=1 and a102=405. If ∣∣a21−a22 – InterviewSolution

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1.	Let a1,a2,a3,... be in A.P. with a1=1 and a102=405. If ∣∣a21−a22+a23−a24+⋯+a2101−a2102∣∣=5∏i=1pbii where pi are prime numbers and bi∈N for all i=1,2,…5, then the value of 5∑i=1pi+5∑i=1bi is
Answer» Let $a_{1}, a_{2}, a_{3}, . . .$ be in A.P. with $a_{1} = 1$ and $a_{102} = 405.$ If $∣ ∣ a_{1}^{2} - a_{2}^{2} + a_{3}^{2} - a_{4}^{2} + \dots + a_{101}^{2} - a_{102}^{2} ∣ ∣ = 5 \prod i = 1 p_{i}^{b_{i}}$ where $p_{i}$ are prime numbers and $b_{i} \in N$ for all $i = 1, 2, \dots 5,$ then the value of $5 \sum i = 1 p_{i} + 5 \sum i = 1 b_{i}$ is

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