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If a1,a2,a3,….a4001 are terms of an AP such that 1a1a2+1a2a3+…+1a4000a4001=10 and a1+a4001=50 then a1−a4001 is equal to

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Detailed Solution

1da4001−a1a1a4001=10⇒4000a1a4001=10 ∵a4001=a1+4000d⇒1da4001-a1=4000⇒a1a4001=400Also given a1+a4001=50∴a1−a40012=a1+a40012−4a1a4001⇒a1−a4001=30

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