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$Let a_{1}, a_{2}, a_{3}, \dots a_{4001} are in A.P. such that \frac{1}{a_{1} a_{2}} + \frac{1}{a_{2} a_{3}} + \dots . + \frac{1}{a_{4000} a_{4001}} = 10 and$ $a_{2} + a_{4000} = 50 then |a_{1} - a_{4001}| is equal to$

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Correct option is B

4000a1a4001=10⇒a1a4001=400 also a1+a4001=50⇒a1−a40012=a1+a40012-4a1a4001 =2500-1600 =900⇒a1−a4001=30

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Let a1,a2,a3,…a4001 are in A.P. such that 1a1a2+1a2a3+….+1a4000a4001=10 and a2+a4000=50 then a1−a4001 is equal to

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$Let a_{1}, a_{2}, a_{3}, \dots a_{4001} are in A.P. such that \frac{1}{a_{1} a_{2}} + \frac{1}{a_{2} a_{3}} + \dots . + \frac{1}{a_{4000} a_{4001}} = 10 and$ $a_{2} + a_{4000} = 50 then |a_{1} - a_{4001}| is equal to$