If a1,a2,a3,a4 are in AP, then 1√a1+√a2+1√a2+√a3+1√a3+√a4 is equal to
A
√a4−√a1a3−a2
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B
a4−a1a3−a2
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C
a3−a2√a4−√a1
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D
a1−a4a3−a1
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E
a5−a0a1−a4
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Solution
The correct option is A√a4−√a1a3−a2 1√a1+√a2+1√a2+√a3+1√a3+√a4 =√a1−√a2(√a1+√a2)(√a1−√a2)+√a2−√a3(√a2+√a3)(√a2−√a3)+√a3−√a4(√a3+√a4)(√a3−√a4) =√a1−√a2a1−a2+√a2−√a3a2−a3+√a3−√a4a3−a4 ∵a1,a2,a3,a4 are in AP ∴a1−a2=a2−a3=a3−a4 Thus 1√a1+√a2+1√a2+√a3+1√a3+√a4=√a1−√a2a1−a2+√a2−√a3a1−a2+√a3−√a4a1−a2 =−√a4+√a1a1−a2=√a4−√a1a2−a3=√a4−√a1a3−a2