KEAM 2016 Math Paper

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Question : 20 of 120

Marks: +1, -0

If a1,a2,a3,a4 are in A.P., then

1

√a1√a2

+

1

√a2√a3

+

1

√a3√a4

=

√a1−√a2
a3−a2
a4−a1
a3−a2
a3−a2
√a4−√a1
a1−a4
a3−a1
a5−ao
a1−a4

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