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Question

If a1,a2,a3,....an are in A.P. then 1a1+a2+1a2+a3+...+1an1+an =

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Solution

1a1+a2=1a1+a2×a1a2a1a2
=a1a2a1a2=a1a2d
Similarly =1d(a1a2)
1a2+a3=a2a3d=1d(a2a3)
1a3+a4=a3a4d=1d(a3a4)
1a4+a5=a4a5d=1d(a4a5)
1an1=an=an1a2d=1d(a1a6)
1d(a1an)

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