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Question

If a1,a2,a3.....an are in A.P. Where ai>0 for all i, then the value of
1a1+a2+1a2+a3+.........+1an1+an=


A

n1a1+an

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B

n1a1+an

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C

n1a1an

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D

n1a1an

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Solution

The correct option is A

n1a1+an


As given a2a1 = a3 a2=.....=anan1=d
Where d is the common difference of the given A.P.
Also an=a1=(n1)d.
Then by rationalising each term,
1a2+a1+1a3+a2+.........+1an+an1
=a2a1a2a1+a3a2a3a2.....+anan1anan1

=1d{a2a1+a3a2+.....anan1}
1d{ana1}= 1d (ana1ana1)
1d{(n1)dan+a1}=n1ana1
Trick : Put a1=a2=a3=>d=0


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