If a1,a2,a3,............an are in A.P and1a1an+1a2an−1+1a3an*2+.......+1ana1=K(1a1+1a2+1a3+..................+1an) . Then K is
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a
2a1+an
b
na1+an
c
1a1+an
d
n−1a1+an
answer is A.
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Detailed Solution
We know that in an A.P a1+an=a2+an−1=a3+an−2=.............(i) [see the properties of A.P]∴1a1an+1a2an−1+1a3an−2+..........+1ana1 =1a1+an[a1+ana1an+a1+ana2an−2+a1+ana3an−2+..........+a1+anana1] =1a1+an[a1+ana1an+a1+an−1a2an−2+a1+an−2a3an−2+..........+an+an1ana1] [using (i)]=1a1+an[(1a1+1an)+(1a2+1an−1)+(1a3+1an−2)+..........+(1a1+1an)] =2a1+an[1a1+1a2+1a3+.........+1an] (each term occurs twice).