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Question

If a,a1,a2,.....,a2n,b are in arithmetic progression and a,g1,g2,....g2n,b are in geometric progression and h is the harmonic mean of a and b, then a1+a2ng1g2n+a2+a2n1g2g2n1+.....+an+an+1gngn+1 is equal to

A
2nh
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B
n/h
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C
nh
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D
2n/h
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Solution

The correct option is D 2n/h
Since a,a1,a2,....a2n,b are in arithmetic progression
a,a1,a2,....a2n are A.M's between a and b
a1+a2n=a2+a2n1=....=a+b
similarly a,g1,g2,....g2n are in G.P
g1g2n=g2g2n1=....=gngn+1=ab
a1+a2ng1g2n+a2+a2n1g2g2n1+....an+an+1gngn+1
=a+bab+a+bab+......+a+bab=n(a+b)ab=2nh
where h is H.M between a and b
h=2aba+b

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