If a,a1,a2,a3,a4,......,a2n,b are in AP and a,g1,g2,g3,g4,........,g2n,b are in GP and h is the HM of a and b, then a1+a2ng1g2n+a2+a2n−1g2g2n−1+....+an+an+1gngn+1
A
2nh
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B
2nh
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C
nh
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D
nh
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Solution
The correct option is A2nh The denominators g1g2n,g2g2n−1,....,gngn+1 are equal to ab as it is the property of G.P Also all the numerators a1+a2n,a2+a2n−1,...,an+an+1 are all equal to a+b as it is the property of A.P So the entire expression is equal to n×(a+b)ab i.e, 2nh