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Question

If H1,H2,,H20 be 20 harmonic means between 2 and 3, then the value of H1+2H12+H20+3H203 is

A
20
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B
21
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C
40
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D
41
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Solution

The correct option is C 40
We know that,
13=12+21dd=16×21(1)
Now, given
H1+2H12+H20+3H203=12+1H1121H1+13+1H20131H20
=12+12+d12d12+13+13d1313+d(1H1=12+d, 1H20=13d)
=1+dd+23dd=23d1dd=132dd=13d2
Using equation (1),
=6×2132=40

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