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20 Harmonic Means Between 2 and 3: HM Identity | JEE

JEE Maths question with a full step-by-step solution.

Question
If H1,H2,…,H20H_1, H_2, \ldots, H_{20} are 2020 harmonic means between 22 and 33, then H1+2H1−2+H20+3H20−3=\frac{H_1+2}{H_1-2}+\frac{H_{20}+3}{H_{20}-3}=
A2020
B2121
C4040correct
D3838
Solution
Step 1: The reciprocals 12,1H1,…,1H20,13\dfrac12,\dfrac{1}{H_1},\ldots,\dfrac{1}{H_{20}},\dfrac13 form an A.P. of 2222 terms:
d=13−1221=−1126.d=\frac{\frac13-\frac12}{21}=-\frac{1}{126}.
 Step 2: Then
1H1=12−1126=3163⇒H1=6331,1H20=13+1126=43126⇒H20=12643.\frac{1}{H_1}=\frac12-\frac{1}{126}=\frac{31}{63}\Rightarrow H_1=\frac{63}{31},\qquad \frac{1}{H_{20}}=\frac13+\frac{1}{126}=\frac{43}{126}\Rightarrow H_{20}=\frac{126}{43}.
 Step 3:
H1+2H1−2=125/311/31=125,H20+3H20−3=255/43−3/43=−85,\frac{H_1+2}{H_1-2}=\frac{125/31}{1/31}=125,\qquad \frac{H_{20}+3}{H_{20}-3}=\frac{255/43}{-3/43}=-85,
 so the sum is 125−85=40.125-85=40. Correct answer: (3)
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