The Harmonic mean of 3, 7, 8, 10, 14 is

\frac{3 + 7 + 8 + 10 + 14}{5}

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\frac{5}{3 + 7 + 8 + 10 + 14}

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\frac{\frac{1}{3} + \frac{1}{7} + \frac{1}{8} + \frac{1}{10} + \frac{1}{14}}{5}

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\frac{5}{\frac{1}{3} + \frac{1}{7} + \frac{1}{8} + \frac{1}{10} + \frac{1}{14}}

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Solution

The correct option is D $\frac{5}{\frac{1}{3} + \frac{1}{7} + \frac{1}{8} + \frac{1}{10} + \frac{1}{14}}$
We know H.M of any distribution is given by, $\frac{Number of terms}{Sum of reciprocal of each terms}$

Hence H.M of $3, 7, 8, 10, 14$ is $= \frac{5}{\frac{1}{3} + \frac{1}{7} + \frac{1}{8} + \frac{1}{10} + \frac{1}{14}}$

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Similar questions

Add using number line:

(1) 5 and 3

(2) 6 and 2

(3) (−3) and (+5)

(4) 8 and (−12)

(5) (−7) and 5

(6) 10 and (−14)

(7) (−5) and (−2)

(8) (+6) and (−9)

(9) (−4) and (−1)

(10) 1 and (−5)

Simplify:
$(\frac{- 4}{5} \times \frac{15}{8}) + (\frac{- 1}{3} \times \frac{- 9}{7}) - (\frac{2}{1} \times \frac{25}{14})$

{(2, 1), (5, 1), (8, 1), (11, 1), (14, 1), (17, 1)}

{(2, 1), (4, 2), (6, 3), (8, 4), (10, 5), (12, 6), (14, 7)}

{(1, 3), (1, 5), (2, 5)}

∣ ∣ ∣ ∣ \begin{matrix} x + 1 & x + 2 & x + 4 x + 3 & x + 5 & x + 8 x + 7 & x + 10 & x + 14 \end{matrix} ∣ ∣ ∣ ∣ =

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