Question

find the harmonic mean of 1; 1/3 ; 1/5; ......1/2n-1


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Answer 1

GIVEN :–

• A Harmonic series => 1 , ⅓ , ⅕ , ........ , 1/(2n - 1)

TO FIND :–

• Harmonic mean = ?

SOLUTION :–

• We know that If two term a , b in H.P. (Harmonic progression) , Then Harmonic mean –

$\\ \: \: \longrightarrow \: \: { \boxed { \bold{H.M. = \dfrac{2}{ \dfrac{1}{a} + \dfrac{1}{b} } = \dfrac{2ab}{a + b} }}}$

• Similarly H.M. for n terms –

$\\ \: \: \longrightarrow \: \: { \boxed { \bold{H.M. = \dfrac{n}{ \dfrac{1}{ x_{1} } + \dfrac{1}{ x_{2}} + ....... +\dfrac{1}{ x_{n}} } }}}$

• Now put the values –

$\\ \: \: \implies \: \: { \bold{H.M. = \dfrac{n}{ \dfrac{1}{ 1 } + \dfrac{1}{ (\frac{1}{3} )} + \dfrac{1}{ (\frac{1}{5} )} + ....... +\dfrac{1}{ 2n - 1} } }}$

$\\ \: \: \implies \: \: { \bold{H.M. = \dfrac{n}{1 +3+5+ ....... +2n - 1} \: \: \: \: - - - eq.(1) }}$

• Let's solve the series which is in Denominator –

$\\ \: \: \implies \: \: { \bold{1 +3+5+ ....... +2n - 1}}$

• This series is an A.P. And we know that sum of A.P. is –

$\\ \: \: \implies \: \: { \bold{ s_{n} = \frac{n_1}{2}(a + l)}}$

• Here –

$\\ \: \: \: \: \: { \huge{.}}\: \: { \bold{ a = first \: \: term}}$

$\\ \: \: \: \: \: { \huge{.}}\: \: { \bold{ l = last \: \: term}}$

$\\ \: \: \: \: \: { \huge{.}}\: \: { \bold{ n_1= total \: \: number \: \: term}}$

• First Let's find 'n' –

$\\ \: \: \implies \: \: { \bold{ l = a + (n_1 - 1)d}}$

$\\ \: \: \implies \: \: { \bold{ 2n - 1 = 1 + (n_1 - 1)2}}$

$\\ \: \: \implies \: \: { \bold{ 2n - 1 = 2n_1 - 1}}$

$\\ \: \: \implies \: \: { \bold{ n_1 = n}}$

• So that sum –

$\\ \: \: \implies \: \: { \bold{ s_{n} = \frac{n}{2}(1 + 2n - 1)}}$

$\\ \: \: \implies \: \: { \bold{ s_{n} = \frac{n}{2}( 2n )}}$

$\\ \: \: \implies \: \: { \bold{ s_{n} = {n}^{2} }}$

• Put sum in eq.(1) –

$\\ \: \: \implies \: \: { \bold{H.M. = \dfrac{n}{( {n}^{2} )} }}$

$\\ \: \: \implies \: \: \large { \boxed{ { \bold{H.M. = \dfrac{1}{n} }}}}$