# The formula to find the${n^{th}}$term of harmonic progression is.

$

{\text{a}}{\text{. }}\dfrac{1}{{a - \left( {n - 1} \right)d}} \\

{\text{b}}{\text{. }}\dfrac{1}{{a + \left( {n + 1} \right)d}} \\

{\text{c}}{\text{. }}\dfrac{1}{{a + \left( {n - 1} \right)d}} \\

{\text{d}}{\text{. }}\dfrac{1}{{a - \left( {n + 1} \right)d}} \\

$

Last updated date: 28th Mar 2023

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Answer

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Hint: - Harmonic Progression is the reciprocal of the Arithmetic Progression.

As we know that the${n^{th}}$term of an A.P is${t_n} = a + \left( {n - 1} \right)d$

So, we know that Harmonic Progression$\left( {H.P} \right)$ is the reciprocal of $\left( {A.P} \right)$

Therefore ${n^{th}}$of H.P is$ = \dfrac{1}{{{t^n}}}$

$ \Rightarrow {H_n} = \dfrac{1}{{a + \left( {n - 1} \right)d}}$

So, this is the required answer which is option c.

Note: - In such types of questions the key concept we have to remember is that${n^{th}}$term of harmonic progression is the reciprocal of arithmetic progression so, if we remember the formula of${n^{th}}$term of $\left( {A.P} \right)$ then we easily calculate the ${n^{th}}$term of$\left( {H.P} \right)$.

As we know that the${n^{th}}$term of an A.P is${t_n} = a + \left( {n - 1} \right)d$

So, we know that Harmonic Progression$\left( {H.P} \right)$ is the reciprocal of $\left( {A.P} \right)$

Therefore ${n^{th}}$of H.P is$ = \dfrac{1}{{{t^n}}}$

$ \Rightarrow {H_n} = \dfrac{1}{{a + \left( {n - 1} \right)d}}$

So, this is the required answer which is option c.

Note: - In such types of questions the key concept we have to remember is that${n^{th}}$term of harmonic progression is the reciprocal of arithmetic progression so, if we remember the formula of${n^{th}}$term of $\left( {A.P} \right)$ then we easily calculate the ${n^{th}}$term of$\left( {H.P} \right)$.

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