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# 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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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$= \dfrac{1}{{{t^n}}}$
$\Rightarrow {H_n} = \dfrac{1}{{a + \left( {n - 1} \right)d}}$
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$\left( {H.P} \right)$.