Question

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}} \\
$

Answer 1

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$\left( {H.P} \right)$.