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Question: The formula to find the\({n^{th}}\)term of harmonic progression is. \( {\text{a}}{\text{. }}\d...

The formula to find thenth{n^{th}}term of harmonic progression is.
a. 1a(n1)d b. 1a+(n+1)d c. 1a+(n1)d d. 1a(n+1)d  {\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}} \\\

Explanation

Solution

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

As we know that thenth{n^{th}}term of an A.P istn=a+(n1)d{t_n} = a + \left( {n - 1} \right)d
So, we know that Harmonic Progression(H.P)\left( {H.P} \right) is the reciprocal of (A.P)\left( {A.P} \right)
Therefore nth{n^{th}}of H.P is=1tn = \dfrac{1}{{{t^n}}}
Hn=1a+(n1)d\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 thatnth{n^{th}}term of harmonic progression is the reciprocal of arithmetic progression so, if we remember the formula ofnth{n^{th}}term of (A.P)\left( {A.P} \right) then we easily calculate the nth{n^{th}}term of(H.P)\left( {H.P} \right).