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**Hint:**An Arithmetic Progression or Arithmetic sequence is a sequence of numbers such that the difference between the consecutive terms is constant.

A Harmonic Progression (H.P) is a progression formed by taking reciprocals of an Arithmetic Progression (A.P).

$H.P = \dfrac{1}{{A.P}}$

A sequence is a set of numbers which are written in some particular order. Basically sequences are of two types: finite ($1,3,5,7$) and infinite (${S_n} = {a_1} + {a_2} + ....... + {a_n}$) .

A series is a sum of the terms in a sequence. If there are $n$terms in the sequence and we evaluate the sum then we often write ${S_n}$ for the result, so that ${S_n} = {a_1} + {a_2} + ....... + {a_n}$.

${S_n} = {a_1} + {a_2} + ....... + {a_n}$.

**Complete step-by-step answer:**

Given that ${a_1},{a_2},{a_3},..........,{a_n}$are in H.P

As we know that $H.P = \dfrac{1}{{A.P}}$

$\therefore \dfrac{1}{{{a_1}}},\dfrac{1}{{{a_2}}},\dfrac{1}{{{a_3}}},.........,\dfrac{1}{{{a_n}}}$are in$A.P.$……………..(i)

We are given that$f(K) = \sum\limits_{r = 1}^n {{a_r} - {a_k}} $

As$\sum\limits_{r = 1}^n {{a_r}} = {S_n}$, so we can rewrite the above equation as:

$f(k) = \sum\limits_{r = 1}^n {{a_r} - {a_k}} = {S_n} - {a_k}.$

$ \Rightarrow \dfrac{{f(k)}}{{{a_k}}} - 1\forall k = 1,2,3,.....,n$……………….(ii)

As from (i) $\dfrac{1}{{{a_1}}},\dfrac{1}{{{a_2}}},\dfrac{1}{{{a_3}}},.........,\dfrac{1}{{{a_n}}}$are in$A.P.$

$ \Rightarrow \dfrac{{{a_1} + {a_2} + ...... + {a_n}}}{{{a_1}}},\dfrac{{{a_1} + {a_2} + ...... + {a_n}}}{{{a_2}}},.......,\dfrac{{{a_1} + {a_2} + ...... + {a_n}}}{{{a_n}}}$are in$A.P.$

Now we subtract 1 from each term:

$ \Rightarrow \dfrac{{{a_1} + {a_2} + ...... + {a_n}}}{{{a_1}}} - 1,\dfrac{{{a_1} + {a_2} + ...... + {a_n}}}{{{a_2}}} - 1,.......,\dfrac{{{a_1} + {a_2} + ...... + {a_n}}}{{{a_n}}} - 1$are in$A.P.$

By taking L.C.M of each term separately and on further solving

\[ \Rightarrow \dfrac{{{a_2} + ...... + {a_n}}}{{{a_1}}},\dfrac{{{a_1} + ...... + {a_n}}}{{{a_2}}},.......,\dfrac{{{a_1} + {a_2} + ...... + {a_{n - 1}}}}{{{a_n}}}\] are in$A.P.$……….(iii)

Now we can write ${a_2} + {a_{3 + }}........ + {a_n} = f(1),{a_1} + {a_3} + ........ + {a_n} = f(2),...........,{a_1} + {a_2} + ....... + {a_{n - 1}} = f(n)$

So we can write the equation (iii) as $\dfrac{{f(1)}}{{{a_1}}},\dfrac{{f(2)}}{{{a_2}}},..........,\dfrac{{f(n)}}{{{a_n}}}$ are in$A.P.$

As$H.P = \dfrac{1}{{A.P}}$

Therefore, $\dfrac{{{a_1}}}{{f(1)}},\dfrac{{{a_2}}}{{f(2)}},.........,\dfrac{{{a_n}}}{{f(n)}}$are in $H.P.$

**Hence option C is the correct answer.**

**Note:**Alternative Method:

We are given that: $f(K) = \sum\limits_{r = 1}^n {{a_r} - {a_k}} $

It can be further written as that $f(K) = [\sum\limits_{r = 1}^n {{a_r}] - {a_k} = {S_n} - {a_k}} $

Now, $\dfrac{{f(k)}}{{{a_k}}} = \dfrac{{{S_n}}}{{{a_k}}} - 1$

Also, it is given that ${a_1},{a_2},{a_3},..........,{a_n}$are in H.P

So, $\dfrac{1}{{{a_1}}},\dfrac{1}{{{a_2}}},\dfrac{1}{{{a_3}}},.........,\dfrac{1}{{{a_n}}}$ are in $A.P.$

Multiply each term with ${S_n}$ we get,

$\dfrac{{{S_n}}}{{{a_1}}},\dfrac{{{S_n}}}{{{a_2}}},\dfrac{{{S_n}}}{{{a_3}}},.........,\dfrac{{{S_n}}}{{{a_n}}}$ are in $A.P.$

Subtracting 1 from each term we get,

$\dfrac{{{S_n}}}{{{a_1}}} - 1,\dfrac{{{S_n}}}{{{a_2}}} - 1,\dfrac{{{S_n}}}{{{a_3}}} - 1,.........,\dfrac{{{S_n}}}{{{a_n}}} - 1$ are in$A.P.$

$\dfrac{{f(1)}}{{{a_1}}},\dfrac{{f(2)}}{{{a_2}}},..........,\dfrac{{f(n)}}{{{a_n}}}$ are in $A.P.$

As $H.P = \dfrac{1}{{A.P}}$

Therefore, $\dfrac{{{a_1}}}{{f(1)}},\dfrac{{{a_2}}}{{f(2)}},.........,\dfrac{{{a_n}}}{{f(n)}}$ are in $H.P.$

Hence the option C is the correct answer.

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