Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# If ${a_1},{a_2},{a_3},..........,{a_n}$are in H.P and$f(k) = \sum\limits_{r = 1}^n {{a_r} - {a_k}}$, then $\dfrac{{{a_1}}}{{f(1)}},\dfrac{{{a_2}}}{{f(2)}},\dfrac{{{a_3}}}{{f(3)}},.........,\dfrac{{{a_n}}}{{f(n)}}$ are in:A) A.P.B) G.P.C) H.P.D) None of them.

Last updated date: 13th Jun 2024
Total views: 402k
Views today: 10.02k
Verified
402k+ views
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 ${S_n} = {a_1} + {a_2} + ....... + {a_n}$.
${S_n} = {a_1} + {a_2} + ....... + {a_n}$.

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.