Question

What is the sum of 12 odd numbers in H.P. series $\dfrac{1}{1},\dfrac{1}{3},\dfrac{1}{5},\dfrac{1}{7},\dfrac{1}{9},.....$?
A. $\dfrac{1}{12}$
B. $\dfrac{1}{14}$
C. $2.24$
D. $\dfrac{1}{46}$

Answer 1

We convert the given H.P. series into an A.P. series. We find out the initial number and the common difference of the series. We also have the number of the terms we are trying to find. The relation between an A.P. and H.P. The term is inverse. We find the solution using that.

Complete step-by-step answer:
Harmonic progression is actually inverse of A.P. This means if the inverse terms of all the numbers of a series is in A.P. then the numbers of the original series are in H.P.
For example: if the terms a, b, c, d is in H.P. then $\dfrac{1}{a},\dfrac{1}{b},\dfrac{1}{c},\dfrac{1}{d}$ are in A.P.
For this problem it’s given the series $\dfrac{1}{1},\dfrac{1}{3},\dfrac{1}{5},\dfrac{1}{7},\dfrac{1}{9},.....$ is in H.P. We add the first term 1 as the extra at the end. For now, we take $\dfrac{1}{3},\dfrac{1}{5},\dfrac{1}{7},\dfrac{1}{9},.....$
We convert the given H.P. into A.P.
So, the inverse of every term in that series will be terms of an A.P.
So, the inverse of the terms will be $3,5,7,9,.....$
We need to find the initial term ‘a’ and the common difference ‘d’ of the series.
So, let's name the terms as ${{t}_{1}}=3,{{t}_{2}}=5,{{t}_{3}}=7,{{t}_{4}}=9,.....$
The series starts with ${{t}_{1}}=3$. So, $a=3$.
Now the difference between any two terms is d which means $d={{t}_{2}}-{{t}_{1}}=5-3=2$.
Now we find the general formula for the sum of A.P.
If we are finding the ${{n}^{th}}$ term which is ${{S}_{n}}$, then ${{S}_{n}}=\dfrac{1}{d}\ln \left[ \dfrac{2a+\left( 2n-1 \right)d}{2a-d} \right]$.
For our given problem we put the $a=3,d=2$ values in the equation ${{S}_{n}}=\dfrac{1}{d}\ln \left[ \dfrac{2a+\left( 2n-1 \right)d}{2a-d} \right]$ and get
$\begin{aligned} & {{S}_{n}}=1+\dfrac{1}{d}\ln \left[ \dfrac{2a+\left( 2n-1 \right)d}{2a-d} \right] \\\ & {{S}_{n}}=1+\dfrac{1}{2}\ln \left[ \dfrac{6+2\left( 2\times 11-1 \right)}{6-2} \right]=1+\dfrac{1}{2}\ln 12=2.24 \\\ \end{aligned}$
Therefore, the sum of 12 odd numbers in H.P. series $\dfrac{1}{1},\dfrac{1}{3},\dfrac{1}{5},\dfrac{1}{7},\dfrac{1}{9},.....$ is $2.24$.

So, the correct answer is “Option C”.

Note: We need to remember the relation of the sum is the approximate values. We can have that close result only when $n>>d$. There is no direct formula to find things for H.P, especially the sum of the series. So, in all cases we convert terms in its A.P. series.