Question

What is the sum of $12$ odd numbers in HP $\frac{1}{1},\frac{1}{3},\frac{1}{5},\frac{1}{7},\frac{1}{9}.....?$

• A. $\frac{1}{12}$
• B. $\frac{1}{14}$
• C. $\frac{1}{144}$
• D. $\frac{1}{140}$

Answer 1

The correct option is C $\frac{1}{144}$

We know the formula for sum of $n^{th}$ term in arithmetic progression.
The reciprocal of arithmetic progression is harmonic progression.
First term of the given arithmetic series $=1$
Second term of the given arithmetic series $=3$
Third term of the given arithmetic series $=5$
Fourth term of the given arithmetic series $=7$
Now, Second term - First term $=3-1=2$
Third term - Second term $=5-3=2$
Therefore, common difference of the given arithmetic series is $2$.
The number of terms of the given A. P. series $\left(n\right)=12$
We know that the sum of first $n$ terms of the Arithmetic Progress, whose first term $=a$ and common difference $=d$ is
$S_n=\frac{n}{2}[2a+(n-1\left)d\right]$
$S_{12}=\frac{12}{2}[2\times 1+(12-1\left)2\right]$
$S_{12}=6[2+22]$
$S_{12}=144$
Sum of HP $=\frac{1}{144}$