Question

What is the sum of first $14$ terms of the H.P. $12,15,18,\mathrm{21....}?$

Answer 1

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 $=12$
Second term of the given arithmetic series $=15$
Third term of the given arithmetic series $=18$
Fourth term of the given arithmetic series $=21$
Now, Second term - First term $=15-12=3$
Third term - Second term $=18-15=3$
Therefore, common difference of the given arithmetic series is $3.$
The number of terms of the given A. P. series $\left(n\right)=14$
We know that the sum of first $n$ terms of the Arithmetic Progress, whose first term $=a$ and common difference $=d$ is
$S_n=\frac{n}{2}[2a+(n-1\left)d\right]$
$S_{14}=\frac{14}{2}[2\times 12+(14-1\left)3\right]$
$S_{14}=7[24+39]$
$S_{14}=441$
Sum of HP $=\frac{1}{441}$