Here you will learn what is harmonic progression (hp) and sum of harmonic progression (hp series).
Let’s begin –
Sum of Harmonic Progression
A sequence is said to be in H.P. if the reciprocal of its terms are in A.P. If the sequence \(a_1\) + \(a_2\) + \(a_3\) + ……… + \(a_n\) is an HP then \(1\over a_1\), \(1\over a_2\)……… \(1\over a_n\),is an AP. Here we do not have the formula for the sum of the n terms of an HP.
The general form of a harmonic progression is \(1\over a\), \(1\over {a+d}\), \(1\over {a+2d}\), \(1\over {a+(n-1)d}\).
Note :
No term of any H.P can be zero.
If a, b, c are in H.P, then b = \(2ac\over{a+c}\)
Example : The sum of three numbers are in H.P.. is 37 and the sum of their reciprocals is 1/4. Find the numbers.
Solution : Three numbers in H.P. can be taken as
\(1\over {a-d}\), \(1\over a\), \(1\over {a+d}\)
then \(1\over {a-d}\)+\(1\over a\)+\(1\over {a+d}\) = 37 …….(i)
and a – d + a + a + d = \(1\over 4\) \(\implies\) a = \(1\over 12\)
from(i), \(12\over {1-12d}\)+12+\(12\over {1+12d}\) = 37 \(\implies\)
\(12\over {1-12d}\)+\(12\over {1+12d}\) = 25
=> \(24\over {1-144d^2}\) = 25 \(\implies\) 1-\(144d^2\) = \(24\over 25\) \(\implies\)
\(d^2\) = \(1\over{25\times144}\)
\(\therefore\) d = \(\pm\)\(1\over 60\)
\(\therefore\) a-d, a, a+d are \(1\over 15\), \(1\over 12\), \(1\over 10\) or \(1\over 10\), \(1\over 12\), \(1\over 15\)
Hence, three numbers in H.P are 15,12,10 or 10,12,15.
Hope you learnt what is harmonic progression (hp) and sum of harmonic progression. To learn more practice more questions and get ahead in competition. Good Luck!
