A sequence is said to be in `H.P.` if the reciprocals of its terms are in `A.P.`

e.g. if `a_1, a_2,a_3,................` are in `H.P.` then `1/a_1, 1/a_2,1/a_3............` are in `A.P.`

A standard `H.P.` is `1/a+1/(a+d)+1/(a+2d)+..............+1/(a+(n-1)d`

For every `HP` there is a corresponding `A.P.`

Terms of harmonic series are the outcomes of an `A.P.`


`text(General term/)``n^(th)` `text(term/last term of)` `H.P.`

`T_n=1/(a+(n-1)d)`

where `a` and `d` are respectively the first term and the common difference of the corresponding `A.P.` and `n =` position of the term which we required.

`text( nth Term of HP from Beginning :)`

Let a be the first term, d be the common difference of an AP. Then, nth term of an AP from beginning `= a+ (n- 1) d`

Hence, the nth term of HP from beginning `= 1/(a+ (n- 1) d)`


`text( nth Term of HP from End :)`

Let l be the last term, d be the common difference of an AP. Then,
nth term of an AP from end `= l- (n - 1) d`
Hence, the nth term of HP from end `= 1/(l- (n - 1) d)`

`text(Note :)`

`(i)` There is no general formula for finding the sum to `n` terms of `H.P.`

`(ii)` If `a,b, c` are in `H.P.` `=> 1/a,1/b,1/c` are in `A.P`

`:. 2/b=1/a+1/c=> b=(2ac)/(a+c)=> a,b,c` are in `H.P.`

also `1/b-1/a=1/c-1/b` i.e. `(a-b)/(ab)=(b-c)/(bc)` i.e. `a/c=(a-b)/(b-c)`

`text(Harmonic Mean) (H .M.) :`


If `a, b, c` are in `H.P.` then middle term is called the `text(harmonic mean)` between them. Hence if `H` is the
harmonic mean `(H .M.)` between `a` and `b` then `a, H, b` are in `H.P.` and `H = (2ab)/(a+b)`