Question

If `p^th` , `q^th` , and `r^th` terms of a H.P are a,b,c respectively, then prove that #(q-r) / (a) + (r-p) / (b) + (p-q)/(c) = 0?

Answer 1

Please see below.

Explanation:

As `p^(th),q^(th)` and `r^(th)` terms of a H.P are `a,b,c`

`p^(th),q^(th)` and `r^(th)` terms are `1/a,1/b,1/c` of corresponding A.P.

Let the first term of this A.P. be `f` and common difference be `d`

then `1/a=f+(p-1)d` ............(1)

`1/b=f+(q-1)d` ............(2)

and `1/c=f+(r-1)d` ............(3)

Subtracting (2) from (1); (3) from (2) and (1) from (3), we ger

`1/a-1/b=d(p-q)` or `p-q=(b-a)/(abd)` ............(A)

`1/b-1/c=d(q-r)` or `q-r=(c-b)/(bcd)` ............(B)

and `1/c-1/a=d(r-p)` or `r-p=(a-c)/(acd)` ............(C)

hence `(q-r)/a+(r-p)/b+(p-q)/c`

= `(c-b)/(abcd)+(a-c)/(abcd)+(b-a)/(abcd)`

= `(c-b+a-c+b-a)/(abcd)`

= `0`