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?

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Answer 1 · 2017-12-17T12:57:06

Please see below.

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#