If #a#, #b#, #c# are in A.P.; #ap#, #bq#, #cr# are in G.P and #p#, #q#, #r# are in H.P., then prove that #(p/r)+(r/p)=(a/c)+(c/a)#?

Question

3850 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-02-29T05:56:13

#(p/r)+(r/p)=(a/c)+(c/a)#

See proof below.

If #a#, #b#, #c# are in A.P; we have #2b=a+c# -------- (A)

If #ap#, #bq#, #cr# are in G.P; then #(bq)^2=apxxcr# or #(rp)/q^2=b^2/(ac)# ------- (B)

and as #p#, #q#, #r# are in H.P, #1/p#, #1/q#, #1/r# are in A.P. and
#2/q=1/p+1/r=(p+r)/(rp)# or #(p+r)=(2rp)/q# ---------- (C)

Now #(p/r)+(r/p)=(p^2+r^2)/(rp)=((p+r)^2-2pr)/(rp)=(p+r)^2/(rp)-2#

Using relation (C) for #(p+r)#, we get

#(p/r)+(r/p)=((2rp)/q)^2xx1/(rp)-2# or

#(p/r)+(r/p)=((4rp)/q^2)-2# and using (B) this becomes

#(p/r)+(r/p)=((4b^2)/(ac))-2=(2b)^2/(ac)-2=(a+c)^2/(ac)-2# i.e.

#(p/r)+(r/p)=((a+c)^2-2ac)/(ac)=(a^2+c^2)/(ac)# or

#(p/r)+(r/p)=(a/c)+(c/a)#