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 $(p/r)+(r/p)=(a/c)+(c/a)$ ?

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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

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

If aa, bb, cc are in A.P.; apap, bqbq, crcr are in G.P and pp, qq, rr are in H.P., then prove that (p/r)+(r/p)=(a/c)+(c/a)(p/r)+(r/p)=(a/c)+(c/a)?