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

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

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Answer 1 · 2016-02-29T05:56:13

$(\frac{p}{r}) + (\frac{r}{p}) = (\frac{a}{c}) + (\frac{c}{a})$ $(p/r)+(r/p)=(a/c)+(c/a)$

See proof below.

Explanation:

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

If $a p$ $ap$ , $b q$ $bq$ , $c r$ $cr$ are in G.P; then ${(b q)}^{2} = a p \times c r$ $(bq)^2=apxxcr$ or $\frac{r p}{q^{2}} = \frac{b^{2}}{a c}$ $(rp)/q^2=b^2/(ac)$ ------- (B)

and as $p$ $p$ , $q$ $q$ , $r$ $r$ are in H.P, $\frac{1}{p}$ $1/p$ , $\frac{1}{q}$ $1/q$ , $\frac{1}{r}$ $1/r$ are in A.P. and
$\frac{2}{q} = \frac{1}{p} + \frac{1}{r} = \frac{p + r}{r p}$ $2/q=1/p+1/r=(p+r)/(rp)$ or $(p + r) = \frac{2 r p}{q}$ $(p+r)=(2rp)/q$ ---------- (C)

Now $(\frac{p}{r}) + (\frac{r}{p}) = \frac{p^{2} + r^{2}}{r p} = \frac{{(p + r)}^{2} - 2 p r}{r p} = \frac{{(p + r)}^{2}}{r p} - 2$ $(p/r)+(r/p)=(p^2+r^2)/(rp)=((p+r)^2-2pr)/(rp)=(p+r)^2/(rp)-2$

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

$(\frac{p}{r}) + (\frac{r}{p}) = {(\frac{2 r p}{q})}^{2} \times \frac{1}{r p} - 2$ $(p/r)+(r/p)=((2rp)/q)^2xx1/(rp)-2$ or

$(\frac{p}{r}) + (\frac{r}{p}) = (\frac{4 r p}{q^{2}}) - 2$ $(p/r)+(r/p)=((4rp)/q^2)-2$ and using (B) this becomes

$(\frac{p}{r}) + (\frac{r}{p}) = (\frac{4 b^{2}}{a c}) - 2 = \frac{{(2 b)}^{2}}{a c} - 2 = \frac{{(a + c)}^{2}}{a c} - 2$ $(p/r)+(r/p)=((4b^2)/(ac))-2=(2b)^2/(ac)-2=(a+c)^2/(ac)-2$ i.e.

$(\frac{p}{r}) + (\frac{r}{p}) = \frac{{(a + c)}^{2} - 2 a c}{a c} = \frac{a^{2} + c^{2}}{a c}$ $(p/r)+(r/p)=((a+c)^2-2ac)/(ac)=(a^2+c^2)/(ac)$ or

$(\frac{p}{r}) + (\frac{r}{p}) = (\frac{a}{c}) + (\frac{c}{a})$ $(p/r)+(r/p)=(a/c)+(c/a)$

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

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Explanation:

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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 (pr)+(rp)=(ac)+(ca)(p/r)+(r/p)=(a/c)+(c/a)?

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