If three unequal numbers p,;q,;r are in H.P. | vedclass

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(d) By hypothesis, $q = \frac{{2pr}}{{p + r}}$

$ \Rightarrow $$\frac{q}{2} = \frac{{pr}}{{p + r}} = K$(say)

$ \Rightarrow $ $q = 2K,\;pr = (p +

Vedclass · Accepted Answer

$1 \mp \sqrt 3 : - 2:1 \pm \sqrt 3 $

Vedclass · Answer

(d) By hypothesis, $q = \frac{{2pr}}{{p + r}}$

$ \Rightarrow $$\frac{q}{2} = \frac{{pr}}{{p + r}} = K$(say)

$ \Rightarrow $ $q = 2K,\;pr = (p + r)K.$

Also ${p^2},\;{q^2},\;{r^2}$ are in $A.P.$

$	herefore $$2{q^2} = {p^2} + {r^2} = {(p + r)^2} - 2pr$

$ \Rightarrow $ $8{K^2} = {(p + r)^2} - 2(p + r)K$

$ \Rightarrow $${(p + r)^2} - 2(p + r)K - 8{K^2} = 0$

$ \Rightarrow $$p + r = 4K,\; - 2K$

When $p + r = 4K$, then $pr = 4{K^2}$

$	herefore $${(p - r)^2} = {(p + r)^2} - 4pr = 16{K^2} - 16{K^2} = 0$

$ \Rightarrow $$p = r$

But this is not possible $(\because \;\;p 
e r)$

$p + r = - 2K$$ \Rightarrow $$pr = - 2K\;.\;K = - 2{K^2}$

Now ${(p - r)^2} = {(p + r)^2} - 4pr = 4{K^2} - 4( - 2{K^2}) = 12{K^2}$

$ \Rightarrow $$p - r = \pm 2\sqrt 3 K,$ also $p + r = - 2K$

$	herefore $$2p = ( - 2 \pm 2\sqrt 3 )K$

$ \Rightarrow $$p = ( - 1 \pm \sqrt 3 )K$

and $2r = - 2K \mp 2\sqrt 3 K$

$ \Rightarrow $$r = ( - 1 \mp \sqrt 3 )K$

$	herefore $$p:q:r = ( - 1 \pm \sqrt 3 )K:2K:( - 1 \mp \sqrt 3 )K$

$ = - 1 \pm \sqrt 3 :2: - 1 \mp \sqrt 3 = 1 \mp \sqrt 3 : - 2:1 \pm \sqrt 3 $.

If three unequal numbers $p,\;q,\;r$ are in $H.P.$ and their squares are in $A.P.$, then the ratio $p:q:r$ is

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