For three positive integers p , q , r , x ^{ | vedclass

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Question

$pq ^2=\log _{ x } \lambda$

$qr =\log _{ y } \lambda$

$p ^2 r =\log _{ z } \lambda$

$\log _{ y } x =\frac{ qr }{ pq ^2}=\frac{ r }{ pq } \ldo

Vedclass · Accepted Answer

$2$

Vedclass · Answer

$pq ^2=\log _{ x } \lambda$

$qr =\log _{ y } \lambda$

$p ^2 r =\log _{ z } \lambda$

$\log _{ y } x =\frac{ qr }{ pq ^2}=\frac{ r }{ pq } \ldots(1)$

$\log _{ x } z =\frac{ pq ^2}{ p ^2 r }=\frac{ q ^2}{ pr } \ldots(2)$

$\log _{ z } y =\frac{ p ^2 r }{ qr }=\frac{ p ^2}{ q } \ldots \ldots(3)$

$3, \frac{3 r }{ pq }, \frac{3 p ^2}{ q }, \frac{7 q ^2}{ pr } 	ext { in A.P }$

$\frac{3 r }{ pq }-3=\frac{1}{2}$

$r =\frac{7}{6} pq.....(4)$

$r = pq +1$

$pq =6 \ldots(5)$

$r =7 \ldots \ldots(6)$

$\frac{3 p ^2}{ q }=4$

After solving $p =2$ and $q =3$

For three positive integers $p , q , r , x ^{ pq p ^2}= y ^{ qr }= z ^{ p ^2 r }$ and $r=p q+1$ such that $3,3 \log _y x, 3 \log _z y, 7 \log _x z$ are in A.P. with common difference $\frac{1}{2}$. Then $r - p - q$ is equal to

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