For three positive integers p , q , r , x p q 2 = y q r = 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 1/2. The r - p - q is equal to [2023]

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Solution

Q.
For three positive integers $p, q, r, x^{p q^{2}} = y^{q r} = z^{p^{2} r}$ 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 1/2. The $r - p - q$ is equal to [2023]

1 6

2 2

3 12

4 - 6

Ans.
(2)

$p q^{2} \log x = q r \log y = p^{2} r \log z$

$Now, \frac{3 \log x}{\log y} - 3 = \frac{1}{2} \Rightarrow \frac{\log x}{\log y} = \frac{7}{6} = \frac{q r}{p q^{2}} = \frac{r}{p q}$

$\Rightarrow 7 p q = 6 r \Rightarrow 7 (r - 1) = 6 r \Rightarrow r = 7$

$and \frac{3 \log y}{\log z} - 3 = 1 \Rightarrow \frac{\log y}{\log z} = \frac{4}{3} = \frac{p^{2} r}{q r} = \frac{p^{2}}{q} \Rightarrow 3 p^{2} = 4 q$

$Also, \frac{7 \log z}{\log x} - 3 = \frac{3}{2} \Rightarrow \frac{7 \log z}{\log x} = \frac{9}{2} \Rightarrow \frac{7 p q^{2}}{p^{2} r} = \frac{7 q^{2}}{p r}$

$\Rightarrow \frac{7 q^{2}}{p \times 7} = \frac{9}{2} \Rightarrow \frac{q^{2}}{p} = \frac{9}{2} \Rightarrow 2 q^{2} = 9 p \Rightarrow 4 q^{4} = 81 p^{2}$

$⇒ 4 q^{4} = 27 \times 3 p^{2} = 27 \times 4 q \Rightarrow q^{3} = 27 \Rightarrow q = 3$

$We have, r = p q + 1 \Rightarrow 7 = 3 p + 1 \Rightarrow p = 2$

$∴$ $r - p - q = 7 - 2 - 3 = 2$

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