For three positive integers p,q,r,x^pq^2 = y^qr = z^p^2r and r = pq + 1 such that 3,3log _yx,3log _zy,7log _xz are in A.P. with common difference 1/2. The r - p - q is equal to

pq^2=log _ x lambda \ qr=log _ y lambda \ p^2 r=log _2 lambda \ log _ y x=frac qr pq^2=frac r pq ... ... (1) \ log _ x z=frac pq^2 p^2 r=frac q^2 pr ... ... . .(2) \ log _ z y=frac p^2 r qr=frac p^2 q ... ... .(3) \ 3, frac3 r pq, frac3 p^2 q, frac7 q^2 pr in A.P \ frac3 r pq-3=(1 / 2) \ r=(7 / 6) pq ... ... .(4) \ r= pq+1 \ pq=6 ... ... ... ... . .(5) \ r=7 ... ... ... ... ... . .(6) \ frac3 p^2 q=4 \ (frac3 p^2q)(p q)=4 x 6 \ => 3 p^3=mathbf2 4 \ => p^3=8=mathbf2^3 \ => boldsymbolp=mathbf2 \ Therefore boldsymbolq=frac6mathbfp=frac6mathbf2=mathbf3 \ Therefore boldsymbolr-boldsymbolp-boldsymbolq=mathbf7-mathbf2-mathbf3=mathbf7-mathbf5=mathbf2\

For three positive integers p,q,r,x^{pq^{2}} = y^{qr} ... | Sequence & Series (Mathematics)

$\begin{aligned} & \mathrm{pq}^2=\log _{\mathrm{x}} \lambda \\ & \mathrm{qr}=\log _{\mathrm{y}} \lambda \\ & \mathrm{p}^2 \mathrm{r}=\log _2 \lambda \\ & \log _{\mathrm{y}} \mathrm{x}=\frac{\mathrm{qr}}{\mathrm{pq}^2}=\frac{\mathrm{r}}{\mathrm{pq}} \ldots \ldots(1) \\ & \log _{\mathrm{x}} \mathrm{z}=\frac{\mathrm{pq}^2}{\mathrm{p}^2 \mathrm{r}}=\frac{\mathrm{q}^2}{\mathrm{pr}} \ldots \ldots . .(2) \\ & \log _{\mathrm{z}} \mathrm{y}=\frac{\mathrm{p}^2 \mathrm{r}}{\mathrm{qr}}=\frac{\mathrm{p}^2}{\mathrm{q}} \ldots \ldots .(3) \\ & 3, \frac{3 \mathrm{r}}{\mathrm{pq}}, \frac{3 \mathrm{p}^2}{\mathrm{q}}, \frac{7 \mathrm{q}^2}{\mathrm{pr}} \text { in A.P } \\ & \frac{3 \mathrm{r}}{\mathrm{pq}}-3=\frac{1}{2} \\ & \mathrm{r}=\frac{7}{6} \mathrm{pq} \ldots \ldots .(4) \\ & \mathrm{r}=\mathrm{pq}+1 \\ & \mathrm{pq}=6 \ldots \ldots \ldots \ldots . .(5) \\ & \mathrm{r}=7 \ldots \ldots \ldots \ldots \ldots . .(6) \\ & \frac{3 \mathrm{p}^2}{\mathrm{q}}=4 \\ & \left(\frac{3 \mathrm{p}^2}{q}\right)(p q)=4 \times 6 \\ & \Rightarrow 3 p^3=\mathbf{2 4} \\ & \Rightarrow p^3=8=\mathbf{2}^3 \\ & \Rightarrow \boldsymbol{p}=\mathbf{2} \\ & \therefore \boldsymbol{q}=\frac{6}{\mathbf{p}}=\frac{6}{\mathbf{2}}=\mathbf{3} \\ & \therefore \boldsymbol{r}-\boldsymbol{p}-\boldsymbol{q}=\mathbf{7}-\mathbf{2}-\mathbf{3}=\mathbf{7}-\mathbf{5}=\mathbf{2}\end{aligned}$

Explanation

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