Verified Solution • 2025–26 Session moderate

QJEEM23Ma0e06: JEE Main 2023 Mathematics - Class 11 JEEM23Ma0e06 Solution: For three positive integers p, q, r, and r = pq + 1 such that…

For three positive integers p, q, r, xpq2=yqr=zp2r{x^{p{q^2}}} = {y^{qr}} = {z^{{p^2}r}} and r = pq + 1 such that 3, 3 log yx_yx , 3 log zy_zy , 7 log xz_xz are in A.P. with common difference 12\frac{1}{2} . Then r-p-q is equal to

Multiple Choice Options

  • A 12
  • B −- 6
  • C 6
  • D 2

Step-by-Step Verified Solution

xpq2=yqr=zp2rx^{p q^{2}}=y^{q r}=z^{p^{2} r}

3log⁡yx=72,3log⁡zy=4,7log⁡xz=923 \log _{y} x=\frac{7}{2}, 3 \log _{z} y=4,7 \log _{x} z=\frac{9}{2}

amp;⇒x=y76,y=z43,z=x914amp;y76pq2=yqr=y34p2ramp;⇒76pq2=qr=34p2ramp;∴7pq=6r,4q=3p2amp;r=pq+1amp;r=6r7+1⇒r=7amp;pq=6amp;p(3p24)=6amp;p=2,q=3amp;r−p−q=7−5=2\begin{aligned}& \Rightarrow x=y^{\frac{7}{6}}, y=z^{\frac{4}{3}}, z=x^{\frac{9}{14}} \\\\& y^{\frac{7}{6} p q^{2}}=y^{q r}=y^{\frac{3}{4} p^{2} r} \\\\& \Rightarrow \frac{7}{6} p q^{2}=q r=\frac{3}{4} p^{2} r \\\\& \therefore 7 p q=6 r, 4 q=3 p^{2} \\\\& r=p q+1 \\\\& r=\frac{6 r}{7}+1 \Rightarrow r=7 \\\\& p q=6 \\\\& p\left(\frac{3 p^{2}}{4}\right)=6 \\\\& p=2, q=3 \\\\& r-p-q=7-5=2\end{aligned}

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