For three positive integers p,q,r, suppose x^pq^2=y^qr=z^p^2r and r=pq+1, such that 3, 3_y…

Question

For three positive integers $p,q,r$, suppose $x^{pq^2}=y^{qr}=z^{p^2r}$ and $r=pq+1$, such that $3, 3\log_y x, 3\log_z y, 7\log_x z$ are in A.P. with common difference $\dfrac12$. Then $r-p-q$ is equal to:

Accepted Answer

Setting $x^{pq^2}=y^{qr}=z^{p^2r}=\lambda$: $\log_yx = \dfrac{qr}{pq^2}=\dfrac{r}{pq}$, $\log_zy=\dfrac{p^2}{q}$, $\log_xz=\dfrac{q^2}{pr}$.

The A.P. condition on term 2 gives $\dfrac{3r}{pq}-3=\dfrac12 \Rightarrow \dfrac{r}{pq}=\dfrac76$. With $r=pq+1$: $pq+1=\dfrac76pq \Rightarrow pq=6, r=7$.

Using the term-3 condition, $3p^2/q - 3r/(pq) = 	frac12$, together with $pq=6$, gives $p=2$, $q=3$.

$r-p-q = 7-2-3 = 2.$

Solutions · 1