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

JEE Main · 2023 · 24 Jan 2023 · Shift 2 · Sequences and Series

For three positive integers

p, q, r, x^{p q^{2}} = y^{q r} = z^{p^{2} r}

and

r = pq + 1

such that

3, 3 lo g_{y} x, 3 lo g_{z} y, 7 lo g_{x} z

are in A.P. with common difference

\frac{1}{2}

. Then

r - p - q

is equal to

Answer: (A)

Set

x^{p q^{2}} = y^{q r} = z^{p^{2} r} = λ

, so that

lo g_{y} x = \frac{q r}{p q ^{2}} = \frac{r}{pq}

lo g_{z} y = \frac{p ^{2} r}{q r} = \frac{p ^{2}}{q}

and

lo g_{x} z = \frac{p q ^{2}}{p ^{2} r} = \frac{q ^{2}}{p r}

. The four terms

3, \frac{3 r}{pq}, \frac{3 p ^{2}}{q}, \frac{7 q ^{2}}{p r}

are in A.P. with common difference

\frac{1}{2}

, so the second term is

\frac{7}{2}

\frac{3 r}{pq} = \frac{7}{2}

gives

r = \frac{7 pq}{6}

, which together with

r = pq + 1

forces

pq = 6

and

r = 7

. The third term is

4

\frac{3 p ^{2}}{q} = 4

, and with

q = \frac{6}{p}

this is

\frac{p ^{3}}{2} = 4

, so

p = 2

and

q = 3

. Hence

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

JEE Main · 2023 · 24 Jan 2023 · Shift 2 · Single correct option