JEE Advanced 2022 — Paper 1 — Question 17

Relationship between $p, q,$ and $r$ Given $p, q, r$ are the $10^{th}, 100^{th},$ and $1000^{th}$ terms of an H.P. Let the first term of the corresponding A.P. be $A$ and common difference be $D$.

\frac{1}{p} = A + 9D, \quad \frac{1}{q} = A + 99D, \quad \frac{1}{r} = A + 999D \end{aligned}$$ Calculating differences: $$\begin{aligned} \frac{1}{q} - \frac{1}{p} &= 90D \\ \frac{1}{r} - \frac{1}{q} &= 900D \\ \implies \frac{1}{r} - \frac{1}{q} &= 10 \left( \frac{1}{q} - \frac{1}{p} \right) \\ \implies \frac{10}{p} - \frac{11}{q} + \frac{1}{r} &= 0 \quad \dots \text{(Identity 1)} \end{aligned}$$ Analyzing the System of Equations The system is: $x + y + z = 110$ $x + 100y + 1000z = 0$ $\frac{x}{p} + \frac{y}{q} + \frac{z}{r} = 0$ (Dividing the given 3rd equation by $pqr$) Substituting values for $y$ and $z$ from Equation (1) and (2) into (Identity 1) or solving via Cramer's Rule/Determinants ($\Delta$): The determinant of the coefficient matrix is:

\Delta = \begin{vmatrix} 1 & 1 & 1 \ 1 & 100 & 1000 \ 1/p & 1/q & 1/r \end{vmatrix}

$$Expanding along the third row:$$

\Delta = \frac{1}{p}(1000 - 100) - \frac{1}{q}(1000 - 1) + \frac{1}{r}(100 - 1)

$$$$

\Delta = \frac{900}{p} - \frac{999}{q} + \frac{99}{r} = 9 \left( \frac{100}{p} - \frac{111}{q} + \frac{11}{r} \right)

Evaluating the Matches (I) If $\frac{q}{r} = 10$, substituting into $\Delta$ and Identity 1 shows the system aligns with solution (Q). (II) If $\frac{p}{r} \neq 100$, the planes do not intersect at a common point/line consistent with the constant $110$, leading to (S) no solution. (III) If $\frac{p}{q} \neq 10$, the system remains inconsistent with the non-homogeneous part, leading to (S) no solution. (IV) If $\frac{p}{q} = 10$, the third equation becomes a linear combination of the first two, resulting in (R) infinitely many solutions.