Question

Let p, q, r be non-zero real numbers that are, respectively, the 10^th, 100^th and 1000^th terms of a harmonic progression. Consider the system of linear equations
x + y + z = 1
10x + 100y + 1000z = 0
qr x + pr y + pq z = 0
 

List-I		List-II
(I)	If \(\frac{q}{r}=10\), then the system of linear equations has	(P)	x = 0, \(y=\frac{10}{9},z=-\frac{1}{9}\) as a solution
(II)	If \(\frac{p}{r}≠100\), then the system of linear equations has	(Q)	\(x=\frac{10}{9},y=\frac{-1}{9},z=0\) as a solution
(III)	If \(\frac{p}{q}≠10,\) then the system of linear equations has	(R)	infinitely many solutions
(IV)	If \(\frac{p}{q}=10,\) then the system of linear equations has	(S)	no solution
		(T)	at least one solution

• A. (I) → (T); (II) → (R); (III) → (S); (IV) → (T)
• B. (I) → (Q); (II) → (S); (III) → (S); (IV) → (R)
• C. (I) → (Q); (II) → (R); (III) → (P); (IV) → (R)
• D. (I) → (T); (II) → (S); (III) → (P); (IV) → (T)

Answer 1

The Correct Option is B

\(x + y + z = 1 ..... (1)\) 
\(10x + 100y + 1000z = 0 ..... (2)\)
\(\frac{x}{p} + \frac{y}{q} + \frac{z}{r} = 0\)

\(\frac{1}{p} = A + 9d, \quad \frac{1}{q} = A + 99d, \quad \frac{1}{r} = A + 999d\)

\(⇒\) From equation (2) and (3), we get \((A-d)x+(A-d)y+(A-d)z=0\)
\(⇒\) If A≠d, then no solution
Option I: If \(\frac{q}{r} = 10 ⇒ a = d\)
And eq. (1) and eq. (2) represents non-parallel planes and eq. (2) and eq. (3) represents same plane

\(⇒\) Infinitely many solutions
I → P, Q, R, T
Option II : \(\frac{p}{r}≠100\) \(⇒\) \(a≠d\)
No solution
II → S

Option III:
\(\frac{p}{q}≠10,\)\(⇒\)\(a≠d\)
No solution
III → S

Option IV: If \(\frac{p}{q}=10,\)\(⇒\)\(a=d\)
Infinitely many solutions

IV → P, Q, R, T

The correct answer is option (B): (I) → (Q); (II) → (S); (III) → (S); (IV) → (R)