Question

Let p , q , r be non-zero real numbers that are, respectively, the 10th, 100th and 1000th terms of a harmonic progression. Consider the system of linear equations
x + y + z = 1
10 x + 100 y + 1000 z = 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)
$y=\frac{10}{9},z=-\frac{1}{9}$
as a solution
(II)	If $\frac{p}{r}≠100$,
then the system of linear equations has	(Q)
as a solution
(III)	If $\frac{p}{q}≠10,$
then the system of linear equations has	(R)
(IV)	If $\frac{p}{q}=10,$
then the system of linear equations has	(S)
	(T)

• 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

Answer: (B)

(I) → (Q); (II) → (S); (III) → (S); (IV) → (R)

Answer 2

The correct answer is option (B): (I) → (Q); (II) → (S); (III) → (S); (IV) → (R)
$x + y + z = 1 ..... (1)$
$10x + 100y + 1000z = 0 ..... (2)$
$\frac{x}{p} + \frac{y}{q} + \frac{z}{r} = 0$

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

$⇒$ 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