Let p, q, r be nonzero real numbers that are, | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Since $p, q, r$ are the $10^{	ext{th}}, 100^{	ext{th}}, 1000^{	ext{th}}$ terms of a harmonic progression,their reciprocals $\frac{1}{p}, \frac{

Vedclass · Accepted Answer

$(I) \rightarrow (Q); (II) \rightarrow (S); (III) \rightarrow (S); (IV) \rightarrow (R)$

Vedclass · Answer

(B) Since $p, q, r$ are the $10^{	ext{th}}, 100^{	ext{th}}, 1000^{	ext{th}}$ terms of a harmonic progression,their reciprocals $\frac{1}{p}, \frac{1}{q}, \frac{1}{r}$ are in arithmetic progression. Let the $AP$ be $a + (n-1)d$. Then $\frac{1}{p} = a + 9d, \frac{1}{q} = a + 99d, \frac{1}{r} = a + 999d$.Dividing the third equation $qrx + pry + pqz = 0$ by $pqr$,we get $\frac{x}{p} + \frac{y}{q} + \frac{z}{r} = 0$.Substituting the $AP$ terms,the system becomes:$x+y+z=1$$x+10y+100z=0$$\frac{x}{p} + \frac{y}{q} + \frac{z}{r} = 0$Calculating the determinant $D$ of the system: $D = \begin{vmatrix} 1 & 1 & 1 \ 1 & 10 & 100 \ \frac{1}{p} & \frac{1}{q} & \frac{1}{r} \end{vmatrix}$.Using properties of $AP$,$D = 0$ if and only if the ratios of the terms match the progression. Specifically,the system has infinitely many solutions if $D=D_x=D_y=D_z=0$. For $(I)$,if $\frac{q}{r}=10$,the system is consistent with infinitely many solutions $(R)$,and since it has infinitely many solutions,it has at least one solution $(T)$.For $(II)$,if $\frac{p}{r} 
eq 100$,the system is inconsistent,leading to no solution $(S)$.For $(III)$,if $\frac{p}{q} 
eq 10$,the system is inconsistent,leading to no solution $(S)$.For $(IV)$,if $\frac{p}{q}=10$,the system has infinitely many solutions $(R)$,and thus at least one solution $(T)$.Matching these,$(I) ightarrow (R, T), (II) ightarrow (S), (III) ightarrow (S), (IV) ightarrow (R, T)$. Option $(B)$ is the correct match.

$List-I$	$List-II$
$(I)$ 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 $\frac{p}{r} \neq 100$,then the system of linear equations has	$(Q)$ $x=\frac{10}{9}, y=-\frac{1}{9}, z=0$ as a solution
$(III)$ If $\frac{p}{q} \neq 10$,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

Similar Questions

Solve the system of linear equations using the matrix method: $2x + 3y + 3z = 5$,$x - 2y + z = -4$,$3x - y - 2z = 3$.

Let $A = \begin{bmatrix} 1 & 1 & 1 \\ 0 & 1 & 3 \\ 1 & -2 & 1 \end{bmatrix}$,$B = \begin{bmatrix} 6 \\ 11 \\ 0 \end{bmatrix}$ and $X = \begin{bmatrix} a \\ b \\ c \end{bmatrix}$. If $AX = B$,then the value of $2a + b + 2c$ is:

If $x = a$,$y = b$,$z = c$ is a solution of the system of linear equations $x + 8y + 7z = 0$,$9x + 2y + 3z = 0$,and $x + y + z = 0$ such that the point $(a, b, c)$ lies on the plane $x + 2y + z = 6$,then $2a + b + c$ equals

If $p, q, r$ are $3$ real numbers satisfying the matrix equation $[p, q, r] \begin{bmatrix} 3 & 4 & 1 \\ 3 & 2 & 3 \\ 2 & 0 & 2 \end{bmatrix} = [3, 0, 1]$,then $2p + q - r$ equals

The number of real values $\lambda$,such that the system of linear equations $2x - 3y + 5z = 9$,$x + 3y - z = -18$,and $3x - y + (\lambda^2 - |\lambda|)z = 16$ has no solution,is :-

Vedclass Test Series

Exam Paper Generator

Online Exam Module