Question

$\text{List I}$ has four entries and $\text{List II}$ has five entries. Each entry of $\text{List I}$ is to be matched with one or more than one entries of $\text{List II}.$

$\begin{array}{cccc}& \text{List I}& & \text{List II}\\ \text{(A)}& \text{The vertices of a triangle are}(1,a),& \text{(P)}& a+b=13\\ & (2,b)\text{and}(c^2,-3).\text{If the centroid of}& & \\ & \text{the triangle lies on the}x\text{-axis, then}& & \\ \text{(B)}& \text{If}a,b,c(a<b<c)\text{are the roots of}& \text{(Q)}& ab=12\\ & x^3-6x^2+11x-6=0\text{, then}& & \\ \text{(C)}& \text{If}a,b,c\text{lie between 2 and 18 such that}& \text{(R)}& (a,b)=(-5,8)\\ & \text{(i)}a+b+c=25& & \\ & \text{(ii)}2,a,b\text{are in A.P.}& & \\ & \text{(iii)}b,c,18\text{are in G.P., then}& & \\ \text{(D)}& \text{If}a,b,c\text{are in G.P. with common ratio}& \text{(S)}& a+b=8\\ & r(r>1)\text{such that}abc=216\text{and}& & \\ & ab+bc+ca=156\text{, then}& & \\ & & \text{(T)}& a+b=3\end{array}$

Which of the following is the only CORRECT combination?

• A. $\text{(C)}\to \text{(P)},\text{(Q)}$
• B. $\text{(C)}\to \text{(P)},\text{(R)}$
• C. $\text{(D)}\to \text{(R)},\text{(S)}$
• D. $\text{(D)}\to \text{(Q)},\text{(S)}$

Answer 1

The correct option is D $\text{(D)}\to \text{(Q)},\text{(S)}$

$\left(C\right)$
$a,b,c$ are three numbers between $2$ and $18$
$a+b+c=25\cdots \left(1\right)$
$2,a,b$ are in A.P.
$\Rightarrow 2a=2+b\cdots \left(2\right)$
$b,c,18$ are in G.P.
$\Rightarrow c^2=18b\cdots \left(3\right)$

From $\left(1\right),\left(2\right)$ and $\left(3\right),$
$\frac{1}{2}(2+\frac{c^2}{18})+\frac{c^2}{18}+c=25\Rightarrow c=12$
$\therefore b=\frac{c^2}{18}=8\Rightarrow a=5$
$\text{(C)}\to \text{(P)}$


$\left(D\right)$
$a,b,c$ are in G.P., common ratio $=r(r>1)$
Let $a=\frac{b}{r},c=br$
$abc=216\Rightarrow b^3=216\Rightarrow b=6$

$ab+bc+ca=156$
$\Rightarrow b^2(\frac{1}{r}+r+1)=156\Rightarrow 3r^2-10r+3=0$
$\Rightarrow r=\frac{1}{3}$ or $3$
But $r>1.\therefore r=3$
$\Rightarrow a=2,c=18$
$\text{(D)}\to \text{(Q)},\text{(S)}$