MCQ
Three positive numbers form an increasing $G.P.$ If the middle term in this $G.P.$ is doubled, the new numbers are in $A.P.$ then the common ratio of the $G.P.$ is:
- A$2 - \sqrt 3 $
- ✓$2 + \surd 3$
- C$\sqrt 2 + \surd 3$
- D$3 + \surd 2$
If $a r$ is doubled then,
$\Longrightarrow 2 a r=\frac{a r^{2}+a}{2}$
$\Longrightarrow 4 a r=a r^{2}+a$
$\Longrightarrow 4 r=r^{2}+1$
$\Longrightarrow r^{2}-4 r+1$
On solving we get
$r=2 \pm \sqrt{3}$
since it is an increasing
$G . P^{\prime} r^{\prime}=2+\sqrt{3}$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
| $column 1$ | $column 2$ | $column 3$ |
| $(I)$ $x^2+y^2=a^2$ | $(i)$ $m y=m^2 x+a$ | $(P)$ $\left(\frac{a}{m^2}, \frac{2 a}{m}\right)$ |
| $(II)$ $x^2+a^2 y^2=a^2$ | $(ii)$ $y=m x+a \sqrt{m^2+1}$ | $(Q)$ $\quad\left(\frac{-m a}{\sqrt{m^2+1}}, \frac{a}{\sqrt{m^2+1}}\right)$ |
| $(III)$ $y^2=4 a x$ | $(iii)$ $y=m x+\sqrt{a^2 m^2-1}$ | $(R)$ $\quad\left(\frac{-a^2 m}{\sqrt{a^2 m^2+1}}, \frac{1}{\sqrt{a^2 m^2+1}}\right)$ |
| $(IV)$ $x^2-a^2 y^2=a^2$ | $(iv)$ $y=m x+\sqrt{a^2 m^2+1}$ | $(S)$ $\quad\left(\frac{-a^2 m}{\sqrt{a^2 m^2-1}}, \frac{-1}{\sqrt{a^2 m^2-1}}\right)$ |
($1$) The tangent to a suitable conic (Column $1$) at $\left(\sqrt{3}, \frac{1}{2}\right)$ is found to be $\sqrt{3} x+2 y=4$, then which of the following options is the only CORRECT combination?
$[A] (II) (iii) (R)$ $[B] (IV) (iv) (S)$ $[C] (IV) (iii) (S)$ $[D] (II) (iv) (R)$
($2$) If a tangent to a suitable conic (Column $1$) is found to be $y=x+8$ and its point of contact is $(8,16$ ), then which of the following options is the only CORRECT combination?
$[A] (III) (i) (P)$ $[B] (III) (ii) (Q)$ $[C] (II) (iv) (R)$ $[D] (I) (ii) (Q)$
($3$) For $a=\sqrt{2}$, if a tangent is drawn to a suitable conic (Column $1$ ) at the point of contact $(-1,1)$, then which of the following options is the only CORRECT combination for obtaining its equation?
$[A] (II) (ii) (Q)$ $[B] (III) (i) (P)$ $[\mathrm{C}]$ $(I) (1) (P)$ $[D] (I) (ii) (Q)$