MCQ
The first term of a $G.P.$ whose second term is $2$ and sum to infinity is $8$, will be
- A$6$
- B$3$
- ✓$4$
- D$1$
$ \Rightarrow $ $8 = \frac{2}{{r(1 - r)}}\left( {\;a = \frac{2}{r}} \right)$
$ \Rightarrow $ $4r(1 - r) = 1 $
$\Rightarrow 4r - 4{r^2} - 1 = 0$
$ \Rightarrow $ $4{r^2} - 4r + 1 = 0$
$\Rightarrow \left( {r - \frac{1}{2}} \right)(4r - 2) = 0$
$\Rightarrow r = \frac{1}{2}$
So first term $a = 4$.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Match the Statements / Expressions in Column $I$ with the Statements / Expressions in Column $II$ and indicate your answer by darkening the appropriate bubbles in the $4 \times 4$ matrix given in the $ORS.$
| Column $I$ | Column $II$ |
| $(A)$ $ \mathrm{L}_1, \mathrm{~L}_2, \mathrm{~L}_3$ are concurrent, if | $(p)$ $\mathrm{k}=-9$ |
| $(B)$ One of $\mathrm{L}_1, \mathrm{~L}_2, \mathrm{~L}_3$ is parallel to at least one of the other two, if | $(q)$ $\mathrm{k}=-\frac{6}{5}$ |
| $(C)$ $\mathrm{L}_1, \mathrm{~L}_2, \mathrm{~L}_3$ form a triangle, if | $(r)$ $\mathrm{k}=\frac{5}{6}$ |
| $(D)$ $ \mathrm{L}_1, \mathrm{~L}_2, \mathrm{~L}_3$ do not form a triangle, if | $(s)$ $\mathrm{k}=5$ |