MCQ
${7^{th}}$ term of the sequence $\sqrt 2 ,\;\sqrt {10} ,\;5\sqrt 2 ,\;.......$ is
- A$125\sqrt {10} $
- B$25\sqrt 2 $
- C$125$
- ✓$125\sqrt 2 $
Common ratio $r = \sqrt 5 $,
first term $a = \sqrt 2 $, then ${7^{th}}$ term
${t_7} = \sqrt 2 {(\sqrt 5 )^{7 - 1}} = \sqrt 2 {(\sqrt 5 )^6}$
$= \sqrt 2 {(5)^3} = 125\sqrt 2 $.
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$[A]$ $\tan \left(\frac{\alpha}{2}\right)+\sqrt{3} \tan \left(\frac{\beta}{2}\right)=0$
$[B]$ $\sqrt{3} \tan \left(\frac{\alpha}{2}\right)+\tan \left(\frac{\beta}{2}\right)=0$
$[C]$ $\tan \left(\frac{\alpha}{2}\right)-\sqrt{3} \tan \left(\frac{\beta}{2}\right)=0$
$[D]$ $\sqrt{3} \tan \left(\frac{\alpha}{2}\right)-\tan \left(\frac{\beta}{2}\right)=0$
$(i)$ Let $\alpha_1$ be the total number of ways in which the committee can be formed such that the committee has $5$ members, having exactly $3$ boys and $2$ girls.
$(ii)$ Let $\alpha_2$ be the total number of ways in which the committee can be formed such that the committee has at least $2$ members, and having an equal number of boys and girls.
$(iii)$ Let $\alpha_3$ be the total number of ways in which the committee can be formed such that the committee has $5$ members, at least $2$ of them being girls.
$(iv)$ Let $\alpha_4$ be the total number of ways in which the committee can be formed such that the committee has $4$ members, having at least $2$ girls and such that both $M _1$ and $G _1$ are $NOT$ in the committee together.
| $LIST I $ | $LIST I $ |
| $P$ The value of $\alpha_1$ is | $1$ $136$ |
| $Q$ The value of $\alpha_2$ is | $2$ $189$ |
| $R$ The value of $\alpha_3$ is | $3$ $192$ |
| $S$ The value of $\alpha_4$ is | $4$ $200$ |
| $5$ $381$ | |
| $6$ $461$ |
The correct option is: