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.
$(\lambda-1) x+(3 \lambda+1) y+2 \lambda z=0$
$(\lambda-1) x+(4 \lambda-2) y+(\lambda+3) z=0$
$2 x+(3 \lambda+1) y+3(\lambda-1) z=0$
has non-zero solutions, is
$S_n(x)=\sum_{k=1}^n \cot ^{-1}\left(\frac{1+k(k+1) x^2}{x}\right)$
where for any $x \in R , \cot ^{-1} x \in(0, \pi)$ and $\tan ^{-1}(x) \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$. Then which of the following
statements is (are) $TRUE$?
$(A)$ $S _{10}( x )=\frac{\pi}{2}-\tan ^{-1}\left(\frac{1+11 x ^2}{10 x }\right)$, for all $x >0$
$(B)$ $\lim _{n \rightarrow \infty} \cot \left(S_n(x)\right)=x$, for all $x>0$
$(C)$ The equation $S_3(x)=\frac{\pi}{4}$ has a root in $(0, \infty)$
$(D)$ $\tan \left( S _{ n }( x )\right) \leq \frac{1}{2}$, for all $n \geq 1$ and $x >0$