The equation $\sin 3\theta = 4 \sin \theta \sin 2\theta \sin 4\theta$ in the interval $0 \le \theta \le \pi$ has:

If $2 \tan ^2 \theta-5 \sec \theta=1$ has exactly $7$ solutions in the interval $\left[0, \frac{n \pi}{2}\right]$,for the least value of $n \in N$,then $\sum_{k=1}^{n} \frac{k}{2^{k}}$ is equal to :

In a $\Delta ABC$,$a = a_1 = 2$,$b = a_2$,$c = a_3$ such that $a_{p+1} = \frac{5^p}{3^{2-p}} a_p \left( 2^{2-p} - \frac{4p-2}{5^p} a_p \right)$ where $p = 1, 2$,then:

If $x = \frac{n\pi}{2}$ satisfies the equation $\sin \frac{x}{2} - \cos \frac{x}{2} = 1 - \sin x$ and the inequality $\left| \frac{x}{2} - \frac{\pi}{2} \right| \le \frac{3\pi}{4}$,then:

Let $PQRS$ be a quadrilateral in a plane,where $QR = 1$,$\angle PQR = \angle QRS = 70^{\circ}$,$\angle PQS = 15^{\circ}$ and $\angle PRS = 40^{\circ}$. If $\angle RPS = \theta^{\circ}$,$PQ = \alpha$ and $PS = \beta$,then the interval$(s)$ that contain$(s)$ the value of $4 \alpha \beta \sin \theta^{\circ}$ is/are
$(A)$ $(0, \sqrt{2})$
$(B)$ $(1, 2)$
$(C)$ $(\sqrt{2}, 3)$
$(D)$ $(2 \sqrt{2}, 3 \sqrt{2})$

In $\triangle ABC$,$AD$ and $BE$ are medians drawn from $A$ and $B$. If $AD = \frac{7}{2}$,$\angle DAB = \frac{\pi}{8}$ and $\angle ABE = \frac{\pi}{4}$,then the area (in sq. units) of $\triangle ABC$ is