Question

If first three terms $\sin \theta ,\sqrt{2}(\sin \theta +1),6(\sin \theta +1)$ are in G.P. then the fifth term is

• A. $162$
• B. $82\sqrt{2}$
• C. $16$
• D. none of these

Answer 1

The correct option is A $162$

Let $a=\sin \theta ,b=\sqrt{2}(\sin \theta +1),c=6(\sin \theta +1)$

In GP, $b^2=ac$

$\Rightarrow 2(\sin \theta +1{)}^2=6\sin \theta (\sin \theta +1)$

$\Rightarrow (\sin \theta +1{)}^2=3(\sin \theta +\sin \theta )$

$\Rightarrow 1+{\sin }^2\theta +2\sin \theta =3{\sin }^2\theta +3\sin \theta$

$\Rightarrow 1=2{\sin }^2\theta +\sin \theta$

$\Rightarrow 2{\sin }^2\theta +\sin \theta -1=0$

$\Rightarrow 2{\sin }^2\theta +\sin \theta -\sin \theta -1=0$

$\Rightarrow 2\sin \theta (\sin \theta +1)-1(\sin \theta +1)=0$

$\Rightarrow (2\sin \theta -1)(\sin \theta +1)=0$

$\sin \theta =\frac{1}{2},-1$

$\sin \theta =\frac{1}{2}$

Then GP is $\frac{1}{2},\sqrt{2}\left(\frac{3}{2}\right),6\left(\frac{3}{2}\right)+...............$

$=\frac{1}{2},\frac{3}{\sqrt{2}},\mathrm{9.............}$

$a=\frac{1}{2},r=\frac{\frac{3}{\sqrt{2}}}{\frac{1}{2}}=3\sqrt{2}$

Fifth term$=a{r}^4$

$=\frac{1}{2}(3\sqrt{2}{)}^4$

$=\frac{1}{2}(9\times 9\times 2\times 2)=162$