Maths

  $LetG.P.be{a}_1=a,a_2=ar,a_3=a{r}^2$ ${}_{}{}_{}{}_{}{}^{}$ , ….

$?3a_2+a_3=2a_4$

^{$\Rightarrow 3ar+a{r}^2=2a{r}^3$}

$\Rightarrow 2a{r}^2-r-3=0$

$\therefore a_2+a_4+2a_5=a\left(r+r^3+2r^4\right)$

$\frac{}{}$ $=\frac{8}{3}\left(\frac{3}{2}+\frac{27}{8}+\frac{81}{8}\right)$  = 40

Kindly consider the following figure

36 = 2 * 2 * 3 * 3

Number should be odd multiple of 2 and does not having factor 3 and 9

Odd multiple of 2 are

102, 106, 110, 114….998 (225 no.)

No. of multiplies of 3 are

102, 114, 126 ….990 (75 no.)

Which are also included multiple of 9

Hence,

Required = 225 – 75 = 150

Equation of perpendicular bisector of AB is

$y-\frac{3}{2}=-\frac{1}{5}\left(x-\frac{5}{2}\right)\Rightarrow x+5y=10$

Solving it with equation of given circle,

${\left(x-5\right)}^2+{\left(\frac{10-x}{5}-1\right)}^2=\frac{13}{2}$

$\Rightarrow x-5=\pm \frac{5}{2}\Rightarrow x=\frac{5}{2}or\frac{15}{2}$

But

$x\ne \frac{5}{2}$

because AB is not the diameter.

So, centre will be

$\left(\frac{15}{2},\frac{1}{2}\right)$

Now,

$r^2={\left(\frac{15}{2}-2\right)}^2+{\left(\frac{1}{2}+1\right)}^2=\frac{65}{2}$

$?X=\left[\begin{array}{ccc}\hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right]$

$\therefore X^2=\left[\begin{array}{ccc}\hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right]$

$\therefore Y=\alpha l+\beta X+\gamma X^2=\left[\begin{array}{ccc}\hfill \alpha \hfill & \hfill \beta \hfill & \hfill \gamma \hfill \\ \hfill 0\hfill & \hfill \alpha \hfill & \hfill \beta \hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill \alpha \hfill \end{array}\right]$

$\begin{array}{l}\therefore \alpha =5,\beta =10,y=15\\ \therefore {\left(\alpha -\beta +y\right)}^2=100\end{array}$

$z^2+z+1=0\Rightarrow z=w,w^2$

$\left|{\displaystyle \sum _{n=1}^{15}{\left(z^n+{\left(-1\right)}^n\frac{1}{z^n}\right)}^2}\right|=\left|{\displaystyle \sum _{n=1}^{15}\left(z^{2n}+\frac{1}{z^{2n}}+2{\left(-1\right)}^n\right)}\right|=\left|{\displaystyle \sum _{n=1}^{15}{w}^{2n}+\frac{1}{w^{2n}}+2{\left(-1\right)}^n}\right|$

$=\left|\frac{w^2\left(1-1\right)}{1-w^2}+\frac{\frac{1}{w^2}\left(1-1\right)}{1-\frac{1}{w^2}}-2\right|=2$

  $?p+q=3$ $$ ….(i)

and $p^4+q^4$ ${}^{}{}^{}$ = 369 ….(ii)

${\left\{{\left(p+q\right)}^2-2pq\right\}}^2-2p^2{q}^2=369$

or ${\left(9-2pq\right)}^2-2{\left(pq\right)}^2=369$ ${}^{}{}^{}$

or ${\left(pq\right)}^2-18pq-144=0$ ${}^{}$

$\therefore pq=-6or24$

But pq 24 is not possible

$\therefore pq=-6$

$Hence,{\left(\frac{1}{p}+\frac{1}{q}\right)}^{-2}={\left(\frac{pq}{p+q}\right)}^2={\left(-2\right)}^2=4$

$f\left(x-y\right)=2^{x}f\left(y\right)+4^{y}f\left(x\right)$ ……. (i)

$$ $f\left(y+x\right)=2^{y}f\left(x\right)+4^{x}f\left(y\right)$  ……. (ii)

(i) – (ii)

$\therefore \frac{f\text{'}\left(4\right)}{f\text{'}\left(2\right)}=\frac{k\left(16\mathcal{l}n2-256\mathcal{l}n4\right)}{k\left(4\mathcal{l}n2-16\mathcal{l}n4\right)}$

$=\frac{\left(16-512\right)\mathcal{l}n2}{\left(4-32\right)\mathcal{l}n2}=\frac{496}{28}$

$14\frac{f\text{'}\left(4\right)}{f\text{'}\left(2\right)}=248$

Clearly r must be equal to $\sim$ $$ p

$?\sim p\vee \sim p=\sim p$

$and\left(p\wedge q\right)\vee \sim p=p$

$$ $\therefore \sim p\Rightarrow p=$  tautology.

$co{s}^{-1}\left(\frac{3}{10}cos\left(ta{n}^{-1}\left(\frac{4}{3}\right)\right)+\frac{2}{5}sin\left(ta{n}^{-1}\left(\frac{4}{3}\right)\right)\right)$

$=co{s}^{-1}\left(\frac{3}{10}.\frac{3}{5}+\frac{2}{5}.\frac{4}{5}\right)$

$=co{s}^{-1}\left(\frac{1}{2}\right)=\frac{\pi }{3}$