Maths

47. Given, a= 729

$a_7=a{r}^6=64$

$\Rightarrow$ 729. r⁶ = 64

$\Rightarrow$ $r^6=\frac{64}{729}$

$\Rightarrow$ $r^6={\left(\frac{2}{3}\right)}^6$

$\Rightarrow$ $r=+\frac{2}{3}$ <1

When $r=\frac{2}{3}$

= $s_7=\frac{a\left(1-r^7\right)}{4-r}=\frac{729\left(1-{\left(\frac{2}{3}\right)}^7\right)}{1-\frac{2}{3}}=\frac{729\left[1-\frac{128}{2187}\right]}{\frac{3-2}{2}}$

= $729*\frac{3}{1}*\left[\frac{2187-128}{2187}\right]$

= $729*\frac{3*2059}{2187}$

= 2059

When $r=\frac{2}{3}$

$s_7=\frac{a\left(1-r^7\right)}{1-r}=\frac{729\left[1-{\left(\frac{-2}{3}\right)}^7\right]}{1-\left(\frac{-2}{3}\right)}=\frac{729\left[1+\frac{128}{2187}\right]}{1+\frac{2}{3}}$

= $\frac{729\left[\frac{2187+128}{2187}\right]}{\frac{3+2}{3}}$

= $729*\frac{3}{5}*\left[\frac{2315}{2187}\right]$

= 463

46. Let a be the first term and r be the common ratio of the G.P.

Then $a_1+a_2+a_3=16$

$a+ar+a{r}^2+16^{}$

$a\left(1+r+r^2\right)=16$ ………. I

Also $a_4+a_5+a_6=128$

$a{r}^3+a{r}^4+a{r}^5=128$

$a{r}^3\left(1+r+r^2\right)=128$ …… II

Dividing II and I we get,

$\frac{a{r}^3\left(1+r+r^2\right)}{a\left(1+r+r^2\right)}=\frac{128}{16}$

r³= 8

r³ = 2³

 r = 2 >1

So, putting r = 2 in I we get,

$a\left(1+2+2^2\right)=16$

$\Rightarrow$ $a\left(1+2+4\right)=16$

$\Rightarrow$ $a\left(7\right)=16$

$\Rightarrow$ $a=\frac{16}{7}$

And $s_x=\frac{a\left(r^n-1\right)}{r-1}$

$\Rightarrow$ $\frac{\frac{16}{7}\left(2^n-1\right)}{2-1}$

$\Rightarrow$ $\frac{16}{7}\left(2^n-1\right)$

45. Given, a=3

$r=\frac{3^3}{3}=3$>1

If s_n=120

Then $\frac{a\left(r^n-1\right)}{r-1}=120$

$\Rightarrow$ $\frac{3\left(3^n-1\right)}{3-1}=120$

$\Rightarrow$ $\frac{3\left(3^n-1\right)}{2}=120$

$\Rightarrow$ $3^n-1=120*\frac{2}{3}$

$\Rightarrow$3ⁿ= 80+1

$\Rightarrow$ 3ⁿ= 81

= 3ⁿ = 34

$\therefore r=4$

So, 120 is the sum of first 4 terms of the G.P

44. Let the three terms of the G.P be $\frac{a}{r},a,ar.$

So, $\frac{a}{r}a.ar=1$

= $a^3=1$

= $a=1$

And $\frac{a}{r}+a+ar=\frac{39}{10}$

= $\frac{1}{r}+1+r=\frac{39}{10}$ (as a = 1)

= $\frac{1+r+r^2}{r}=\frac{39}{10}$

= $\left(r^2+r+1\right)*10=39*r$

= $10r^2+10r+10=39r$

= $10r^2+10r-39r+10=0$

= $10r^2-29r+10=0$

= $10r^2-25r-4r+10=0$

= $5r\left(2r-5\right)-2\left(2r-5\right)=10$

$=\left(2x-5\right)\left(5r-2\right)=0$

= $\left(2x-5\right)=0$ or $\left(5r-2\right)=0$

= $r=\frac{5}{2}$ or $r=\frac{2}{5}$

When $a=1,$ $r=\frac{5}{2}$ the terms are,

 $\frac{1}{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.},1,$ $1*\frac{5}{2}=\frac{2}{5},1,\frac{5}{2}$

When $a=1,$ $r=\frac{2}{5}$ the terms are,

$\frac{1}{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$5$}\right.},$ 1, $1*\frac{2}{5}=\frac{5}{2},$ 1, $\frac{2}{5}$

43. ${\displaystyle \sum _{}^{11}}\left(2+3^x\right)=\left(2+3^1\right)+\left(2+3^2\right)+.........+\left(2+3^{11}\right)$

$x=1$  [2+2+ ….+ (11lines)] + (31+ 32+……+311)

= $11*2+\frac{3\left(3^{11}-1\right)}{3-1}\left[?s_n=\frac{a\left(r^n-1\right)}{r-1},r\ge 1\right]$

= $22+\frac{3\left(3^{11}-1\right)}{2}$

41. Here,  $a_1=1$

$r=\frac{-9}{1}=-a$

So,  $s_n=\frac{a_1\left[1-r^n\right]}{1-r}$

= $\frac{1\left[1-{\left(-a\right)}^n\right]}{1-\left(-a\right)}$

= ${\frac{1-\left(a\right)}{1+a}}^n$

$ta{n}^{-1}\frac{x}{y}-ta{n}^{-1}\frac{x-y}{x+y}\left\{-ta{n}^{-1}x-ta{n}^{-1}y=\frac{x-y}{1+xy}\right\}$

$=ta{n}^{-1}\left\{\frac{\left(\frac{x}{y}\right)-\left(\frac{x-y}{x+y}\right)}{1+\frac{x}{y}·\left(\frac{x-y}{x+y}\right)}\right\}$

$=ta{n}^{-1}\left\{\frac{\frac{x(x+y)-(x-y)·y}{y(x+y)}}{\frac{y(x+y)+x(x-y)}{y(x+y)}}\right\}$

$=ta{n}^{-1}\left(\frac{x^2+xy-xy+y^2}{xy+y^2-x^2-xy}\right)$

$=ta{n}^{-1}\frac{x^2+y^2}{x^2+y^2}$

$=ta{n}^{-1}1$

$=ta{n}^{-1}\left(tan\frac{\pi }{4}\right)$

$=\frac{\pi }{4}.$

Option C is correct.

$si{n}^{-1}(1-x)-2si{n}^{-1}x=\frac{\pi }{2}\to (1)$

(M) Let $x=sin\theta .Then,\theta =si{n}^{-1}x.$

Putting this in $q^n(1)$ we get

$si{n}^{-1}(1-x)-2·\theta =\frac{\pi }{2}$

$\Rightarrow si{n}^{-1}(1-x)=\frac{\pi }{2}+2\theta$

$\Rightarrow 1-x=sin\left(\frac{\pi }{2}+2\theta \right)\left\{sin\left(\frac{\pi }{2}+x\right)=cosx\right\}$

$\Rightarrow 1-x=cos2\theta$

$\Rightarrow 1-x=1-2si{n}^2\theta ·\left\{cos2x=1-2si{n}^{2}x\right\}$

$\Rightarrow 1-x=1-2x^2·\{sin\theta =x\}$

$\begin{array}{l}\Rightarrow 2x^2-x=0\\ \Rightarrow x(2x-1)=0\end{array}$

$\Rightarrow so,x=0x2x-1=0x2x=1\to x=\frac{1}{2}.$

Putting $x=0$ in qⁿ (1) .

L.H.S $=si{n}^{-1}(1-0)-2si{n}^{-1}0=si{n}^{-1}sin\frac{x}{2}-0=\frac{\pi }{2}=R.H.S.$

$x=\frac{1}{2}\mathrm{q(1)}$

$L.H.S=si{n}^{-1}\left(1-\frac{1}{2}\right)-2si{n}^{-1}\frac{1}{2}=si{n}^{-1}\left(\frac{1}{2}\right)-2si{n}^{-1}\frac{1}{2}$

$=si{n}^{-1}\frac{1}{2}-2si{n}^{-1}\frac{1}{2}$

$=-si{n}^{-1}\frac{1}{2}=-si{n}^{-1}\left(sin\frac{\pi }{6}\right)=-\frac{\pi }{6}$ $\ne$

So, $=0.$

Option (c) is correct.

Given, (M)

$ta{n}^{-1}\left(\frac{1-x}{1+x}\right)=\frac{1}{2}ta{n}^{-1}x$

$x=tan\mathrm{\theta .Then}\theta =ta{n}^{-1}\mathrm{x\; Bo\; we\; have,}$

$ta{n}^{-1}\left(\frac{1-tan\theta }{1+tan\theta }\right)=\frac{1}{2}ta{n}^{-1}(tan\theta )$

$\Rightarrow ta{n}^{-1}\left(\frac{tan\frac{\pi }{4}-tan\theta }{1+tan\frac{x}{4}tan\theta }\right)=\frac{1}{2}\theta \left\{?tan\frac{\pi }{4}=1\right\}.$

$\Rightarrow ta{n}^{-1}\left\{tan\left(\frac{x}{4}-\theta \right)\right\}=\frac{\theta }{2}\{\frac{?tanx-tany}{1+tanxtan}=tan(x-y)$

$\Rightarrow \frac{\pi }{4}-\theta =\frac{\theta }{2}$

$\Rightarrow \frac{\theta }{2}+\theta =\frac{x}{4}$

$\Rightarrow \frac{3\theta }{2}=\frac{\pi }{4}$

$\Rightarrow \theta =\frac{\pi }{4}*\frac{2}{3}=\frac{\pi }{6}$