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- Question 9 & 10 Exercise 4.5
- c{a_1(r^6-1)}{r-1}&=9 \dfrac{a_1(r^3-1)}{r-1} \\ \Rightarrow r^6-1-9(r^3-1) \\ \Rightarrow r^6-1&=9 r^3-9 \\ \Rightarrow r^6&=9 r^3-8 \\ \Rightarrow r^6-9 r^3+8&=0, \\ \Rightarrow r^6-r^3-8 r^3+8&=0 \\ \Rightarrow r^3(r^3-1)-8(r^3
- Question 1 Exercise 4.5
- ^{n-1} \text { or }(2)^{n-1}=\dfrac{3.2^9}{3} \\ \Rightarrow(2)^{n-1}&=2^9 \\ \Rightarrow n-1&=9 \text { or } n=10 \\ \text {. Now }\quad S_n&=\dfrac{a_1(r^n-1)}{r-1},\e... \begin{align}S_{10}&=\dfrac{3[2^{10}-1]}{2-1} \\ \Rightarrow \quad S_{10}&=3(2^{10}-1)\end{align}\\ is the req... \ (\dfrac{1}{2})^{n-1}&=\dfrac{1}{8 \times 16}\\ \Rightarrow(\dfrac{1}{2})^{n-1}&=(\dfrac{1}{2})^7\\ \Rightarr
- Question 2 Exercise 4.5
- n-1}$, therefore\\ \begin{align}64&=(-2)^{n-1}\\ \Rightarrow(-2)^{n-1}&=(-2)^6 \\ \Rightarrow n-1&=6 \\ \Rightarrow n&=7\\ S_7&=\dfrac{a_1[r^{\prime \prime}-1]}{r-1}\\ \text{then}\\ S_7&=\dfrac{1[(-2)^7-1]}{-2-1}\\ \Rightarrow S_7&=\dfrac{-128-1}{-3}\\ s_7&=\dfrac{129}{3}\end
- Question 5 & 6 Exercise 4.3
- , thus\\ \begin{align}a-3 d+a-d+a+d+a+3 d&=20 \\ \Rightarrow 4 a&=20\\ \Rightarrow a&=5 .\end{align} $Condition-2$\\ The sum of their square is $120$, therefore\\ \begin{... lign}(a-3 d)^2+(a-d)^2+(a+d)^2+(a+2 d)^2&=120 \\ \Rightarrow a^2-6 a d+9 d^2+a^2-2 a d+d^2+a^2+2 a d+d^2+a^2+6 a d+9 d^2&=120 \\ \Rightarrow 4 a^2+20 d^2&=120 \\ \Rightarrow 20 d^2&=120-4 \c
- Question 7 & 8 Exercise 4.5
- rac{1}{2}[1-(\dfrac{1}{2})^n]}{1-\dfrac{1}{2}}\\ \Rightarrow S_n&=\dfrac{\dfrac{1}{2}[1-\dfrac{1}{2^n}]}{\dfrac{1}{2}} \\ \Rightarrow S_n&=1-\dfrac{1}{2^n}\end{align}\\ is the require... gin{align}\dfrac{a}{r} \cdot a \cdot a r&=1728\\ \Rightarrow a^3&=1728\\ \Rightarrow \quad a&=12,\\ \text{putting}\text{in} (1)\\ \dfrac{12}{r}+12+12 r&=38\\ \Rightarrow
- Question 3 & 4 Exercise 4.3
- it becomes,\\ \begin{align} 350&=25+(n-1)(5) \\ \Rightarrow 5 n-5+25&=350 \\ \Rightarrow 5 n&=350-20=330 \\ \Rightarrow n&=66, \text { now for the sum } \\ S_n&=\dfrac{n}{2}(a_1+a_n), \text { that becomes } \\ S_{66}&=\dfrac{66}{2}(25+350) \\ \Rightarrow S_{66}&=33(375)=12375 .\end{align} =====Question
- Question 5 & 6 Exercise 4.5
- (r^{10}-1)}{r-1}&=244 \dfrac{a_1(r^5-1)}{r-1} \\ \Rightarrow r^{10}-1&=244(r^5-1) \\ \Rightarrow r^{10}-244 r^5 \cdots 1+244&=0 \\ \Rightarrow r^{10}-244 r^5+243&=0 \\ \Rightarrow r^{10}-r^5-243 r^5+243&=0 \\ \Rightarrow r^5(r^5-1)-243(r^5-1)&=0 \\ \Ri
- Question 10 Exercise 4.4
- M=G \cdot M+18$ or $A \cdot M-G \cdot M=18$\\ $$\Rightarrow \dfrac{a+b}{2}-\sqrt{a b}=18$$ Multiplying both s... \\ \begin{align}(b+48+b)-2 \sqrt{b(b+48)}&=36 \\ \Rightarrow(2 b+48)-2 \sqrt{b(b+48)}&=36 \\ \Rightarrow(b+24)-\sqrt{b(b-48)}&=18 \\ \Rightarrow b-\sqrt{b(b+48)}&=18-24=-6 \\ \Rightarrow-\sqrt{b(b+48)}&=-b-6 \\ \Ri
- Question 7 & 8 Exercise 4.3
- d] \text{is:}\\ S_n&=\dfrac{n}{2}[2.1+(n-1) 6]\\ \Rightarrow S_n&=\dfrac{n(6 n-4)}{2}\\ &=n(3 n-2)\end{align}\... \cdot S_n^{\prime}&=\dfrac{n}{2}[2.3+(n-1) 6]\\ \Rightarrow S_n^{\prime}&=\dfrac{6 n^2}{2}=3 n^2\end{align}\\... S_n^{\prime \prime}&=\dfrac{n}{2}[2.5+(n-1) 6]\\ \Rightarrow S_n^{\prime \prime}&=\dfrac{n(6 n+4)}{2}=n(3 n+2)... {3 n}&=S_n+S_n^{\prime}-(S_n^{\prime \prime}) \\ \Rightarrow S_{3 n}&=n(3 n-2)+3 n^2-n(3 n+2) \\ \Rightarrow S
- Question 9 & 10 Exercise 4.3
- egin{align}a_n&=a_1+(n-1) d \text { becomes } \\ \Rightarrow a_1+(n-1) d&=693 \\ \Rightarrow 306+(n-1) \cdot 9&=693 \\ \Rightarrow 9 n&=396 \\ \Rightarrow n&=44.\end{align} $\therefore$ Required sum is:\\ \begin{align} S_{44}&=\dfrac{44}{2}
- Question 4 & 5 Exercise 4.4
- n{align}\dfrac{1}{64}&=16(\dfrac{1}{2})^{n-1} \\ \Rightarrow(\dfrac{1}{2})^{n-1}&=\dfrac{1}{64 \times 16}=\dfrac{1}{1024} \\ \Rightarrow(\dfrac{1}{2})^{n-1}&=\dfrac{1}{2^{10}}=(\dfrac{1}{2})^{10} \\ \Rightarrow n-1&=10 \text { or } n=11\end{align} Hence the to... \therefore \dfrac{a_2}{a_1}&=\dfrac{a_3}{a_2} \\ \Rightarrow \dfrac{x-3}{x+7}&=\dfrac{x-8}{x-3} \\ \Rightarrow
- Question 12 Exercise 4.4
- \sqrt{a b}\quad \because G \cdot M=\sqrt{a b} \\ \Rightarrow \dfrac{a^{n+1}+b^{n+1}}{a^n+b^n}&=a^{\dfrac{1}{2}} b^{\dfrac{1}{2}} \\ \Rightarrow a^{n+1}+b^{n+1}&=(a^n+b^n) a^{\dfrac{1}{2}} b^ \dfrac{1}{2} \\ \Rightarrow a^{n+\dfrac{1}{2}+\dfrac{1}{2}}+b^{n+\dfrac{1}{2}... ac{1}{2}}+a^{\dfrac{1}{2}} b^{n+\dfrac{1}{2}} \\ \Rightarrow a^{\dfrac{1}{2}} a^{n+\dfrac{1}{2}}+b^{\dfrac{1}{
- Question 8 Exercise 4.2
- -b}{b} \\ \text{Let}\quad S&=\dfrac{a+b+c}{2} \\ \Rightarrow a+b+c&=2 S\\ \text{then} \Rightarrow a+b-c&=2(S-c) \text {, }\\ a+c-b&=2(S-b), \quad\text{and}\\ b+c-a&=2(S-a... b-b S+a b}{a b}&=\dfrac{b S-b c-c S+b c}{b c} \\ \Rightarrow \dfrac{(a-b) S}{a b}&=\dfrac{(b-c) S}{b c}\end{align} Dividing both sides by $S$\\ \begin{align}\Rightarrow \dfrac{a-b}{a b}&=\dfrac{b-c}{b c} \\ \Rightarrow
- Question 9 Exercise 4.4
- herefore, \begin{align}a_1 r^6&=\dfrac{81}{2} \\ \Rightarrow \dfrac{32}{9} r^6&=\dfrac{81}{2}\\ \Rightarrow r^6&=\dfrac{81}{2} \times \dfrac{9}{32}=\dfrac{3^4}{2} \times \dfrac{3^2}{2^5}\\ \Rightarrow r^6&=(\dfrac{3}{2})^6\\ \Rightarrow r&=\dfrac{3}{2}.\end{align}\\ \begin{align}G_1&=a_1 r=\dfrac{32}{9} \time
- Question 2 & 3 Exercise 4.4
- n}\dfrac{a_1 r^4}{a_1 r^2}&=\dfrac{243}{27}=9 \\ \Rightarrow r^2&=9 \text { or } r= \pm 3 .\end{align} Putting... r}&=-\dfrac{\sqrt{2}}{2}=-\dfrac{1}{\sqrt{2}}\\ \Rightarrow r&=-\dfrac{1}{\sqrt{2}}.\end{align} Putting in (i), then\\ \begin{align}\Rightarrow a_1(-\dfrac{1}{\sqrt{2}})&=2 \\ \Rightarrow a_1&=-2 \sqrt{2}\end{align} Now we know that\\ \begin{align}a_n&=