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- 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 14 Exercise 4.2
- ext{ and } a_6=41.$$ Now \begin{align}& a_5=11\\ \Rightarrow &a_1+4 d=41 \\ \Rightarrow &6+4 d=41 \\ \Rightarrow &d=\dfrac{41-6}{4}\\ &=\dfrac{35}{4}.\end{align} Now \begin{align} A_1&=a+d=6+\dfrac... } a_6=32.$$ Now \begin{align} & a_6=a_1+5 d \\ \Rightarrow &17+5 d=32 \\ \Rightarrow &d=\dfrac{32-17}{5}\\ &
- Question 12 & 13 Exercise 4.2
- b^{\prime}}{2}\\ &=\dfrac{(a+b)^2+(a-b)^2}{2} \\ \Rightarrow A&=\dfrac{a^2+b^2+2 a b+a^2+b^2-2 a b}{2} \\ & =\
- 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 3 and 4 Exercise 4.1
- -a_n.$$ For $n=1$ \begin{align}a_{1+1}&=5-a_1\\ \Rightarrow a_2&=5-3=2\end{align} For $n=2$ \begin{align}a_{2+1}&=5-a_2\\ \Rightarrow a_3&=5-2=3\end{align} For $n=3$ \begin{align}a_{3+1}&=5-a_3\\ \Rightarrow a_4&=5-3=2\end{align} For $n=4$ \begin{align}a_{4+1}&=5-a_4\\ \Rightarrow a_5&=5-2=3\end{align} Hence the first five terms
- Question 13 & 14 Exercise 4.5
- (i), we have \begin{align}1+y&=\dfrac{3}{3-x} \\ \Rightarrow \quad 3-x&=\dfrac{3}{1+y}\\ \Rightarrow \quad x&=3-\dfrac{3}{1+y} \\ \Rightarrow \quad x&=\dfrac{3+3 y-3}{1+y} \\ \Rightarrow \quad x&=\dfrac{3 y}{1+y}\end{align} which is required result.
- Question 11 & 12 Exercise 4.5
- in{align}a_1&=4(a_1 r+a_1 r^2+a_1 r^3+\cdots) \\ \Rightarrow a_1&=4 a_1(r+r ^2+r^3+\ldots) \\ \Rightarrow \dfrac{1}{4}&=\dfrac{r}{1-r} \quad \because r+r^2+r^3+\ldots=\dfrac{r}{1-r} \\ \Rightarrow 4 r&=1-r\\ \Rightarrow 5 r&=1\\ r&=\dfrac{1}{5}\end{align}\\ putting in (ii), we get\\ \begin{align}6 a_1&=1
- 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 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 4 Exercise 4.5
- &=\dfrac{0.63}{1-0.01}\\ &=\dfrac{0.63}{0.99} \\ \Rightarrow S_{\infty}&=\dfrac{7}{11} \ldots \ldots \ldots \l
- 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 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 3 Exercise 4.5
- gin{align}\dfrac{a_1 r^2}{a_1 r}&=\dfrac{1}{2}\\ \Rightarrow r&=\dfrac{1}{2} \text {, }\end{align} putting thi... (i), we have\\ \begin{align}\dfrac{a_1}{2}&=2\\ \Rightarrow a_1&=4 \text {. }\\ a_2&=a_1 r=4 \cdot \dfrac{1}
- Question 15 & 16 Exercise 4.5
- ign}S_{30}&=\dfrac{1[2^{30}-1]}{2-1}=2^{30}-1 \\ \Rightarrow S_{30}&=R s .1073741823 \end{align} $$\text{Rs.}=... 7$, then \begin{align}&729000=64000 r^{7-1} \\ & \Rightarrow r^6=\dfrac{729000}{64000} \\ & \Rightarrow r^6=\dfrac{729}{64}=\dfrac{3^6}{2^6} \\ & \Rightarrow r^6=(\dfrac{3}{2})^6 \\ & \Rightarrow r=\dfrac{3}{2}\en
- 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