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- Question 1 Exercise 4.5
- In the given geometric series: $a_1=3, \quad r=\dfrac{6}{3}=2$ and $a_n=3.2^9$.\\ We first find $n$ and... n{align}3.2^9&=3(2)^{n-1} \text { or }(2)^{n-1}=\dfrac{3.2^9}{3} \\ \Rightarrow(2)^{n-1}&=2^9 \\ \Righta... \text { or } n=10 \\ \text {. Now }\quad S_n&=\dfrac{a_1(r^n-1)}{r-1},\end{align} becomes in the given case\\ \begin{align}S_{10}&=\dfrac{3[2^{10}-1]}{2-1} \\ \Rightarrow \quad S_{10}&=3(
- Question 9 Exercise 4.4
- (i)===== Insert five geometric means between $3 \dfrac{5}{9}=\dfrac{32}{9}\quad$ and $\quad40 \dfrac{1}{2}=\dfrac{81}{2}$. ====Solution==== Let $G_1, G_2, G_3, G_4$ and $G_5$ be the five geometric means b
- Question 11 Exercise 4.4
- \ \because a_1&=a \\ \Rightarrow \quad r^{n+1}&=\dfrac{b}{a} . \\ \Rightarrow \quad r&=(\dfrac{b}{a})^{\dfrac{1}{n+1}}\end{align} Now \begin{align}G_1&=a_2=a r=a(\dfrac{b}{a})^{\dfrac{1}{n+1}}\\ G_2&=a_3=a_1 r^2=a(\dfr
- Question 12 Exercise 4.4
- =====Question 12===== For what value of $n, . \dfrac{a^{n+1}+b^{n+1}}{a^n+b^n}$ is geometric mean betw... not zero simultaneously. ====Solution==== Let $\dfrac{a^{n+1}+b^{n-1}}{a^n+b^n}$ be geometric mean between $a$ and $b$, then\\ \begin{align}\dfrac{a^{n+1}+b^{n+1}}{a^n+b^n}&=\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^{\d
- Question 6 Exercise 4.1
- efined for Pascal sequences is $$P_0=1, P_{r+1}=\dfrac{n-r}{r+1} P_r, \text{ where } r=0,1,2,3,\ldots.$$... ve Pascal sequence as follows: $$P_0=1, P_{r+1}=\dfrac{5-r}{r+1} P_r, \text{ where } r=0,1,2,3,\ldots.$$ For $r=0$ \begin{align}&P_{0+1}=\dfrac{5-0}{0+1} P_0\\ \implies &P_1=5.\end{align} For $r=1$ \begin{align}&P_{1+1}=\dfrac{5-1}{1+1} P_1\\ \implies &P_2=2\cdot 5=10.\end{a
- Question 1 Exercise 4.4
- metric ric sequence given that $a_1=8, \quad r=-\dfrac{1}{2}$ ====Solution==== The geomerric sequence is... _1 r^3, a_1 r^4, \ldots$,\\ so for $$a_1=8 ; r=-\dfrac{1}{2}$$ we have\\ \begin{align}&8,8(-\dfrac{1}{2}), 8(-\dfrac{1}{2})^2, 8(-\dfrac{1}{2})^3,8(-\dfrac{1}{2})^4, \ldots\\ \Rightarrow &8,-4,2,-1, \df
- Question 8 Exercise 4.2
- ) Peshawar, Pakistan. =====Question 8===== If $\dfrac{b+c-a}{a}, \dfrac{c+a-b}{b}, \dfrac{a+b-c}{c}$ are in A.P, then prove $\dfrac{1}{a}, \dfrac{1}{b}, \dfrac{1}{c}$ are in A.P. GOOD ====Solut
- Question 7 & 8 Exercise 4.5
- S_n$ of the first $n$ terms of the sequence $\{(\dfrac{1}{2})^n\}$. ====Solution==== The sequence is:\\ $$\{(\dfrac{1}{2})^n\}=\dfrac{1}{2}, \dfrac{1}{2^2}, \dfrac{1}{2^3}, \ldots$$\\ where $$a_1=\dfrac{1}{2}$$\\ and $$r=\dfrac{\dfrac{1}
- Question 13 & 14 Exercise 4.5
- eshawar, Pakistan. =====Question 13===== If $y=\dfrac{x}{3}+\dfrac{x^2}{3^2}+\dfrac{x^3}{3^3}+\ldots$ where $0<x<3$, then show that $x=\dfrac{3 y}{1+y}$. =====Solution===== Adding 1 to both s
- Question 4 Exercise 4.5
- Question 4(i)===== Convert each decimal to common fraction $0 . \overline{8}$ ====Solution==== We can wr... We can find the infinite sum as:\\ $$S_{\infty}=\dfrac{a_1}{1-r}=\dfrac{0.8}{1-0.1}=\dfrac{0.8}{0.9}=\dfrac{8}{9}$$\\ Hence putting $S_{\infty}$ in (i), we get $$0 . \overline
- Question 5 Exercise 4.1
- g series in expanded form, $\sum_{j=1}^{\infty} \dfrac{1}{2^j}$ ====Solution==== \begin{align}\sum_{j=1}^{\infty} \dfrac{1}{2^j}&=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+\ldots\\ &=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\d
- Question 16 Exercise 4.2
- mplies &8=5+6d\\ \implies &6d=8-5\\ \implies &d=\dfrac{3}{6}=\dfrac{1}{2}. \end{align} Now \begin{align} A_1&=a+d=5+\dfrac{1}{2}=\dfrac{11}{2},\\ A_2&=a+2 d=5+2 \cdot \dfrac{1}{2}=6,\\ A_3&=a+3 d=5+3 \cdot \dfrac{1}{2}=\dfrac{
- Question 1 and 2 Exercise 4.1
- lassify into finite and infinite sequences. $1,-\dfrac{1}{3}, \dfrac{1}{9},-\dfrac{1}{27}, \ldots,-\dfrac{1}{2187}$ ====Solution==== This finite sequence. =====Question 2(i)==== Find fi
- Question 3 and 4 Exercise 4.1
- m of the sequence as suggested by the pattern. $\dfrac{1}{2}, \dfrac{2}{3} \dfrac{3}{4}, \dfrac{4}{5}, \ldots$ ====Solution==== We can reform the given sequence to pick the pattern of t
- Question 2 Exercise 4.3
- d.$$ Thus $$a_{17}=2+(17-1)(3)=50.$$ Also $$S_n=\dfrac{n}{2}[a_1+a_n]$$ Thus \begin{align}S_{17}&=\dfrac{17}{2}(a_1+a_17) \\ &=\dfrac{17}{2}(2+50)=442.\end{align} Hence $a_{17}=50$ and $S_{17}=442$. GOOD ===... find $a_{21}$ and $d$. As \begin{align}&S_{21}=\dfrac{21}{2}(a_1+a_{21}) \\ \implies &210=\dfrac{21}{2}