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- Question 6 Exercise 4.1
- xt{ where } r=0,1,2,3,\ldots.$$ For $r=0$ \begin{align}&P_{0+1}=\dfrac{5-0}{0+1} P_0\\ \implies &P_1=5.\end{align} For $r=1$ \begin{align}&P_{1+1}=\dfrac{5-1}{1+1} P_1\\ \implies &P_2=2\cdot 5=10.\end{align} For $r=2$ \begin{align}&P_{2+1}=\dfrac{5-2}{2+1}
- Question 3 and 4 Exercise 4.1
- ce to pick the pattern of the sequence as: \begin{align} &(-1)^2 \cdot 2 \cdot 1, (-1)^3 \cdot 2 \cdot 2,... \cdot 3, (-1)^{4+1} \cdot 2 \cdot 4, \ldots \end{align} Hence the general term of the sequence is $(-1)^... ce to pick the pattern of the sequence as: \begin{align}(-1)^2,(-1)^3,(-1)^4,(-1)^5, \ldots, (-1)^{n+1}, \ldots \end{align} Hence the general term of the sequence is $(-1)^
- Question 5 & 6 Exercise 4.3
- $Condition-1$\\ Their sum is $20$ , thus\\ \begin{align}a-3 d+a-d+a+d+a+3 d&=20 \\ \Rightarrow 4 a&=20\\ \Rightarrow a&=5 .\end{align} $Condition-2$\\ The sum of their square is $120$, therefore\\ \begin{align}(a-3 d)^2+(a-d)^2+(a+d)^2+(a+2 d)^2&=120 \\ \Righ... 0 \\ \Rightarrow d^2&=1 \text { or } d= \pm 1\end{align} When $a=5$ and $d=1$ then the numbers are\\ \beg
- Question 1 Exercise 4.5
- ies.\\ We know that $$a_n=a_1 r^{n-1}$$,\\ \begin{align}3.2^9&=3(2)^{n-1} \text { or }(2)^{n-1}=\dfrac{3.... t {. Now }\quad S_n&=\dfrac{a_1(r^n-1)}{r-1},\end{align} becomes in the given case\\ \begin{align}S_{10}&=\dfrac{3[2^{10}-1]}{2-1} \\ \Rightarrow \quad S_{10}&=3(2^{10}-1)\end{align}\\ is the required sum. =====Question 1(ii)=====
- Question 4 Exercise 4.5
- dots$$\\ That can be written in the form\\ \begin{align}0 . \overline{8}&=0.8+0.08+0.008 \div 0.0008+ \ld... .8)+\ldots \ldots \ldots \ldots .(\mathrm{i})\end{align}\\ It is geometric series with $$a_1=0.8, \quad r... $1 . \overline{63}$ ====Solution==== Since \begin{align}1 . \overline{63}&=1+0.63+0.0063+0.000063 +\ldots... 3)-(0.01)^2 0.63+\ldots \ldots \text { (i) }\end{align} The serics in braces is infinite gcometric serie
- Question 14 Exercise 4.2
- We have $$a_1=6 \text{ and } a_6=41.$$ Now \begin{align}& a_5=11\\ \Rightarrow &a_1+4 d=41 \\ \Rightarrow... htarrow &d=\dfrac{41-6}{4}\\ &=\dfrac{35}{4}.\end{align} Now \begin{align} A_1&=a+d=6+\dfrac{35}{4} \\ &=\dfrac{59}{4}=14\dfrac{3}{4},\end{align} \begin{align} A_2&=a+2 d\\ &=6+2 \cdot \dfrac{35
- Question 2 Exercise 4.3
- $$ 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 =====Q... $ and we have to find $a_{21}$ and $d$. As \begin{align}&S_{21}=\dfrac{21}{2}(a_1+a_{21}) \\ \implies &21... mes 210}{21}=20 \\ \implies &a_{21}=20+40=60.\end{align} Also $a_{21}=a_1+20 d$, then\\ \begin{align}&20d
- Question 5 Exercise 4.1
- rm, $\sum_{j=1}^6(2 j-3)$ ====Solution==== \begin{align}\sum_{j=1}^6(2 j-3)&=(2.1-3)+(2.2-3)+(2.3-3)+(2.4... \implies \sum_{j=1}^6(2 j-3)&=-1+1+3+5+7+9 .\end{align} =====Question 5(ii)===== Write each of the foll... um_{k=1}^5(-1)^k 2^{k-1}$ ====Solution==== \begin{align}\sum_{k=1}^5(-1)^k 2^{k-1}& =(-1)^1 2^{1-1}+(-1)^... implies \sum_{k=1}^5(-1)^k 2^k& =-1+2-4+8-16.\end{align} =====Question 5(iii)===== Write each of the fol
- Question 3 & 4 Exercise 4.3
- and $a_n=350$\\ To find $n$, we know that \begin{align}a_n&=a_1+(n-1) d\end{align} in the given case it becomes,\\ \begin{align} 350&=25+(n-1)(5) \\ \Rightarrow 5 n-5+25&=350 \\ \Rightar... 5+350) \\ \Rightarrow S_{66}&=33(375)=12375 .\end{align} =====Question 4===== The sum of three numbers i
- Question 7 & 8 Exercise 4.3
- given series into three arithmetic series \begin{align}&(1+7+13+\ldots)+(3+9+15+\ldots)- \\ & (5+11+17+\ldots) \ldots \ldots \ldots . . .(1)\end{align} Now the series each one in parenthesis is arithm... 1=1, d=7-1=6$$ then sum of the $n$ terms\\ \begin{align}S_n&=\dfrac{n}{2}[2 a_1+(n-1) d] \text{is:}\\ S_n... htarrow S_n&=\dfrac{n(6 n-4)}{2}\\ &=n(3 n-2)\end{align}\\ Now for $3+9+15+\ldots$, with $a_1=3, d=9-3=6$
- Question 2 & 3 Exercise 4.4
- uad\text{and}\quad a_5=243$$ and we know\\ \begin{align}a_3&=a_1 r^2=27\\ a_5&=a_1 r^4=243.\end{align} Dividing (ii) by (i), we get\\ \begin{align}\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 this in (i), then\\ $$a_1(9)=27 \quad \
- Question 2 Exercise 4.5
- $\\ We know $a_n=a_1 r^{n-1}$, therefore\\ \begin{align}64&=(-2)^{n-1}\\ \Rightarrow(-2)^{n-1}&=(-2)^6 \\... S_7&=\dfrac{-128-1}{-3}\\ s_7&=\dfrac{129}{3}\end{align} =====Question 2(ii)===== Some of the components... know $$a_9=a_1 r^8$$\\ therefore we have\\ \begin{align}1&=a_1(\dfrac{1}{2})^{9-1}\\ &=a_1 \dfrac{1}{2^8}... \ S_9&=\dfrac{a_1[1-r^{\prime \prime}]}{1-r},\end{align} becomes in the given case\\ \begin{align}\Righta
- Question 7 Exercise 4.2
- $d$ be common difference of A.P. As given \begin{align} &a_6+a_4=6 \\ \implies & a_1+5d+a_1+3d=6\\ \impl... es & 2a_1+8d=6\\ \implies & a_1+4d=3 --- (1) \end{align} Also, we have given \begin{align} &a_6-a_4=\dfrac{2}{3} \\ \implies & a_1+5d-a_1-3d=\dfrac{2}{3}\\ \imp... 2d=\dfrac{2}{3}\\ \implies & d=\dfrac{1}{3} \end{align} Using the value of $d$ in (1), we get \begin{ali
- Question 11 & 12 Exercise 4.5
- 1 r^{q-1}=b$ and $a_r=a_1 r^{r-1}$. Then\\ \begin{align}a^{q-r}&=(a_1 r^{p-1})^{q-r} . \\ b^{r-p}&=(a_1 r... text { and } \\ c^{p-q}&=(a_1 r^{r-1})^{p-q}.\end{align}\\ Multiplying the above three equations\\ \begin{align}a^{q-r} b^{r-p} c^{p-q}& =(a_1 r^{p-1})^{q-r} \cd... \text { Thus } \\ a^{q-r} b^{r-p} c^{p-q}&=1.\end{align} =====Question 12===== Find an infinite geometri
- Question 1 Exercise 4.3
- d$ be common difference of given A.P. Then \begin{align}&a_1=9 \\ &d=7-9=-2 \\ &n=20. \end{align} We know that \begin{align}&a_n=a_1+(n-1)d \\ \implies &a_20=9+(20-1)(-2)=-29. \end{align} Assume $S_n$ represents the sum of first $n$ ter