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- Question 6 Exercise 4.1
- r, \text{ where } r=0,1,2,3,\ldots.$$ For $r=0$ \begin{align}&P_{0+1}=\dfrac{5-0}{0+1} P_0\\ \implies &P_1=5.\end{align} For $r=1$ \begin{align}&P_{1+1}=\dfrac{5-1}{1+1} P_1\\ \implies &P_2=2\cdot 5=10.\end{align} For $r=2$ \begin{align}&P_{2+1}=\dfrac{5-2}{2+1} P_2\\ \implies &P_3=1\cdot 10=10\end{align} For $r=3$ \begin{align}&P_{3+1}=\dfrac{5-3}{3+1} P_3\\ \implies &P
- Question 3 and 4 Exercise 4.1
- sequence to pick the pattern of the sequence as: \begin{align} &(-1)^2 \cdot 2 \cdot 1, (-1)^3 \cdot 2 \c... sequence to pick the pattern of the sequence as: \begin{align}(-1)^2,(-1)^3,(-1)^4,(-1)^5, \ldots, (-1)^{... on==== Given $$a_1=3, a_{n+1}=5-a_n.$$ For $n=1$ \begin{align}a_{1+1}&=5-a_1\\ \Rightarrow a_2&=5-3=2\end{align} For $n=2$ \begin{align}a_{2+1}&=5-a_2\\ \Rightarrow a_3&=5-2=3\end
- Question 5 & 6 Exercise 4.3
- d,$$ $Condition-1$\\ Their sum is $20$ , thus\\ \begin{align}a-3 d+a-d+a+d+a+3 d&=20 \\ \Rightarrow 4 a&... \\ 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 \\... ign} When $a=5$ and $d=1$ then the numbers are\\ \begin{align} a-3d&=5-3=2, \\ a-d&=5-1=4, \\ a+d&=5+1=6 ... gn} When $a=5$ and $d=-1$ then the numbers are\\ \begin{align}a-3 d&=5-3(-1)=8, \\ a-d&=5-(-1)=6, \\ a+d&
- Question 1 Exercise 4.5
- of series.\\ We know that $$a_n=a_1 r^{n-1}$$,\\ \begin{align}3.2^9&=3(2)^{n-1} \text { or }(2)^{n-1}=\df... 1)}{r-1},\end{align} becomes in the given case\\ \begin{align}S_{10}&=\dfrac{3[2^{10}-1]}{2-1} \\ \Righta... d $n$ and then the sum of series. We know that\\ \begin{align}a_n&=a_1 r^{n-1} \text {, }\\ \therefore \d... text {, }\end{align} becomes in the given case\\ \begin{align}S_8&=\dfrac{8[1-(\dfrac{1}{2})^8]}{1-\dfrac
- Question 4 Exercise 4.5
- 888 \ldots$$\\ That can be written in the form\\ \begin{align}0 . \overline{8}&=0.8+0.08+0.008 \div 0.000... ction $1 . \overline{63}$ ====Solution==== Since \begin{align}1 . \overline{63}&=1+0.63+0.0063+0.000063 +... 1$.\\ Therefore the sum exists and is given by\\ \begin{align}S_{\infty}&=\dfrac{a_1}{1-r}\\ &=\dfrac{0.6... ction $2 . \overline{15}$ ====Solution==== Since \begin{align}2 . \overline{15}&=2+0.15+0.0015+0.000015+\
- Question 14 Exercise 4.2
- A.P. We have $$a_1=6 \text{ and } a_6=41.$$ Now \begin{align}& a_5=11\\ \Rightarrow &a_1+4 d=41 \\ \Righ... dfrac{41-6}{4}\\ &=\dfrac{35}{4}.\end{align} Now \begin{align} A_1&=a+d=6+\dfrac{35}{4} \\ &=\dfrac{59}{4}=14\dfrac{3}{4},\end{align} \begin{align} A_2&=a+2 d\\ &=6+2 \cdot \dfrac{35}{4} \\ &=\dfrac{47}{2}=23\dfrac{1}{2}\end{align} and \begin{align}A_3&=a+3 d=6+3 \cdot \dfrac{35}{4}\\ &=\dfr
- Question 2 Exercise 4.3
- 3)=50.$$ Also $$S_n=\dfrac{n}{2}[a_1+a_n]$$ Thus \begin{align}S_{17}&=\dfrac{17}{2}(a_1+a_17) \\ &=\dfrac... $n=21$ and we have to find $a_{21}$ and $d$. As \begin{align}&S_{21}=\dfrac{21}{2}(a_1+a_{21}) \\ \impli... 40=60.\end{align} Also $a_{21}=a_1+20 d$, then\\ \begin{align}&20d=60-(-40)=100 \\ \implies &d=\dfrac{100... S_n=\dfrac{n}{2}[2 a_1+(n-1) d].$$ Thus, we have \begin{align} & 225=\dfrac{n}{2}[2 \cdot(-7)+(n-1) \cdot
- Question 5 Exercise 4.1
- ded form, $\sum_{j=1}^6(2 j-3)$ ====Solution==== \begin{align}\sum_{j=1}^6(2 j-3)&=(2.1-3)+(2.2-3)+(2.3-3... m, $\sum_{k=1}^5(-1)^k 2^{k-1}$ ====Solution==== \begin{align}\sum_{k=1}^5(-1)^k 2^{k-1}& =(-1)^1 2^{1-1}... _{j=1}^{\infty} \dfrac{1}{2^j}$ ====Solution==== \begin{align}\sum_{j=1}^{\infty} \dfrac{1}{2^j}&=\dfrac{... 1}{16}+\ldots \end{align} or we can simply write \begin{align}\sum_{j=1}^{\infty} \dfrac{1}{2^j}=\dfrac{1
- Question 3 & 4 Exercise 4.3
- , d=5$ and $a_n=350$\\ To find $n$, we know that \begin{align}a_n&=a_1+(n-1) d\end{align} in the given case it becomes,\\ \begin{align} 350&=25+(n-1)(5) \\ \Rightarrow 5 n-5+25&=... by first condition their sum is equal to $36$\\ \begin{align}(a-d)+a+(a+d)&=36 \\ \Rightarrow 3 a&=36 \\... the sum of their cubes is $6336$,\\ so we have\\ \begin{align}(a-d)^3+a^3+(a+d)^3&=6336 \\ \Rightarrow a^
- Question 7 & 8 Exercise 4.3
- rm the given series into three arithmetic series \begin{align}&(1+7+13+\ldots)+(3+9+15+\ldots)- \\ & (5+1... e $$a_1=1, d=7-1=6$$ then sum of the $n$ terms\\ \begin{align}S_n&=\dfrac{n}{2}[2 a_1+(n-1) d] \text{is:}... then sum of the $\mathrm{n}$ terms, let denote\\ \begin{align}S_n^{\prime}&=\dfrac{n}{2}[2 a_1+(n-1) d] \... e}=\dfrac{n}{2}[2 a_1+(n-1) d]$$ in this case\\ \begin{align}S_n^{\prime \prime}&=\dfrac{n}{2}[2.5+(n-1)
- Question 2 & 3 Exercise 4.4
- =27 \quad\text{and}\quad a_5=243$$ and we know\\ \begin{align}a_3&=a_1 r^2=27\\ a_5&=a_1 r^4=243.\end{align} Dividing (ii) by (i), we get\\ \begin{align}\dfrac{a_1 r^4}{a_1 r^2}&=\dfrac{243}{27}=9... nd and third term respectively. ====Solution==== \begin{align}\text{The second term is}=a_2&=a_1 r=2...(i... ..(ii)\end{align} Dividing (ii) by (i), we get\\ \begin{align}\dfrac{a_1 r^2}{a_1 r}&=-\dfrac{\sqrt{2}}{2
- Question 2 Exercise 4.5
- n $S_n$\\ We know $a_n=a_1 r^{n-1}$, therefore\\ \begin{align}64&=(-2)^{n-1}\\ \Rightarrow(-2)^{n-1}&=(-2... \ We know $$a_9=a_1 r^8$$\\ therefore we have\\ \begin{align}1&=a_1(\dfrac{1}{2})^{9-1}\\ &=a_1 \dfrac{1... }]}{1-r},\end{align} becomes in the given case\\ \begin{align}\Rightarrow S_9&=\dfrac{2^8[1-(\dfrac{1}{2}... _n=-63, a_n=-96$ ====Solution==== We know that\\ \begin{align}S_n&=\dfrac{a_1(r^{\prime \prime}-1)}{r-1}\
- Question 7 Exercise 4.2
- m and $d$ be common difference of A.P. As given \begin{align} &a_6+a_4=6 \\ \implies & a_1+5d+a_1+3d=6\\... a_1+4d=3 --- (1) \end{align} Also, we have given \begin{align} &a_6-a_4=\dfrac{2}{3} \\ \implies & a_1+5d... end{align} Using the value of $d$ in (1), we get \begin{align} &a_1+4\left(\dfrac{1}{3}\right)=3\\ \impli... _1=3-\dfrac{4}{3}=\dfrac{5}{3}. \end{align} Also \begin{align} a_2&=a_1+d\\ &=\dfrac{5}{3}+\dfrac{1}{3}=2
- Question 1 Exercise 4.3
- and $d$ 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)(... presents the sum of first $n$ terms of A.P. Then \begin{align} S_n&=\dfrac{n}{2}[a_1+a_n], \\ \implies S_... and $d$ be common difference of given A.P. Then \begin{align}&a_1=3 \\ &d=\dfrac{8}{3}-3=-\dfrac{1}{3}
- Question 11 & 12 Exercise 4.5
- a_q=a_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}&=... align}\\ Multiplying the above three equations\\ \begin{align}a^{q-r} b^{r-p} c^{p-q}& =(a_1 r^{p-1})^{q-... e sum of all the terms following it, therefore\\ \begin{align}a_1&=4(a_1 r+a_1 r^2+a_1 r^3+\cdots) \\ \Ri... rac{1}{5}\end{align}\\ putting in (ii), we get\\ \begin{align}6 a_1&=1-\dfrac{1}{5}=\dfrac{4}{5}\\ \Right