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- Question 9 & 10 Exercise 4.3
- ots, 693$$.\\ Here, $a=306$,\\ $$d=(315-306) = 9 \text { and } a_n=693 .$$\\ Let the number of terms be $n$. Then\\ \begin{align}a_n&=a_1+(n-1) d \text { becomes } \\ \Rightarrow a_1+(n-1) d&=693 \\ \R... 1000$, \\ therefore we have $n=4$\\ \begin{align}\text{Let the first person receives}&=R s . a\\ \text{Then the seçond receives}&=\operatorname{Rs}(a-20)\\ \
- Question 2 & 3 Exercise 4.4
- he sequence. ====Solution==== Here $$a_3=27 \quad\text{and}\quad a_5=243$$ and we know\\ \begin{align}a_... _1 r^2}&=\dfrac{243}{27}=9 \\ \Rightarrow r^2&=9 \text { or } r= \pm 3 .\end{align} Putting this in (i), then\\ $$a_1(9)=27 \quad \text{then} \quad a_1=3$$ $$\text{Hence}\quad a_1=3, r= \pm 3$$ =====Question 3===== Find the seventh term o
- Question 5 & 6 Exercise 4.3
- \\ \Rightarrow 20 d^2&=20 \\ \Rightarrow d^2&=1 \text { or } d= \pm 1\end{align} When $a=5$ and $d=1$ t... align} a-3d&=5-3=2, \\ a-d&=5-1=4, \\ a+d&=5+1=6 \text { and } \\ a+3 d&=5+3=8.\end{align} When $a=5$ an... =5-3(-1)=8, \\ a-d&=5-(-1)=6, \\ a+d&=5+(-1)=, 4 \text { and } \\ a+3 d&=5-3=2\\ & 2,4,6,8; 8,6,4,2\end{... are in A.P. If\\ \begin{align}x_1+x_7+x_{10}&=-6 \text { and } \\ x_3+x_8+x_{12}&=-11, \text { find } \\
- Question 5 & 6 Exercise 4.5
- then\\ $$S_{10}=\dfrac{a_1(r^{10}-1)}{r-1} \quad \text{and}\quad S_5=\dfrac{a_1(r^5-1)}{r-1}$$\\ Putting... \\ \Rightarrow(r^5-1)(r^5-243)&=0 \\ \Rightarrow \text { Either } r^5-10&=0 \text { or } r^5-243=0 \\ \Rightarrow r^5&=1 \text { or } r^5=3^5 \\ \Rightarrow r&=1 \text { or } r=3 . \\ \
- Question 4 Exercise 4.5
- verline{8}&=0.8+0.08+0.008 \div 0.0008+ \ldots\\ \text { or } 0 . \overline{8}&=0.8+(0.1)(0.8) +(0.1)^2(... \overline{63}&=1+0.63+0.0063+0.000063 +\ldots \\ \text { or } 1 . \overline{63}&=1+[0.63+ (0.01)(0.63)-(0.01)^2 0.63+\ldots \ldots \text { (i) }\end{align} The serics in braces is infini... =\dfrac{7}{11} \ldots \ldots \ldots \ldots . . . \text { (ii) }\end{align} Putting (ii) in (i), we get\\
- Question 1 Exercise 4.5
- =a_1 r^{n-1}$$,\\ \begin{align}3.2^9&=3(2)^{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},\end{align} becomes in the given case\... es. We know that\\ \begin{align}a_n&=a_1 r^{n-1} \text {, }\\ \therefore \dfrac{1}{16}&=8(\dfrac{1}{2})^
- Question 7 & 8 Exercise 4.5
- ightarrow a^3&=1728\\ \Rightarrow \quad a&=12,\\ \text{putting}\text{in} (1)\\ \dfrac{12}{r}+12+12 r&=38\\ \Rightarrow \dfrac{1}{r}+1+r&=\dfrac{38}{12}=\dfrac... htarrow(3 r-2)(2 r-3)&=0 \\ \Rightarrow 3 r-2&=0 \text { or } 2 r-3=0 \\ \Rightarrow r=\dfrac{2}{3} \text { or } r=\dfrac{3}{2}\end{align} When $a=12$ and $r=
- Question 8 Exercise 4.2
- {b+c-a}{a}&=\dfrac{a+b-c}{c}-\dfrac{c+a-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)\end{align} then (1), becomes\\ \begin{align}\dfra
- Question 3 & 4 Exercise 4.3
- 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)... d^2&=6336 \\ \Rightarrow 3(12)^3+6(12) d^2&=6336 \text { as } a=12 \\ \Rightarrow 3(1728)+72 d^2&=6336 \... he numbers are\\ \begin{align} a-d&=12-4=8, a=12 \text { and } \\ a+d&=12+4=16 .\end{align} When $a=12$
- Question 11 & 12 Exercise 4.3
- d $a_1=16 \mathrm{ft}$\\ The distance in the $2^{\text {nd }}$ second $a_2=48 \mathrm{ft}$\\ The distanc... the thirtieth day? ====Solution==== \begin{align}\text{Saves at first day}& =R s .1\\ \text{Saves at second day}&=Rs. 2\\ \text{Saves at third day}&=Rs. 3 \text{ and so on}\end{align} So the sequ
- Question 6 Exercise 4.1
- uences is $$P_0=1, P_{r+1}=\dfrac{n-r}{r+1} P_r, \text{ where } r=0,1,2,3,\ldots.$$ =====Question 6(i)=... 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{al... uence is $$P_0=1, P_{r+1}=\dfrac{n-r}{r+1} P_r, \text{ where } r=0,1,2,3,\ldots.$$ When $n=6$, then $$P_0=1, P_{r+1}=\dfrac{6-r}{r+1} P_r, \text{ where } r=0,1,2,3,\ldots.$$ Now for $r=0$ \begi
- Question 13 & 14 Exercise 4.3
- re in the theater? ====Solution==== \begin{align}\text{Total number of rows}& n=40,\\ \text{Seats in a first row} a_1&=20\\ \text{Seat in a second row} a_2&=23\\ \text{Seats in third row} a_3&=26\end{align}\\ and so on upto 40 rows.
- Question 3 Exercise 4.5
- 1 r}&=\dfrac{1}{2}\\ \Rightarrow r&=\dfrac{1}{2} \text {, }\end{align} putting this in (i), we have\\ \b... gin{align}\dfrac{a_1}{2}&=2\\ \Rightarrow a_1&=4 \text {. }\\ a_2&=a_1 r=4 \cdot \dfrac{1}{2}=2,\\ a_3&=a_1 r^2=4 \cdot(\dfrac{1}{2})^2=1 \text {. }\\ a_4&=a_1 r^3=4(\dfrac{1}{2})^3=\dfrac{1}{2... _5&=a_1 r^4=4 \cdot(\dfrac{1}{2})^4=\dfrac{1}{4} \text {. }\end{align} The infinite geometric sequence i
- Question 9 & 10 Exercise 4.5
- three terms is:\\ $$S_3=\dfrac{a_1(r^3-1)}{r-1} \text {. }$$\\ The sum of the $3$ terms is $9$ times th... ghtarrow(r^3-1)(r^3-8)&=0 \\ \Rightarrow r^3&=1 \text { or } r^3=8 \\ \Rightarrow r&=1 \text { or } r=2.\end{align} But $r$ can not be 1 thus $r=3$. =====Q... ave to find $n$ such that\\ $$S_n=40+13 \sqrt{3} \text {. }$$\\ We know that:\\ $$S_n=\dfrac{a_1(r^n-1)}
- Question 11 & 12 Exercise 4.5
- ^{p-1})^{q-r} . \\ b^{r-p}&=(a_1 r^{q-1})^{r-p}, \text { and } \\ c^{p-q}&=(a_1 r^{r-1})^{p-q}.\end{alig... r+q r-p q r+p+p r-q r-p+q} \\ & =1 \cdot r^0=1 . \text { Thus } \\ a^{q-r} b^{r-p} c^{p-q}&=1.\end{align... S_{\infty}=6$. Therefore,\\ $$\dfrac{a_1}{1-r}=6 \text { or } 6 a_1=1-r ...(ii)$$\\ Also we are given th... htarrow a_1&=\dfrac{4}{5 \times 6}=\dfrac{2}{15} \text {. }\end{align}\\ Thus the infinite gcometric ser