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- Question 11 and 12, Exercise 4.1
- Question 11 and 12, Exercise 4.2
- Question 11 and 12, Exercise 4.3
- Question 25 and 26, Exercise 4.3
- Question 12 and 13, Exercise 4.4
- Question 24 and 25, Exercise 4.4
- Question 11, 12 and 13, Exercise 4.5
- Question 12, Exercise 4.6
- Question 11, 12 and 13, Exercise 4.7
- Question 23 and 24, Exercise 4.7
- Question 11 and 12, Exercise 4.8
Fulltext results:
- Question 5, 6 and 7, Exercise 4.4
- -2)= -6 \\ & a_{3}=a_{1} r^{2}=(3)(-2)^{2}=3 (4)= 12 \\ & a_{4}=a_{1} r^{3}=(3)(-2)^{3}=3 (-8) = -24 \end{align*} Hence $a_1=3$, $a_2=-6$, $a_3=12$, $a_4=-24$. GOOD =====Question 6===== Find the ... our terms of the geometric sequence. $\quad a_{1}=12, r=\frac{1}{2}$ ** Solution. ** Given $a_{1}=12$ and $r=\frac{1}{2}$. Use the formula $$a_{n}=a_{1
- Question 5 and 6, Exercise 4.5
- ==== Find the sum of the geometric series. $a_{1}=12, a_{5}=972, r=-3$ ** Solution. ** Given $a_1 = 12$, $a_5 = 972$ and $r = -3$.\\ Now, we can use \(r... $$a_n = a_1 \cdot r^{n-1}$$ \begin{align*} 972 &= 12 (-3)^{n-1}\\ 81 &= (-3)^{n-1}\\ (-3)^{n-1} &= 3^{... uad r \neq 1.$$ Thus, \begin{align*} S_5 &= \frac{12 \left(1 - (-3)^{5}\right)}{1 - (-3)} \\ &= \frac{
- Question 14, Exercise 4.5
- al notation for the infinite geometric series; $0.12121212 \ldots$ ** Solution. ** We can express the decimal as $$0.121212 \ldots = 0.12 + 0.0012 + 0.00001
- Question 25 and 26, Exercise 4.7
- {6 \cdot 7^n}\\ &=\frac{7}{6} + \frac{7(7^n - 1)}{12 \times7^n } - \frac{1 + 3n}{6 \times 7^n}\\ &=\frac{2\times 7^{n+1}+7^{n+1}-7-2-6n}{12\times 7^n} \\ &=\frac{3 \times7^{n+1}-9-6n}{12\times 7^n}\\ &=\frac{3(7^{n+1}-3-2n)}{12 \times7^n}\\ &=\frac{7^{n+1}-3-2n}{4\times 7^n}\\ \end{align
- Question 13 and 14, Exercise 4.1
- the sequence. $a_{n}=(-1)^{n-1}(3.4 n-17.3) ; a_{12}$ ** Solution. ** Given: $$a_n = (-1)^{n-1}(3.4n - 17.3).$$ Then \begin{align*} a_{12} &= (-1)^{12-1}(3.4 \cdot 12 - 17.3) \\ &= (-1)^{11}(40.8 - 17.3) \\ &= (-1)^{11}(23.5) \\ &= -23.5 \end{align*} H
- Question 7 and 8, Exercise 4.3
- ind the sum of series. $9+11+13+15+\cdots$ for $n=12$ ** Solution. ** Given series is arithmetic series with $a_1=9$, $d=11-9=2$, $n=12$.\\ Let $S_n$ represents sum of the arithmetic se... ign} S_n&=\frac{n}{2}[2a_1+(n-1)d] \\ \implies S_{12}&=\frac{12}{2}[2(9)+(12-1)(2)]\\ &=6\times [18+22]\\ &=240. \end{align} Hence $S_{12}=240$. GOOD ==
- Question 20 and 21, Exercise 4.4
- 2=a_1 r= (3)(2) = 6 \\ & a_3=a_1 r^2 = (3)(2)^2 = 12 \\ & a_4=a_1 r^3= (3)(2)^3=24. \end{align*} Hence $6$, $12$, $24$ are required geometric means. <callout ty... 2=a_1 r= (3)(2) = 6 \\ & a_3=a_1 r^2 = (3)(2)^2 = 12 \\ & a_4=a_1 r^3= (3)(2)^3=24. \end{align*} If $a... _1 r= (3)(-2) = -6 \\ & a_3=a_1 r^2 = (3)(-2)^2 = 12 \\ & a_4=a_1 r^3= (3)(-2)^3=-24. \end{align*} Hen
- Question 15, Exercise 4.5
- 0 ft$ \\ First rebound $= 30 \times \frac{2}{5} = 12 ft$ \\ Second rebound $= 12 \times \frac{2}{5} = \frac{24}{5} ft$ \\ Third rebound $= \frac{24}{5} \... distance covered by the ball. Then $$D=30+2\left(12+\frac{24}{5}+\frac{24}{5}+... \right)$$ To find the sum of infinite geometric series $$ 12+\frac{24}{5}+\frac{24}{5}+... $$ We have $a_1=12$
- Question 1, Exercise 4.2
- n - 1)d.$$ Now \begin{align*} a_2&=7+(2-1)(5)=7+5=12\\ a_3 &= 7+ (3-1)(5) = 7 + 10 = 17\\ a_4&=7+(4-1)(5)=7+15=22 \end{align*} Hence $a_1=7$, $a_2=12$, $a_3=17$, $a_4=22$. GOOD =====Question 1(iii)... -1)(-2)=16-2=14\\ a_3 &= 16+ (3-1)(-2) = 16 - 4 = 12\\ a_4&=16+(4-1)(-2)=16-6=10 \end{align*} Hence $a_1=16$, $a_2=14$, $a_3=12$, $a_4=10$. GOOD =====Question 1(iv)===== Find
- Question 20, 21 and 22, Exercise 4.3
- plies & 1752=146n\\ \implies & n=\frac{1752}{146}=12. \end{align} Also we have \begin{align} &a_n=a_1+(n-1)d\\ \implies & 139=7+(12-1)d\\ \implies & 139-7=11d\\ \implies & d=\frac{132}{11}=12. \end{align} Thus \begin{align} &a_2=a_1+d=7+12=19\\ &a_3=a_1+2d=7+2(12)=31.\\ \end{align} Hence $a_1
- Question 3 and 4, Exercise 4.8
- ifference, find the sum of the series: $1+4+13+40+121+ \ldots$ to $n$ term. ** Solution. ** Let $$ S_{n}=1+4+13+40+121+\ldots +T_{n} $$ Also $$ S_{n}=1+4+13+40+\ldots ... , we have \begin{align*} S_{n}-S_{n}& =1+4+13+40+121+\ldots +T_{n} \\ & -\left(1+4+13+40+\ldots +T_... gin{align*} \implies 0=&1+(4-1)+(13-4)+(40-13)+(121-40) \\ & +\ldots+(T_{n}-T_{n-1})-T_{n}. \\ \im
- Question 17, 18 and 19, Exercise 4.3
- ion 17===== Find sum of the arithmetic series. $6+12+18+\ldots+96$. ** Solution. ** Given arithmetic series: $$6+12+18+\ldots+96.$$ So, $a_{1}=6$, $d=12-6=6$, $a_{n}=96$, $n=?$.\\ We have \begin{align} & a_n=a_1+(n-... 1+a_n] \\ \implies S_{24}&=\frac{24}{2}[6+96]\\ &=12\times 102\\ &=1224. \end{align} Hence the sum of
- Question 3 and 4, Exercise 4.5
- sum of geometric series with $a_{1}=5$, $r=3$, $n=12$. ** Solution. ** Given: $a_{1}=5$, $r=3$, $n=12$ The formula to find the sum of $n$ terms of geom... {1 - r}, \quad r\neq 1. \] Thus \begin{align*} S_{12} &= \frac{5\left(1 - 3^{12}\right)}{1 - 3} \\ &= \frac{5\left(1 - 531441\right)}{-2} \\ &= \frac{5(-5
- Question 1 and 2, Exercise 4.6
- the harmonic progression. $\frac{1}{9}, \frac{1}{12}, \frac{1}{15}, \cdots \quad 7$ th term. ** Solution. ** $$\frac{1}{9}, \frac{1}{12}, \frac{1}{15}, \cdots \text{ is in H.P.}$$ $$9, 12, 15, ... \text{ is in A.P.}$$ Here $a_1=9$, $d=12-9=3$, $a_7=?$. Gneral term of A.P is given as $$
- Question 29 and 30, Exercise 4.7
- the series: $$3+\frac{6}{10}+\frac{9}{100}+\frac{12}{1000}+\ldots$$ ** Solution. ** The given arith... is:\\ \[ 3 + \frac{6}{10} + \frac{9}{100} + \frac{12}{1000} + \ldots \] It can be rewritten as:\\ \[... 6 \times \frac{1}{10} + 9 \times \frac{1}{100} + 12 \times \frac{1}{1000} + \ldots \] \(3, 6, 9, 12, \ldots\) are in AP with \(a = 3\) and \(d = 6 -