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- Question 11 and 12, Exercise 4.8
- present the $k$th term of the series. Then \begin{align*} T_k &= \frac{1}{k(k+2)}. \end{align*} Resolving it into partial fractions: \begin{align*} \frac{1}{k(k+2)} = \frac{A}{k} + \frac{B}{k+2} \ldots (1) \end{align*} Multiplying both sides by $k(k+2)$, we get \beg
- Question 7 and 8, Exercise 4.8
- presents the kth term of the series. Then \begin{align*} T_k &=\frac{1}{(3k-2)(3k+1)}. \end{align*} Resolving it into partial fraction: \begin{align*} \frac{1}{(3k-2)(3k+1)} = \frac{A}{3k-2}+\frac{B}{3k+1} \ldots (1) \end{align*} Multiplying with $(3k-2)(3k+1)$ \begin{align*}
- Question 5 and 6, Exercise 4.2
- uence is given as $$a_n=a_1+(n-1)d$$ Given \begin{align*} & a_{17} = -40 \\ \implies &a_1 + 16d = -40 \quad \cdots (1) \end{align*} Also \begin{align*} &a_{28}=-73\\ \implies &a_1 + 27d = -73 \quad \cdots (2) \end{align*} Now, subtract equation (1) from equation (2): \
- Question 20, 21 and 22, Exercise 4.3
- _{n}=876$. First we find $n$ and $d$.\\ As \begin{align} &S_n=\frac{n}{2}[a_1+a_n]\\ \implies & 876=\frac... 752=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
- Question 3 and 4, Exercise 4.8
- ression from the first expression, we have \begin{align*} S_{n}-S_{n}& =1+4+13+40+121+\ldots +T_{n} \\ ... \left(1+4+13+40+\ldots +T_{n-1}+T_{n}\right) \end{align*} \begin{align*} \implies 0=&1+(4-1)+(13-4)+(40-13)+(121-40) \\ & +\ldots+(T_{n}-T_{n-1})-T_{n}. \\ ... \text { up to } (n-1) \text { terms })-T_{n} \end{align*} Then \begin{align*} T_{n} & =1+(3+9+27+81+\ldo
- Question 5 and 6, Exercise 4.8
- ression from the first expression, we have \begin{align*} S_{n}-S_{n}& =3+4+6+10+18+\ldots +T_{n} \\ & -\left(3+4+6+10+\ldots +T_{n-1}+T_{n}\right) \end{align*} \begin{align*} \implies 0=&3+(4-3)+(6-4)+(10-6)+(18-10) \\ & +\ldots+(T_{n}-T_{n-1})-T_{n}. \\ \impli... \text { up to } (n-1) \text { terms })-T_{n} \end{align*} Then \begin{align*} T_{n} & =3+(1+2+4+8+\ldots
- Question 9 and 10, Exercise 4.8
- 1}{(k+1)(k+2)}$ ** Solution. ** Consider \begin{align*} T_k &= \frac{1}{(k+1)(k+2)}. \end{align*} Resolving it into partial fractions: \begin{align*} \frac{1}{(k+1)(k+2)} = \frac{A}{k+1} + \frac{B}{k+2} \ldots (1) \end{align*} Multiplying both sides by $(k+1)(k+2)$, we get
- Question 1 and 2, Exercise 4.8
- ression from the first expression, we have \begin{align*} S_{n}-S_{n}& =3+7+13+21+31+\ldots +T_{n} \\ & ... \left(3+7+13+21+\ldots +T_{n-1}+T_{n}\right) \end{align*} \begin{align*} \implies 0=&3+(7-3)+(13-7)+(21-13) \\ &+(31-21)+\ldots +\left(T_{n}-T_{n-1}\right)-T_{... \text { up to } (n-1) \text { terms })-T_{n} \end{align*} Then \begin{align*} T_{n} & =3+(4+6+8+10+\ldots
- Question 13, 14 and 15, Exercise 4.8
- present the $k$th term of the series. Then \begin{align*} T_k &= \frac{1}{(2k+3)(2k+9)}. \end{align*} Resolving it into partial fractions: \begin{align*} \frac{1}{(2k+3)(2k+9)} = \frac{A}{2k+3} + \frac{B}{2k+9} \ldots (1) \end{align*} Multiplying both sides by $(2k+3)(2k+9)$, we ge
- Question 20 and 21, Exercise 4.4
- rm, we have $$ a_n=ar^{n-1}. $$ This gives \begin{align*} &a_5=a_1 r^4 \\ \implies & 48=3r^4 \\ \implies ... \\ \implies & r^4 = 2^4 \\ \implies & r = 2. \end{align*} Thus \begin{align*} & a_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 me
- Question 16 and 17, Exercise 4.2
- A.P is given as $$a_n=a_1+(n-1)d.$$ Thus \begin{align*} &a_4 = a_1 + 3d \\ \implies & 17=5+3d\\ \implies & 3d=12\\ \implies & \boxed{d=4}.\end{align*} Now \begin{align*} A_1 &= a_2= a_1+d \\ &=5+4=9 \end{align*} and \begin{align*} &A_2= a_3=a_1+2d\\ &= 5 + 2(4) \\ &=13 \e
- Question 3 and 4, Exercise 4.2
- $a_1 = 0.07$, $d=0.05$, $a_{11}=?$.\\ Now \begin{align*} a_n&=a_1+(n-1)d \\ \implies a_{11}&= 0.07+(11-1)(0.05)\\ &=0.07+(10)(0.05)\\ &=0.57 \end{align*} Hence $a_{11}=0.57.$ GOOD =====Question 4=====... e is given as $$a_n = a_1 + (n-1)d$$ Given \begin{align*} & a_3 = 14 \\ \implies & a_1 + 2d = 14 \quad \cdots (1) \end{align*} Also \begin{align*} & a_9 = -1 \\ \implies & a_
- Question 9 and 10, Exercise 4.2
- frac{1}{a}, b, \dfrac{1}{c}$ are in A.P.\\ \begin{align*} d&=b-\frac{1}{a}\cdots (i)\\ \end{align*} Also \begin{align*} d&=\frac{1}{c}-b \cdots (ii) \end{align*} Comparing (i) and (ii) we have\\ \begin{align*} b-\frac{1}{a
- Question 17, 18 and 19, Exercise 4.3
- $d=12-6=6$, $a_{n}=96$, $n=?$.\\ We have \begin{align} & a_n=a_1+(n-1)d \\ \implies & 96=6+(n-1)(6) \\... 6 \\ \implies & 6n=96 \\ \implies & n = 24. \end{align} Now \begin{align} S_n&=\frac{n}{2}[a_1+a_n] \\ \implies S_{24}&=\frac{24}{2}[6+96]\\ &=12\times 102\\ &=1224. \end{align} Hence the sum of given series is $1224$. =====Qu
- Question 14, Exercise 4.5
- ce $|r|=0.1 < 1$, this series has the sum: \begin{align*} S-\infty & = \frac{a_1}{1-r} \\ & = \frac{0.4}{... 1.0.1} = \frac{0.4}{0.9} \\ & = \frac{4}{9}. \end{align*} Hence $S_{\infty} =\dfrac{4}{9}$. =====Questio... ce $|r|=0.1 < 1$, this series has the sum: \begin{align*} S-\infty & = \frac{a_1}{1-r} \\ & = \frac{0.9}{1.0.1} = \frac{0.9}{0.9} \\ & = 1 \end{align*} Hence $S_{\infty}= 1 $. =====Question 14(iii