Search
You can find the results of your search below.
Fulltext results:
- Question 11 and 12, Exercise 4.8
- _k$ represent 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}... ign*} Multiplying both sides by $k(k+2)$, we get \begin{align*} 1 = A(k+2) + Bk \ldots (2) \end{align*} Put $k=0$ in (2), we have \begin{align*} &1=2A + 0 \\ \implies & A = \frac{1}{2}.
- Question 7 and 8, Exercise 4.8
- _k$ represents 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}+\... (1) \end{align*} Multiplying with $(3k-2)(3k+1)$ \begin{align*} 1 = (3k+1)A+(3k-2)B \ldots (2) \end{ali... k-2=0$ $\implies k=\dfrac{2}{3}$ in (2), we have \begin{align*} &1 = \left(3\times\frac{2}{3}+1 \right)A+
- Question 5 and 6, Exercise 4.2
- ic sequence 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 \... *} Now, subtract equation (1) from equation (2): \begin{align*} \begin{array}{ccc} a_1& + 27d &= -73\\ \mathop{}\limits_{-}a_1 &\mathop+\limits_{-} 16d &= \ma
- Question 20, 21 and 22, Exercise 4.3
- 9$, $S_{n}=876$. First we find $n$ and $d$.\\ As \begin{align} &S_n=\frac{n}{2}[a_1+a_n]\\ \implies & 876... 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 & 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)=3... $S_{n}=378$. First we find $a_1$ and $d$.\\ As \begin{align} &S_n=\frac{n}{2}[a_1+a_n]\\ \implies & 378
- Question 1 and 2, Exercise 4.8
- nd expression from the first expression, we have \begin{align*} S_{n}-S_{n}& =3+7+13+21+31+\ldots +T_{n} ... +13+21+\ldots +T_{n-1}+T_{n}\right) \end{align*} \begin{align*} \implies 0=&3+(7-3)+(13-7)+(21-13) \\ &+... } (n-1) \text { terms })-T_{n} \end{align*} Then \begin{align*} T_{n} & =3+(4+6+8+10+\ldots \text { up t... ies: $$ T_{k}=k^2++k+1 $$ Now taking sum, we get \begin{align*} S_{n} & =\sum_{k=1}^{n} T_{k}=\sum_{k=1}^
- Question 3 and 4, Exercise 4.8
- nd expression from the first expression, we have \begin{align*} S_{n}-S_{n}& =1+4+13+40+121+\ldots +T_{n... +13+40+\ldots +T_{n-1}+T_{n}\right) \end{align*} \begin{align*} \implies 0=&1+(4-1)+(13-4)+(40-13)+(121... } (n-1) \text { terms })-T_{n} \end{align*} Then \begin{align*} T_{n} & =1+(3+9+27+81+\ldots \text { up... \frac{3^{k}-1}{2}. $$ Now taking the sum, we get \begin{align*} S_{n} & =\sum_{k=1}^{n} T_{k}=\sum_{k=1}
- Question 5 and 6, Exercise 4.8
- nd expression from the first expression, we have \begin{align*} S_{n}-S_{n}& =3+4+6+10+18+\ldots +T_{n} ... 4+6+10+\ldots +T_{n-1}+T_{n}\right) \end{align*} \begin{align*} \implies 0=&3+(4-3)+(6-4)+(10-6)+(18-10)... } (n-1) \text { terms })-T_{n} \end{align*} Then \begin{align*} T_{n} & =3+(1+2+4+8+\ldots \text { up to... $ T_{k}=2^{k-1}+2. $$ Now taking the sum, we get \begin{align*} S_{n} & =\sum_{k=1}^{n} T_{k}=\sum_{k=1}^
- Question 9 and 10, Exercise 4.8
- dfrac{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} + \f... } Multiplying both sides by $(k+1)(k+2)$, we get \begin{align*} 1 = (k+2)A + (k+1)B \ldots (2) \end{align... Now, put $k+1=0 \implies k=-1$ in equation (2): \begin{align*} 1 &= (-1+2)A + 0 \\ \implies A &= 1. \end
- Question 13, 14 and 15, Exercise 4.8
- _k$ represent 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} +... Multiplying both sides by $(2k+3)(2k+9)$, we get \begin{align*} 1 = (2k+9)A + (2k+3)B \ldots (2) \end{ali... = 0 \implies k = -\frac{3}{2}$ in equation (2): \begin{align*} 1 &= (2 \times \left(-\frac{3}{2}\right)+
- Question 3 and 4, Exercise 4.2
- have $a_1 = 0.07$, $d=0.05$, $a_{11}=?$.\\ Now \begin{align*} a_n&=a_1+(n-1)d \\ \implies a_{11}&= 0.07... equence 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_1 + 8d = -1 \q... } Now, subtract equation (1) from equation (2): \begin{align*} \begin{array}{ccc} a_1 & + 8d &= -1\\ \ma
- Question 16 and 17, Exercise 4.2
- erm of A.P is given as $$a_n=a_1+(n-1)d.$$ Thus \begin{align*} &a_4 = a_1 + 3d \\ \implies & 17=5+3d\\ \... 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... erm of A.P is given as $$a_n=a_1+(n-1)d.$$ Thus \begin{align*} &a_5 = a_1 + 4d \\ \implies & -18=2+4d\\
- Question 20 and 21, Exercise 4.4
- nth term, we have $$ a_n=ar^{n-1}. $$ This gives \begin{align*} &a_5=a_1 r^4 \\ \implies & 48=3r^4 \\ \im... 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 ... nth term, we have $$ a_n=ar^{n-1}. $$ This gives \begin{align*} &a_5=a_1 r^4 \\ \implies & 48=3r^4 \\ \im... 2. \end{align*} Thus, if $a_1=3$ and $r=2$, then \begin{align*} & a_2=a_1 r= (3)(2) = 6 \\ & a_3=a_1 r^2
- Question 1, Exercise 4.2
- metic sequence is: $$a_n = a_1 + (n - 1)d.$$ Now \begin{align*} a_2&=4+(2-1)3=4+3=7\\ a_3 &= 4+ (3-1) 3 =... metic sequence is: $$a_n = a_1 + (n - 1)d.$$ Now \begin{align*} a_2&=7+(2-1)(5)=7+5=12\\ a_3 &= 7+ (3-1)(... $d=-2$.\\ We have $$a_n = a_1 + (n - 1)d.$$ Now \begin{align*} a_2&=16+(2-1)(-2)=16-2=14\\ a_3 &= 16+ (3... $d=-4$.\\ We have $$a_n = a_1 + (n - 1)d.$$ Now \begin{align*} a_2&=38+(2-1)(-4)=38-4=34\\ a_3 &= 38+ (3
- Question 9 and 10, Exercise 4.2
- ce $\dfrac{1}{a}, b, \dfrac{1}{c}$ 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}&=\frac{1}{c}-b\\ b+b&=\frac... ign*} Putting the value of $b$ in (i), we have\\ \begin{align*} d&=\frac{a+c}{2ac}-\frac{1}{a}\\ &=\frac{
- Question 17, 18 and 19, Exercise 4.3
- 1}=6$, $d=12-6=6$, $a_{n}=96$, $n=?$.\\ We have \begin{align} & a_n=a_1+(n-1)d \\ \implies & 96=6+(n-1)... s & 6n=96 \\ \implies & n = 24. \end{align} Now \begin{align} S_n&=\frac{n}{2}[a_1+a_n] \\ \implies S_{2... 1}=34$, $d=30-34=-4$, $a_{n}=2$, $n=?$. We have \begin{align} & a_n=a_1+(n-1)d \\ \implies & 2=34+(n-1)... es & 4n=36 \\ \implies & n = 9. \end{align} Now \begin{align} S_n&=\frac{n}{2}[a_1+a_n] \\ \implies S_{9