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- Question 11 and 12, Exercise 4.8
- es. Then \begin{align*} T_k &= \frac{1}{k(k+2)}. \end{align*} Resolving it into partial fractions: \beg... k(k+2)} = \frac{A}{k} + \frac{B}{k+2} \ldots (1) \end{align*} 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}. \end{align*} Put $k+2=0 \implies k=-2$ in (2), we have
- Question 7 and 8, Exercise 4.8
- en \begin{align*} T_k &=\frac{1}{(3k-2)(3k+1)}. \end{align*} Resolving it into partial fraction: \begi... k+1)} = \frac{A}{3k-2}+\frac{B}{3k+1} \ldots (1) \end{align*} Multiplying with $(3k-2)(3k+1)$ \begin{align*} 1 = (3k+1)A+(3k-2)B \ldots (2) \end{align*} Put $3k-2=0$ $\implies k=\dfrac{2}{3}$ in... }{3}+1 \right)A+0 \\ \implies &A = \frac{1}{3}. \end{align*} Now put $3k+1=0$ $\implies k=-\dfrac{1}{3
- Question 5 and 6, Exercise 4.2
- 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 ... \mathop-\limits_{+}40 \\ \hline & 11d &= -33\\ \end{array}&\\ \implies \boxed{d = -3} \quad \\ \end{align*} Putting the value $d$ in (1) \begin{align*} &
- Question 20, 21 and 22, Exercise 4.3
- & 1752=146n\\ \implies & n=\frac{1752}{146}=12. \end{align} Also we have \begin{align} &a_n=a_1+(n-1)d... es & 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 $a_1=7$, $a_2=19$, $a_3=31$. ... 371\\ \implies & 7a_1=378-371 \\ \implies a_1=1. \end{align} Also we have \begin{align} &a_n=a_1+(n-1)d
- Question 1 and 2, Exercise 4.8
- & -\left(3+7+13+21+\ldots +T_{n-1}+T_{n}\right) \end{align*} \begin{align*} \implies 0=&3+(7-3)+(13-7... ts \text { up to } (n-1) \text { terms })-T_{n} \end{align*} Then \begin{align*} T_{n} & =3+(4+6+8+10+... & =3+(n-1)(n+2) \\ & =3+n^2-n+2n-2 \\ & =n^2+n+1 \end{align*} Thus, the kth term of series: $$ T_{k}=k^... c{n}{6}(2n^2+6n+10) \\ & =\frac{n}{3}(n^2+3n+5) \end{align*} Hence the sum of given series is $\dfrac{
- Question 3 and 4, Exercise 4.8
- & -\left(1+4+13+40+\ldots +T_{n-1}+T_{n}\right) \end{align*} \begin{align*} \implies 0=&1+(4-1)+(13-... ots \text { up to } (n-1) \text { terms })-T_{n} \end{align*} Then \begin{align*} T_{n} & =1+(3+9+27+8... \frac{2+3^{n}-3}{2} \\ \\ & =\frac{3^{n}-1}{2}. \end{align*} Thus, the kth term of the series: $$ T_{k... 3}{2}-n\right) \\ & =\frac{1}{4}(3^{n+1}-3-2n). \end{align*} Hence, the sum of the given series is \(
- Question 5 and 6, Exercise 4.8
- \ & -\left(3+4+6+10+\ldots +T_{n-1}+T_{n}\right) \end{align*} \begin{align*} \implies 0=&3+(4-3)+(6-4)... ots \text { up to } (n-1) \text { terms })-T_{n} \end{align*} Then \begin{align*} T_{n} & =3+(1+2+4+8+\... r-1}\right) \\ & =3+(2^{n-1}-1) \\ & =2^{n-1}+2. \end{align*} Thus, the kth term of the series: $$ T_{k... & =\frac{1(2^n-1)}{2-1} + 2n \\ & =2^n - 1 + 2n. \end{align*} Hence, \( S_n = 2^n - 1 + 2n \). GOOD
- Question 9 and 10, Exercise 4.8
- ider \begin{align*} T_k &= \frac{1}{(k+1)(k+2)}. \end{align*} Resolving it into partial fractions: \beg... k+2)} = \frac{A}{k+1} + \frac{B}{k+2} \ldots (1) \end{align*} Multiplying both sides by $(k+1)(k+2)$, w... et \begin{align*} 1 = (k+2)A + (k+1)B \ldots (2) \end{align*} Now, put $k+1=0 \implies k=-1$ in equatio... gin{align*} 1 &= (-1+2)A + 0 \\ \implies A &= 1. \end{align*} Next, put $k+2=0 \implies k=-2$ in equati
- Question 13, 14 and 15, Exercise 4.8
- en \begin{align*} T_k &= \frac{1}{(2k+3)(2k+9)}. \end{align*} Resolving it into partial fractions: \beg... 9)} = \frac{A}{2k+3} + \frac{B}{2k+9} \ldots (1) \end{align*} Multiplying both sides by $(2k+3)(2k+9)$,... \begin{align*} 1 = (2k+9)A + (2k+3)B \ldots (2) \end{align*} Now, put $2k+3 = 0 \implies k = -\frac{3}... 9)A + 0 \\ 1 &= 6A \\ \implies A &= \frac{1}{6}. \end{align*} Next, put $2k+9 = 0 \implies k = -\frac{9
- Question 3 and 4, Exercise 4.2
- = 0.07+(11-1)(0.05)\\ &=0.07+(10)(0.05)\\ &=0.57 \end{align*} Hence $a_{11}=0.57.$ GOOD =====Question ... 14 \\ \implies & a_1 + 2d = 14 \quad \cdots (1) \end{align*} Also \begin{align*} & a_9 = -1 \\ \implies & a_1 + 8d = -1 \quad \cdots (2) \end{align*} Now, subtract equation (1) from equation... = \mathop{}\limits_{-}14 \\ \hline & 6d &= -15\\ \end{array}&\\ \implies \boxed{d = -\frac{5}{2}} \quad
- Question 16 and 17, Exercise 4.2
- 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 \end{align*} Hence $A_1 = 9$ and $A_2 = 13$. GOOD ===... 4d\\ \implies & 4d=-20\\ \implies & \boxed{d=-5}.\end{align*} Now \begin{align*} A_1 &= a_2= a_1+d \\ &
- Question 20 and 21, Exercise 4.4
- 16 \\ \implies & r^4 = 2^4 \\ \implies & r = 2. \end{align*} Thus \begin{align*} & a_2=a_1 r= (3)(2) =... 2 = (3)(2)^2 = 12 \\ & a_4=a_1 r^3= (3)(2)^3=24. \end{align*} Hence $6$, $12$, $24$ are required geomet... plies & r^4 = (\pm 2)^4 \\ \implies & r = \pm 2. \end{align*} Thus, if $a_1=3$ and $r=2$, then \begin{a... 2 = (3)(2)^2 = 12 \\ & a_4=a_1 r^3= (3)(2)^3=24. \end{align*} If $a_1=3$ and $r=-2$, then \begin{align*
- Question 1, Exercise 4.2
- = 4+ (3-1) 3 = 4 + 6 = 10\\ a_4&=4+(4-1)3=4+9=13 \end{align*} Hence $a_1=4$, $a_2=7$, $a_3=10$, $a_4=13... (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=2... 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=... )(-4) = 38 - 8 = 30\\ a_4&=38+(4-1)(-4)=38-12=26 \end{align*} Hence $a_1=38$, $a_2=34$, $a_3=30$, $a_4=
- Question 9 and 10, Exercise 4.2
- 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{a... }{a}\\ 2b&= \frac{a+c}{ac}\\ b&= \frac{a+c}{2ac} \end{align*} Putting the value of $b$ in (i), we have\... {2ac}\\ &=\frac{a+c-2c}{2ac}\\ &=\frac{a-c}{2ac} \end{align*} As required. GOOD =====Question 10===== D
- Question 17, 18 and 19, Exercise 4.3
- +6n-6 \\ \implies & 6n=96 \\ \implies & n = 24. \end{align} Now \begin{align} S_n&=\frac{n}{2}[a_1+a_n... }&=\frac{24}{2}[6+96]\\ &=12\times 102\\ &=1224. \end{align} Hence the sum of given series is $1224$. =... =38-4n \\ \implies & 4n=36 \\ \implies & n = 9. \end{align} Now \begin{align} S_n&=\frac{n}{2}[a_1+a_n... ac{9}{2}[34+2]\\ &=\frac{9}{2}\times 36\\ &=162. \end{align} Hence the sum of the given series is $162$