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- 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... 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{... k-2=0$ $\implies k=\dfrac{2}{3}$ in (2), we have \begin{align*} &1 = \left(3\times\frac{2}{3}+1 \right
- Question 16 and 17, Exercise 4.2
- ===Question 16===== Find the two arithmetic means between $5$ and $17$. ** Solution. ** Let $A_1$ and $A_2$ be two arithmetic means between $5$ and $17$.\\ Then $5$, $A_1$, $A_2$, $17$ are in A.P.\\ Here $a_1=5$ ... 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\
- 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 = -7... *} Now, subtract equation (1) from equation (2): \begin{align*} \begin{array}{ccc} a_1& + 27d &= -73\\ \mathop{}\limits_{-}a_1 &\mathop+\limits_{-} 16d
- Question 20 and 21, Exercise 4.4
- We have given $a_1=3$ and $a_5=48$. Assume $r$ be common difference, then by general formula for nth term, we have $$ a_n=ar^{n-1}. $$ This gives \begin{align*} &a_5=a_1 r^4 \\ \implies & 48=3r^4 \\ ... 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... We have given $a_1=3$ and $a_5=48$. Assume $r$ be common difference, then by general formula for nt
- 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_{... +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 { u... \ & =3+\frac{n-1}{2}[2(4)+(n-1-1)(2)] \quad\left(\because S_{n}=\frac{n}{2}[2 a+(n-1) d]\right) \\ & =
- 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... +13+40+\ldots +T_{n-1}+T_{n}\right) \end{align*} \begin{align*} \implies 0=&1+(4-1)+(13-4)+(40-13)+(... } (n-1) \text { terms })-T_{n} \end{align*} Then \begin{align*} T_{n} & =1+(3+9+27+81+\ldots \text {... }) \\ & =1+\frac{3(3^{n-1}-1)}{3-1} \quad\left(\because S_{n}=\frac{a(r^n-1)}{r-1}\right) \\ & =1+\
- Question 14 and 15, Exercise 4.2
- =====Question 14===== Find '$b$' if $10$ is A.M between $b$ and $20$. ** Solution. ** Let $a= b$ and $b=20$. Then \begin{align*} &\text{A.M.} = \frac{a + b}{2} \\ \imp... answer given at the end of the book, question can be as follows: * Find '$b$' if $25$ is A.M between $b$ and $20$. * Find '$b$' if $10$ is A.M between
- 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 & ... n=\frac{1752}{146}=12. \end{align} Also we have \begin{align} &a_n=a_1+(n-1)d\\ \implies & 139=7+(12-... \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... $S_{n}=378$. First we find $a_1$ and $d$.\\ As \begin{align} &S_n=\frac{n}{2}[a_1+a_n]\\ \implies &
- 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-... } (n-1) \text { terms })-T_{n} \end{align*} Then \begin{align*} T_{n} & =3+(1+2+4+8+\ldots \text { up... s }) \\ & =3+\frac{1(2^{n-1}-1)}{2-1} \quad\left(\because S_{n}=\frac{a(r^n-1)}{r-1}\right) \\ & =3+(2
- 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{... = 0 \implies k = -\frac{3}{2}$ in equation (2): \begin{align*} 1 &= (2 \times \left(-\frac{3}{2}\righ
- Question 13, Exercise 4.2
- , Pakistan. =====Question 13(i)===== Find A.M. between $7$ and $17$ ** Solution. ** Here $a=7$ and $b=17$.\\ Now \begin{align*} \text{A.M.} &= \frac{a + b}{2}\\ &= \f... = $12$. GOOD =====Question 13(ii)===== Find A.M. between $3+3 \sqrt{2}$ and $7-3 \sqrt{2}$ ** Soluti... Here $a=3+3\sqrt{2}$ and $b=7-3\sqrt{2}$.\\ Now \begin{align*} \text{A.M.} &= \frac{a + b}{2}\\ &= \f
- Question 12, Exercise 4.6
- Pakistan. =====Question 12===== Find four H.Ms. between $\dfrac{1}{3}$ and $\dfrac{1}{11}$. ** Solution. ** Let $H_1, H_2, H_3, H_4$ be four $H.Ms$ between $\dfrac{1}{3}$ and $\dfrac{1}{11}$.\\ Then $$\dfrac{1}{3},H_1, H_2, H_3, H_4, \df... term of A.P is given as $$a_n=a_1+(n-1)d.$$ Thus \begin{align*}\\ &11 =3+(6-1)d\\ \implies &5d = 11-3\
- Question 9 and 10, Exercise 4.8
- dfrac{1}{(k+1)(k+2)}$ ** Solution. ** Consider \begin{align*} T_k &= \frac{1}{(k+1)(k+2)}. \end{align*} Resolving it into partial fractions: \begin{align*} \frac{1}{(k+1)(k+2)} = \frac{A}{k+1} +... } Multiplying both sides by $(k+1)(k+2)$, we get \begin{align*} 1 = (k+2)A + (k+1)B \ldots (2) \end{al... Now, put $k+1=0 \implies k=-1$ in equation (2): \begin{align*} 1 &= (-1+2)A + 0 \\ \implies A &= 1. \
- Question 3 and 4, Exercise 4.2
- have $a_1 = 0.07$, $d=0.05$, $a_{11}=?$.\\ Now \begin{align*} a_n&=a_1+(n-1)d \\ \implies a_{11}&= 0... equence is given as $$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... } Now, subtract equation (1) from equation (2): \begin{align*} \begin{array}{ccc} a_1 & + 8d &= -1\\