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- Question 1 and 2, Exercise 4.1
- Question 3 and 4, Exercise 4.1
- Question 5 and 6, Exercise 4.1
- Question 7 and 8, Exercise 4.1
- Question 9 and 10, Exercise 4.1
- Question 11 and 12, Exercise 4.1
- Question 13 and 14, Exercise 4.1
- Question 15 and 16, Exercise 4.1
- Question 17 and 18, Exercise 4.1
- Question 19 and 20, Exercise 4.1
- Question 21 and 22, Exercise 4.1
- Question 1, Exercise 4.2
- Question 2, Exercise 4.2
- Question 3 and 4, Exercise 4.2
- Question 5 and 6, Exercise 4.2
- Question 7 and 8, Exercise 4.2
- Question 9 and 10, Exercise 4.2
- Question 11 and 12, Exercise 4.2
- Question 13, Exercise 4.2
- Question 14 and 15, Exercise 4.2
- Question 16 and 17, Exercise 4.2
- Question 1 and 2, Exercise 4.3
- Question 3 and 4, Exercise 4.3
- Question 5 and 6, Exercise 4.3
- Question 7 and 8, Exercise 4.3
- Question 9 and 10, Exercise 4.3
- Question 11 and 12, Exercise 4.3
- Question 13 and 14, Exercise 4.3
- Question 15 and 16, Exercise 4.3
- Question 17, 18 and 19, Exercise 4.3
- Question 20, 21 and 22, Exercise 4.3
- Question 23 and 24, Exercise 4.3
- Question 25 and 26, Exercise 4.3
- Question 1 and 2, Exercise 4.4
- Question 3 and 4, Exercise 4.4
- Question 5, 6 and 7, Exercise 4.4
- Question 8 and 9, Exercise 4.4
- Question 10 and 11, Exercise 4.4
- Question 12 and 13, Exercise 4.4
- Question 14 and 15, Exercise 4.4
- Question 16 and 17, Exercise 4.4
- Question 18 and 19, Exercise 4.4
- Question 20 and 21, Exercise 4.4
- Question 22 and 23, Exercise 4.4
- Question 24 and 25, Exercise 4.4
- Question 26 and 27, Exercise 4.4
- Question 28 and 29, Exercise 4.4
- Question 30, Exercise 4.4
- Question 1 and 2, Exercise 4.5
- Question 3 and 4, Exercise 4.5
- Question 5 and 6, Exercise 4.5
- Question 7 and 8, Exercise 4.5
- Question 9 and 10, Exercise 4.5
- Question 11, 12 and 13, Exercise 4.5
- Question 14, Exercise 4.5
- Question 15, Exercise 4.5
- Question 16, Exercise 4.5
- Question 1 and 2, Exercise 4.6
- Question 3 & 4, Exercise 4.6
- Question 5 & 6, Exercise 4.6
- Question 7 & 8, Exercise 4.6
- Question 9 & 10, Exercise 4.6
- Question 11, Exercise 4.6
- Question 12, Exercise 4.6
- Question 1 and 2, Exercise 4.7
- Question 3 and 4, Exercise 4.7
- Question 5 and 6, Exercise 4.7
- Question 7 and 8, Exercise 4.7
- Question 9 and 10, Exercise 4.7
- Question 11, 12 and 13, Exercise 4.7
- Question 14, 15 and 16, Exercise 4.7
- Question 17 and 18, Exercise 4.7
- Question 19 and 20, Exercise 4.7
- Question 19 and 20, Exercise 4.7
- Question 21 and 22, Exercise 4.7
- Question 23 and 24, Exercise 4.7
- Question 25 and 26, Exercise 4.7
- Question 27 and 28, Exercise 4.7
- Question 29 and 30, Exercise 4.7
- Question 1 and 2, Exercise 4.8
- Question 3 and 4, Exercise 4.8
- Question 5 and 6, Exercise 4.8
- Question 7 and 8, Exercise 4.8
- Question 9 and 10, Exercise 4.8
- Question 11 and 12, Exercise 4.8
- Question 13, 14 and 15, Exercise 4.8
Fulltext results:
- Question 2, Exercise 4.2
- he next three terms of each arithmetic sequence. $11,14,17, \ldots$ ** Solution. ** Given: $$11, 14, 17, \ldots$$ Thus $a_1=11$, $d=14-11=3$.\\ Now $$a_n=a_1+(n-1)d.$$ So, we have \begin{align*} a_4 &= 11 + (4-1) \cdot 3 = 11 +
- Question 7 and 8, Exercise 4.7
- 16 - 8 + 3) + (25 - 10 + 3) \\ &= 3 + 2 + 3 + 6 + 11 + 18 \\ &= 43 \end{align*}m( =====Question 8==... + \frac{1}{8(9)} + \frac{1}{9(10)} + \frac{1}{10(11)} \\ &= \frac{1}{2} + \frac{1}{6} + \frac{1}{12} ... ac{1}{9} - \frac{1}{10} + \frac{1}{10} - \frac{1}{11} \\ &= 1 - \frac{1}{11} \\ &= \frac{11}{11} - \frac{1}{11} \\ &= \frac{10}{11} \end{align*} ====
- Question 1 and 2, Exercise 4.6
- cated term of the harmonic progression. $\frac{1}{11}, \frac{1}{9}, \frac{1}{7}, \ldots \quad 10$ th term. ** Solution. ** \begin{align*} &\frac{1}{11}, \frac{1}{9}, \frac{1}{7}, \ldots\quad \text{ is in H.P.}\\ &11, 9, 7, \ldots \quad \text{ is in A.P.}\end{align*} Here $a_1 = 11$, $d = 9 - 11 = -2.$ $a_{10}=?$ The general ter
- Question 9 & 10, Exercise 4.6
- P. =====Question 10===== Find H.M. between 9 and 11 . Also find $A, H, G$ and show that $A H=G^{2}$. ** Solution. ** Here $a = 9, b = 11$ \begin{align*} H &= \frac{2 a b}{a+b} = \frac{2(9)(11)}{9+11} \\ &= \frac{198}{21} =\frac{66}{7} \end{align*} Now\\ \begin{align*} A & = \frac{a+b}{2} = \f
- Question 12, Exercise 4.6
- 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 $$\dfrac{1}{3},H_1, H_2, H_3, H_4, \dfrac{1}{11} \text{ are in H.P.}$$ $$\quad 3,\dfrac{1}{H_1},\dfrac{1}{H_2}, \dfrac{1}{H_3}, \dfrac{1}{H_4},11 \text{ are in A.P.}$$ Here $a_1=3$, $a_n=11$ and
- Question 11 and 12, Exercise 4.1
- ====== Question 11 and 12, Exercise 4.1 ====== Solutions of Question 11 and 12 of Exercise 4.1 of Unit 04: Sequence and S... book Board, Islamabad, Pakistan. =====Question 11===== Find the indicated term of the sequence: $a_... nd the indicated term of the sequence. $a_{n}=5 n+11 ; a_{9}$ ** Solution. ** Do yourself. ====Go
- Question 3 and 4, Exercise 4.2
- $$ From this, we have $a_1 = 0.07$, $d=0.05$, $a_{11}=?$.\\ Now \begin{align*} a_n&=a_1+(n-1)d \\ \implies a_{11}&= 0.07+(11-1)(0.05)\\ &=0.07+(10)(0.05)\\ &=0.57 \end{align*} Hence $a_{11}=0.57.$ GOOD =====Question 4===== The third term
- Question 17, 18 and 19, Exercise 4.3
- 6n=16+50 \\ \implies & 6n= 66 \\ \implies & n = 11. \end{align} Now \begin{align} S_n&=\frac{n}{2}[a_1+a_n] \\ \implies S_{11}&=\frac{11}{2}[10+(-50)]\\ &=\frac{11}{2}\times (-40)\\ &=-220. \end{align} Hence the sum of the given series is
- Question 20, 21 and 22, Exercise 4.3
- \\ \implies & 139-7=11d\\ \implies & d=\frac{132}{11}=12. \end{align} Thus \begin{align} &a_2=a_1+d=7+... plies & 3432=312n\\ \implies & n=\frac{3432}{312}=11. \end{align} Also we have \begin{align} &a_n=a_1+(n-1)d\\ \implies & 306=6+(11-1)d\\ \implies & 306-6=10d\\ \implies & d=\frac{3... == <text align="left"><btn type="primary">[[math-11-nbf:sol:unit04:ex4-3-p9|< Question 17, 18 & 19]]<
- Question 27 and 28, Exercise 4.7
- of the series: $$5+\frac{7}{3}+\frac{9}{9}+\frac{11}{27}+\ldots$$. ** Solution. ** Given arithmetic... tric series is: $$5+\frac{7}{3}+\frac{9}{9}+\frac{11}{27}+\ldots$$ It can be written as: $$ 5\times 1+7\times\frac{1}{3}+9\times\frac{1}{9}+11\times\frac{1}{27}+\ldots $$ The numbers $5,7,9,11,4,\ldots$ are in A.P. with $a=5$ and $d=7-5=2$. T
- Question 7 and 8, Exercise 4.8
- series: $$\frac{1}{1 \times 6}+\frac{1}{6 \times 11}+\frac{1}{11 \times 16}+\ldots$$ ** Solution. ** Let $T_k$ represents the kth term of the series. T... }-\frac{1}{6}\right) + \left(\frac{1}{6}-\frac{1}{11}\right)+\left(\frac{1}{11}-\frac{1}{16}\right)\right.\\ &\left.+\ldots+\left(\frac{1}{5n-9}-\frac{1}{
- Question 13 and 14, Exercise 4.1
- } &= (-1)^{12-1}(3.4 \cdot 12 - 17.3) \\ &= (-1)^{11}(40.8 - 17.3) \\ &= (-1)^{11}(23.5) \\ &= -23.5 \end{align*} Hence $a_{12} =-23.5 $ GOOD ====Go to ... == <text align="left"><btn type="primary">[[math-11-nbf:sol:unit04:ex4-1-p6|< Question 11 & 12]]</btn></text> <text align="right"><btn type="success">[[m
- Question 5 and 6, Exercise 4.2
- he fifth term of an arithmetic sequence is 19 and 11 th term is 43. Find the first term and 87 th term. ** Solution. ** Given: $a_5 = 19$ and $a_{11} = 43$. The nth term of the arithmetic sequence ... \cdots (1) \end{align*} Also \begin{align*} & a_{11} = 43 \\ \implies & a_1 + 10d = 43 \quad \cdots (... == <text align="left"><btn type="primary">[[math-11-nbf:sol:unit04:ex4-2-p3|< Question 3 & 4]]</btn><
- Question 11 and 12, Exercise 4.2
- ====== Question 11 and 12, Exercise 4.2 ====== Solutions of Question 11 and 12 of Exercise 4.2 of Unit 04: Sequence and S... book Board, Islamabad, Pakistan. =====Question 11===== If Rs. $1000$ is saved on August 1, Rs. $300... == <text align="left"><btn type="primary">[[math-11-nbf:sol:unit04:ex4-2-p6|< Question 9 & 10]]</btn>
- Question 14 and 15, Exercise 4.2
- 3$, $y$ are in A.P.\\ Here $a_1 = x$ and $d=13-2=11$. Now \begin{align*} & a_2= a_1+d\\ \implies &2 = x + 11 \\ \implies & \boxed{x=-9} \end{align*} Also \begin{align*} &a_4=a_1+3d\\ \implies &y=x+3(11)\\ \implies &y=-9 + 33 \\ \implies &\boxed{y=24}\... == <text align="left"><btn type="primary">[[math-11-nbf:sol:unit04:ex4-2-p8|< Question 13]]</btn></te