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- Question 29 and 30, Exercise 4.7
- nd sum to infinity of the series: $$1+4 x+7 x^{2}+10 x^{3}+\ldots$$ ** Solution. ** The given arithmetic-geometric series is:\\ \[ 1 + 4x + 7x^2 + 10x^3 + \ldots \] It can be rewritten as:\\ \[ 1 \times 1 + 4 \times x + 7 \times x^2 + 10 \times x^3 + \ldots \] The numbers \(1, 4, 7, 10, \ldots\) are in AP with \(a = 1\) and \(d = 4 - 1
- Question 5 and 6, Exercise 4.1
- he sequence is given, find the first 4 terms; the 10th term, $a_{10}$ and the 15 th term: $a_{15}$.$a_{n}=n^{2}-2 n$ ** Solution. ** Given $$a_n = n^2 - ... 16 - 8 = 8\\ \end{align*} Now \begin{align*} a_{10} &= (10)^2 - 2(10) = 100 - 20 = 80\\ a_{15} &= (15)^2 - 2(15) = 225 - 30 = 195 \end{align*} So, $a_1
- Question 1 and 2, Exercise 4.1
- he sequence is given, find the first 4 terms; the 10th term, $a_{10}$ and the 15 th term: $a_{15}$. $$a_{n}=3 n+1$$ ** Solution. ** Given $$a_{n}=3 n+1$... 3(2) + 1 = 6 + 1 = 7\\ a_3 &= 3(3) + 1 = 9 + 1 = 10\\ a_4 &= 3(4) + 1 = 12 + 1 = 13\\ \end{align*} Now \begin{align*} a_{10} &= 3(10) + 1 = 30 + 1 = 31\\ a_{15} &= 3(15) + 1
- Question 9 and 10, Exercise 4.1
- ====== Question 9 and 10, Exercise 4.1 ====== Solutions of Question 9 and 10 of Exercise 4.1 of Unit 04: Sequence and Series. ... he sequence is given, find the first 4 terms; the 10th term, $a_{10}$ and the 15 th term, $a_{15}$: $a_{n}=(-1)^{n}(n+3)$ ** Solution. ** Do yourself.
- Question 1 and 2, Exercise 4.7
- frac{1}{4} + \frac{1}{6} + \frac{1}{8} + \frac{1}{10}\\ &= \frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \frac{1}{8} + \frac{1}{10}\\ &= \frac{30}{60} + \frac{15}{60} + \frac{10}{60} + \frac{7.5}{60} + \frac{6}{60}\\ &= \frac{68.5}{... }{9} + \frac{1}{11} + \frac{1}{13}\\ &= \frac{35}{105} + \frac{21}{105} + \frac{15}{105} + \frac{11.67
- Question 3 and 4, Exercise 4.1
- he sequence is given, find the first 4 terms; the 10th term, $a_{10}$ and the 15 th term: $a_{n}=\frac{n}{n+1}$ ** Solution. ** Given $$a_n = \frac{n}{n+... frac{4}{5}\\ \end{align*} Now \begin{align*} a_{10} &= \frac{10}{10+1} = \frac{10}{11}\\ a_{15} &= \frac{15}{15+1} = \frac{15}{16} \end{align*} So, $a_
- Question 1 and 2, Exercise 4.3
- =====Question 1===== Find the sum of series: $4+7+10+13+16+19+22+25$ ** Solution. ** Given: $4+7+10+13+16+19+22+25$. As the given series is arithmetic s... ind the sum of series. $a_{1}=2$, $a_{n}=200$, $n=100$ FIXME Statement is logically incorrenct. Find... ithematic series with: $a_{1}=2$, $a_{n}=200$, $n=100$ ** Solution. ** Given $a_{1}=2$, $a_{n}=200$,
- Question 7 and 8, Exercise 4.5
- he geometric series. $a_{1}=16, r=-\frac{1}{2}, n=10$ ** Solution. ** Given $a_1 = 16$, $r = -\frac{1}{2}$ and $n = 10$. The formula to find the sum of $n$ terms of a g... 1 - r}, \quad r \neq 1.$$ Thus, \begin{align*} S_{10} &= \frac{16 \left(1 - \left(-\frac{1}{2}\right)^{10}\right)}{1 - \left(-\frac{1}{2}\right)} \\ &= \fr
- Question 1 and 2, Exercise 4.8
- n}-T_{n-1}\right)-T_{n}. \\ \implies 0=&3+(4+6+8+10+\ldots \text { up to } (n-1) \text { terms })-T_... end{align*} Then \begin{align*} T_{n} & =3+(4+6+8+10+\ldots \text { up to }(n-1) \text { terms }) \\ ... {6}(2n^2+2n+n+1+3n+3+6) \\ & =\frac{n}{6}(2n^2+6n+10) \\ & =\frac{n}{3}(n^2+3n+5) \end{align*} Hence ... d of difference, find the sum of the series: $1+4+10+22+\ldots$ to $n$ term. ** Solution. ** Let $$
- Question 7 and 8, Exercise 4.2
- 6+(n-1)4\\ \implies &70=-6+4n-4\\ \implies &70=4n-10\\ \implies &4n=80\\ \implies & n=20 \end{align*} ... , \dfrac{3}{2}, \dfrac{1}{2}, \ldots$ is $-\dfrac{105}{2}$? ** Solution. ** Given: $\dfrac{5}{2}, \d... ots$ is an arithmetic sequence and $a_n = -\dfrac{105}{2}$. Here $a_1 = \dfrac{5}{2}$, $d = \dfrac{3}... _1 + (n-1)d.$$ This gives \begin{align*} & -\frac{105}{2} = \frac{5}{2} + (n-1)(-1)\\ \implies & -\fra
- Question 11 and 12, Exercise 4.2
- mabad, Pakistan. =====Question 11===== If Rs. $1000$ is saved on August 1, Rs. $3000$ on August $2$... olution. ** The sequence of the saved money is $$1000, 3000, 5000, \dots, \text{ upto 20 terms}.$$ This is in A.P with $a_1 = 1000$, $d=3000-1000=2000$, $S_20=?$ Since $$S_n =\frac{n}{2}[2a_1+(n-1)d],$$ implies \begin{align*} S_{2
- Question 7 and 8, Exercise 4.7
- - 4 + 3) + (9 - 6 + 3)\\ &+ (16 - 8 + 3) + (25 - 10 + 3) \\ &= 3 + 2 + 3 + 6 + 11 + 18 \\ &= 43 \end{... ===Question 8===== Evaluate the sum: $\sum_{k=1}^{10} \frac{1}{k(k+1)}$ ** Solution. ** \begin{align*} \sum_{k=1}^{10} \frac{1}{k(k+1)} &= \frac{1}{1(2)} + \frac{1}{2(... )} + \frac{1}{7(8)} + \frac{1}{8(9)} + \frac{1}{9(10)} + \frac{1}{10(11)} \\ &= \frac{1}{2} + \frac{1}
- Question 5 and 6, Exercise 4.8
- of difference, find the sum of the series: $3+4+6+10+18+34+66+\dots$ to $n$ term. ** Solution. ** Let $$ S_{n}=3+4+6+10+18+\ldots +T_{n} $$ Also $$ S_{n}=3+4+6+10+\ldots +T_{n-1}+T_{n}. $$ Subtracting the second expressio... ssion, we have \begin{align*} S_{n}-S_{n}& =3+4+6+10+18+\ldots +T_{n} \\ & -\left(3+4+6+10+\ldots +T_
- Question 7 and 8, Exercise 4.1
- he sequence is given, find the first 4 terms; the 10th term, $a_{10}$ and the 15 th term: $a_{15}$.$a_{n}=\left(\frac{-1}{2}\right)^{n-1}$ ** Solution. ... = \frac{-1}{8} \end{align*} Now \begin{align*} a_{10} &= \left( \frac{-1}{2} \right)^{10-1} = \left( \frac{-1}{2} \right)^9 = \frac{-1}{512} \\ a_{15} &=
- Question 5 and 6, Exercise 4.3
- mplies S_{20}&=\frac{20}{2}[2(50)+(20-1)(-4)]\\ &=10\times [100-76]\\ &=240. \end{align} Hence $S_{20}=240$. GOOD =====Question 6===== Find the sum of series. $-3+(-7)+(-11)+\cdots +a_{10}$ ** Solution. ** Given series is arithmetic series with $a_1=-3$, $d=-7-(-3)=-4$, $n=10$.\\ Let $S_n$ represents sum of arithmetic series