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- Question 9 and 10, Exercise 4.1
- Question 19 and 20, Exercise 4.1
- Question 9 and 10, Exercise 4.2
- Question 16 and 17, Exercise 4.2
- Question 9 and 10, Exercise 4.3
- Question 20, 21 and 22, Exercise 4.3
- Question 10 and 11, Exercise 4.4
- Question 20 and 21, Exercise 4.4
- Question 9 and 10, Exercise 4.5
- Question 9 & 10, Exercise 4.6
- Question 9 and 10, Exercise 4.7
- Question 19 and 20, Exercise 4.7
- Question 9 and 10, Exercise 4.8
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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 arithm... 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... Find sum to infinity of the series: $$3+\frac{6}{10}+\frac{9}{100}+\frac{12}{1000}+\ldots$$ ** Solut
- Question 5 and 6, Exercise 4.1
- given, find the first 4 terms; the 10th term, $a_{10}$ and the 15 th term: $a_{15}$.$a_{n}=n^{2}-2 n$ ... 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, $a_1 = -1$,$
- Question 1 and 2, Exercise 4.1
- given, find the first 4 terms; the 10th term, $a_{10}$ and the 15 th term: $a_{15}$. $$a_{n}=3 n+1$$ ... 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 = 45 + 1 = 46. \end{align*} Hence $a_1=4$, $a_2=7$,
- 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 $$
- Question 9 and 10, Exercise 4.1
- ====== Question 9 and 10, Exercise 4.1 ====== Solutions of Question 9 and 10 of Exercise 4.1 of Unit 04: Sequence and Series. ... given, find the first 4 terms; the 10th term, $a_{10}$ and the 15 th term, $a_{15}$: $a_{n}=(-1)^{n}(n... ** Solution. ** Do yourself. =====Question 10===== The $n$th term of the sequence is given, fin
- Question 7 and 8, Exercise 4.7
- - 4 + 3) + (9 - 6 + 3)\\ &+ (16 - 8 + 3) + (25 - 10 + 3) \\ &= 3 + 2 + 3 + 6 + 11 + 18 \\ &= 43 \end{... ===Question 8===== Evaluate the sum: $\sum_{k=1}^{10} \frac{1}{k(k+1)}$ ** Solution. ** \begin{align*} \sum_{k=1}^{10} \frac{1}{k(k+1)} &= \frac{1}{1(2)} + \frac{1}{2(... )} + \frac{1}{7(8)} + \frac{1}{8(9)} + \frac{1}{9(10)} + \frac{1}{10(11)} \\ &= \frac{1}{2} + \frac{1}
- Question 5 and 6, Exercise 4.8
- of difference, find the sum of the series: $3+4+6+10+18+34+66+\dots$ to $n$ term. ** Solution. ** Let $$ S_{n}=3+4+6+10+18+\ldots +T_{n} $$ Also $$ S_{n}=3+4+6+10+\ldots +T_{n-1}+T_{n}. $$ Subtracting the second expressio... ssion, we have \begin{align*} S_{n}-S_{n}& =3+4+6+10+18+\ldots +T_{n} \\ & -\left(3+4+6+10+\ldots +T_
- Question 3 and 4, Exercise 4.1
- given, find the first 4 terms; the 10th term, $a_{10}$ and the 15 th term: $a_{n}=\frac{n}{n+1}$ ** S... frac{4}{5}\\ \end{align*} Now \begin{align*} a_{10} &= \frac{10}{10+1} = \frac{10}{11}\\ a_{15} &= \frac{15}{15+1} = \frac{15}{16} \end{align*} So, $a_1 = \frac{1}{
- Question 5 and 6, Exercise 4.3
- mplies S_{20}&=\frac{20}{2}[2(50)+(20-1)(-4)]\\ &=10\times [100-76]\\ &=240. \end{align} Hence $S_{20}=240$. GOOD =====Question 6===== Find the sum of series. $-3+(-7)+(-11)+\cdots +a_{10}$ ** Solution. ** Given series is arithmetic series with $a_1=-3$, $d=-7-(-3)=-4$, $n=10$.\\ Let $S_n$ represents sum of arithmetic series
- Question 17, 18 and 19, Exercise 4.3
- \implies S_{24}&=\frac{24}{2}[6+96]\\ &=12\times 102\\ &=1224. \end{align} Hence the sum of given ser... stion 19===== Find sum of the arithmetic series. $10+4+(-2)+\ldots+(-50)$ ** Solution. ** Given arithmetic series: $$10+4+(-2)+\ldots+(-50).$$ So, $a_{1}=10$, $d=4-10=-6$, $a_{n}=-50$, $n=?$. We have \begin{align} & a_n
- Question 7 and 8, Exercise 4.5
- he geometric series. $a_{1}=16, r=-\frac{1}{2}, n=10$ ** Solution. ** Given $a_1 = 16$, $r = -\frac{1}{2}$ and $n = 10$. The formula to find the sum of $n$ terms of a g... 1 - r}, \quad r \neq 1.$$ Thus, \begin{align*} S_{10} &= \frac{16 \left(1 - \left(-\frac{1}{2}\right)^{10}\right)}{1 - \left(-\frac{1}{2}\right)} \\ &= \fr
- Question 7 and 8, Exercise 4.1
- given, find the first 4 terms; the 10th term, $a_{10}$ and the 15 th term: $a_{15}$.$a_{n}=\left(\fra... = \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} &= ... $, $a_3 = \frac{1}{4}$, $a_4 = \frac{-1}{8}$, $a_{10} = \frac{-1}{512}$, $a_{15} = \frac{1}{16384}$. G
- Question 1, Exercise 4.2
- a_2&=4+(2-1)3=4+3=7\\ a_3 &= 4+ (3-1) 3 = 4 + 6 = 10\\ a_4&=4+(4-1)3=4+9=13 \end{align*} Hence $a_1=4$, $a_2=7$, $a_3=10$, $a_4=13$. GOOD =====Question 1(ii)===== Find t... a_2&=7+(2-1)(5)=7+5=12\\ a_3 &= 7+ (3-1)(5) = 7 + 10 = 17\\ a_4&=7+(4-1)(5)=7+15=22 \end{align*} Hence... (3-1)(-2) = 16 - 4 = 12\\ a_4&=16+(4-1)(-2)=16-6=10 \end{align*} Hence $a_1=16$, $a_2=14$, $a_3=12$,
- Question 8 and 9, Exercise 4.4
- next two terms of the geometric sequence: $$90,30,10 \ldots$$ ** Solution. ** Given sequence is geom... rac{1}{3} \right)^3=90 \times\dfrac{1}{27}=\dfrac{10}{3}\\ & a_{5}=a_{1} r^3=(90)\left(\dfrac{1}{4} \right)^4=90 \times\dfrac{1}{81}=\dfrac{10}{9}\\ \end{align*} Hence $a_4=\dfrac{10}{3}$, $a_5=\dfrac{10}{9}$. GOOD =====Question 9===== Fi
- Question 1 and 2, Exercise 4.5
- the sum of the geometric series. $75+15+3+...$ to 10 terms. ** Solution. ** Given: $75+15+3+...$ to 10 terms.\\ Here $a_1= 75$, $r = \frac{15}{75} = \frac{1}{5}$, $n = 10$. The formula to find the sume of $n$ terms of g... }{1 - r}, \quad r\neq 1.$$ Thus \begin{align*} S_{10} &= \frac{75\left(1 - (\frac{1}{5})^{10}\right)}{