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- Question 7 and 8, Exercise 4.8
- 7===== Find the sum of $n$ term of the series: $$\frac{1}{1 \times 4}+\frac{1}{4 \times 7}+\frac{1}{7 \times 10}+\ldots$$ ** Solution. ** Given: $$\frac{1}{1 \times 4}+\frac{1}{4 \times 7}+\frac{1}{7 \t
- Question 11 and 12, Exercise 4.8
- Evaluate the sum of the series: $\sum_{k=1}^{n} \frac{1}{k(k+2)}$ ** Solution. ** Let $T_k$ represent... h 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{B}{k+2} \ldots (1) \end{align*} Multiplying
- Question 25 and 26, Exercise 4.7
- of the series (arithmetico-geometric series): $1+\frac{4}{7}+\frac{7}{7^{2}}+\frac{10}{7^{3}}+\ldots$ ** Solution. ** The given arithmetic-geometric series is: \[ 1 + \frac{4}{7} + \frac{7}{7^2} + \frac{10}{7^3} + \ldots \
- Question 1, Exercise 4.2
- t four terms of each arithmetic sequence. $a_{1}=\frac{3}{4}, d=\frac{1}{4}$ ** Solution. ** Given: $a_1=\frac{3}{4}$, $d=\frac{1}{4}$.\\ We have $$a_n = a_1 + (n - 1)d.$$ Now \begin{align*} a_2&=\frac{3}{4}+(2-1)\
- Question 27 and 28, Exercise 4.7
- 27===== Find sum to infinity of the series: $$5+\frac{7}{3}+\frac{9}{9}+\frac{11}{27}+\ldots$$. ** Solution. ** Given arithmetic-geometric series is: $$5+\frac{7}{3}+\frac{9}{9}+\frac{11}{27}+\ldots$$ It can b
- Question 1 and 2, Exercise 4.7
- uestion 1===== Evaluate the sum: $\sum_{k=1}^{5} \frac{1}{2 k}$ ** Solution. ** \begin{align*} \sum_{k=1}^{5} \frac{1}{2k} &= \frac{1}{2(1)} + \frac{1}{2(2)} + \frac{1}{2(3)} + \frac{1}{2(4)} + \frac{1}{2(5)}\\ &= \frac{1}{2} + \frac{1
- Question 14, Exercise 4.5
- amabad, Pakistan. =====Question 14(i)===== Find fractional notation for the infinite geometric series;... is infinite geometric series with $a_1=0.4$, $r=\frac{0.04}{0.4}=0.1$.\\ Since $|r|=0.1 < 1$, this series has the sum: \begin{align*} S-\infty & = \frac{a_1}{1-r} \\ & = \frac{0.4}{1.0.1} = \frac{0.4}{0.9} \\ & = \frac{4}{9}. \end{align*} Hence $S_{\infty}
- Question 7 and 8, Exercise 4.7
- estion 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(3)} + \frac{1}{3(4)} + \frac{1}{4(5)} + \frac{1}{5(6)} \\ &\quad + \frac{1}{6(7)}
- Question 9 and 10, Exercise 4.5
- f the geometric series. $a_{1}=343, a_{4}=-1, r=-\frac{1}{7}$ ** Solution. ** Given $a_{1}=343$, $a_{4}=-1$, $r=-\frac{1}{7}$\\ Let $S_n$ represents the sum of geometric series. Then $$ S_n =\frac{a_1-a_n r}{1-r}, \quad r\neq 1.$$ Thus \begin{align*} S_4 & =\frac{343-(-1)\left(-\frac{1}{7}\right)}{1+\frac{1}{7}}
- Question 29 and 30, Exercise 4.7
- -geometric series is given by:\\ \[ S_{\infty} = \frac{a}{1 - r} + \frac{d r}{(1 - r)^{2}} \] Thus, we have:\\ \begin{align*} S_{\infty} &= \frac{1}{1 - x} + \frac{(3 \times x)}{(1 - x)^{2}} \\ &= \frac{1}{1 - x} + \frac{3x}{(1 - x)^2}\\ &= \frac{1
- Question 3 and 4, Exercise 4.8
- \text { up to }(n-1) \text { terms }) \\ & =1+\frac{3(3^{n-1}-1)}{3-1} \quad\left(\because S_{n}=\frac{a(r^n-1)}{r-1}\right) \\ & =1+\frac{3^{n}-3}{2} \\ & =\frac{2+3^{n}-3}{2} \\ \\ & =\frac{3^{n}-1}{2}. \end{align*} Thus, the kth term of
- Question 5 & 6, Exercise 4.6
- he indicated term of the harmonic progression. $\dfrac{1}{27}, \dfrac{1}{20}, \dfrac{1}{13}, \ldots \quad$ nth term. ** Solution. ** \begin{align*} &\frac{1}{27}, \frac{1}{20}, \frac{1}{13}, \ldots \quad
- Question 9 and 10, Exercise 4.8
- estion 9===== Evaluate the sum of the series: $$\frac{1}{1 \cdot 3}+\frac{1}{2 \cdot 5}+\frac{1}{3 \cdot 7}+\ldots \ldots \text{ up to } \infty$$ ** Solution. ** Do yourself... Evaluate the sum of the series: $\sum_{k=3}^{n} \dfrac{1}{(k+1)(k+2)}$ ** Solution. ** Consider \begin
- Question 12, Exercise 4.6
- =====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 $$\dfrac{1}{3},H_1, H_2, H_3, H_4, \dfrac{1}{11} \text{ are in H.P.}
- Question 5 and 6, Exercise 4.1
- 10th term, $a_{10}$ and the 15 th term: $a_{n}=\frac{n^{2}-1}{n^{2}+1}$ ** Solution. ** Given $$a_n = \frac{n^2 - 1}{n^2 + 1}.$$ Then \begin{align*} a_1 &= \frac{1^2 - 1}{1^2 + 1} = \frac{1 - 1}{1 + 1} = \frac{0}{2} = 0\\ a_2 &= \frac{2^2 - 1}{2^2 + 1} = \frac{4 -