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- 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 $$\dfrac{1}{3},H_1, H_2, H_3, H_4, \dfrac{1}{11} \text{ are in H.P.}$
- Question 7 and 8, Exercise 4.2
- =====Question 8===== Which term of the sequence $\dfrac{5}{2}, \dfrac{3}{2}, \dfrac{1}{2}, \ldots$ is $-\dfrac{105}{2}$? ** Solution. ** Given: $\dfrac{5}{2}, \dfrac{3}{2}, \dfrac{1}{2},
- Question 8 and 9, Exercise 4.4
- Given sequence is geometric with $a_1=90$ and $r=\dfrac{30}{90}=\dfrac{1}{3}$.\\ General term of the geometric series is given as $$a_{n}=a_{1} r^{n-1}.$$ Thus \begin{align*} & a_{4}=a_{1} r^3=(90)\left(\dfrac{1}{3} \right)^3=90 \times\dfrac{1}{27}=\dfrac{10}{3}\\ & a_{5}=a_{1} r^3=(90)\left(\dfrac{1}{4} \right)^
- Question 11, Exercise 4.6
- stan. =====Question 11===== Find H.M. between $\dfrac{2}{3}$ and $\dfrac{4}{7}$. ** Solution. ** Here $a=\dfrac{2}{3}$ and $b=\dfrac{4}{7}$, therefore \begin{align*} \text{H.M.}&=\frac{2ab}{a+b} \\ &=\frac{2\times\f
- Question 2, Exercise 4.2
- e next three terms of each arithmetic sequence. $\dfrac{1}{2}, \dfrac{3}{2}, \dfrac{5}{2}, \ldots$ ** Solution. ** The given sequence is $$\frac{1}{2}, \frac{3}{2}, \frac{... Thus, the next three terms of the sequence are $ \dfrac{7}{2}, \dfrac{9}{2}, \dfrac{11}{2} $. =====Quest
- Question 7 and 8, Exercise 4.8
- \ldots (2) \end{align*} Put $3k-2=0$ $\implies k=\dfrac{2}{3}$ in (2), we have \begin{align*} &1 = \left(... {3}. \end{align*} Now put $3k+1=0$ $\implies k=-\dfrac{1}{3}$ in (2), we have \begin{align*} &1 = 0+\lef... } \end{align*} Hence the sum of given series is $\dfrac{n}{3n+1}$. GOOD =====Question 8===== Find the ... \ldots (2) \end{align*} Put $5k-4=0$ $\implies k=\dfrac{4}{5}$ in (2), we have \begin{align*} &1 = \left(
- Question 9 and 10, Exercise 4.2
- Islamabad, Pakistan. =====Question 9===== If $\dfrac{1}{a}, b, \dfrac{1}{c}$ are in A.P. Show that the common difference is $\dfrac{a-c}{2 a c}$. ** Solution. ** Since $\dfrac{1}{a}, b, \dfrac{1}{c}$ are in A.P.\\ \begin{align*} d&=b-
- Question 26 and 27, Exercise 4.4
- the rebounds of the ball, then $$a_1 = 16\times \dfrac{1}{4} = 4\,\, ft.$$ Thus the sequence is geometric with $r=\dfrac{1}{4}$. We have to find $a_6$, where general ter... us \begin{align*} a_{6}&=a_{1} r^5 \\ &=(4)\left(\dfrac{1}{4} \right)^5 \\ & = \dfrac{1}{256} \end{align*} Hence, the ball will rebound $\dfrac{1}{256}\,\, ft$
- Question 5 & 6, Exercise 4.6
- the 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}{... . \end{align*} Hence, the $n$th term in H.P. is $\dfrac{1}{\frac{n+3}{2}}$ or $\dfrac{2}{n+3}$. m(
- Question 27 and 28, Exercise 4.7
- A.P. with $a=5$ and $d=7-5=2$. The numbers $1, \dfrac{1}{3}, \dfrac{1}{9}, \dfrac{1}{27}, \ldots$ are in G.P. with first term as 1 and $r=\dfrac{1/3}{1}=\dfrac{1}{3}$. The sum of infinite arith
- Question 14, 15 and 16, Exercise 4.7
- }{2} \\ & = \frac{n^2 + 3n}{2} \\ \sum T_{n} & = \dfrac{n(n+3)}{2} \end{align*} Thus, the sum of the series is $\sum_{n=1}^{\infty} T_{n}= \dfrac{n(n+3)}{2}$. GOOD =====Question 15===== Find the... of the series is $\sum\limits_{k=1}^{n} T_{k} = \dfrac{n}{6}(2n^2+9n+7)$. GOOD =====Question 16===== Fi
- Question 9 and 10, Exercise 4.8
- Evaluate the sum of the series: $\sum_{k=3}^{n} \dfrac{1}{(k+1)(k+2)}$ ** Solution. ** Consider \begin... \end{align*} Now \begin{align*} &\sum_{k=3}^{n} \dfrac{1}{(k+1)(k+2)} = \sum_{k=3}^n T_k \\ &= \sum_{k=... 4(n+2)} \\ \end{align*} Hence, the required sum $\dfrac{n-2}{4(n+2)}$. GOOD ====Go to ==== <text
- Question 3 and 4, Exercise 4.2
- end{align*} Hence the first four terms are $19, \dfrac{33}{2}, 14, \dfrac{23}{2}$. GOOD ====Go to ==== <text align="left"><btn type="primary">[[math-11-nbf:so
- Question 11 and 12, Exercise 4.8
- \ldots (2) \end{align*} Put $3k-2=0$ $\implies k=\dfrac{2}{3}$ in (2), we have \begin{align*} &1 = \left(... {3}. \end{align*} Now put $3k+1=0$ $\implies k=-\dfrac{1}{3}$ in (2), we have \begin{align*} &1 = 0+\lef
- Question 13, Exercise 4.2
- \ &= \frac{7y}{2} + 4 \end{align*} Hence A.M. = $\dfrac{7y}{2} + 4$. GOOD ====Go to ==== <text align="l