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- Question 6 Exercise 4.1
- r, \text{ where } r=0,1,2,3,\ldots.$$ For $r=0$ \begin{align}&P_{0+1}=\dfrac{5-0}{0+1} P_0\\ \implies &P_1=5.\end{align} For $r=1$ \begin{align}&P_{1+1}=\dfrac{5-1}{1+1} P_1\\ \implies &P_2=2\cdot 5=10.\end{align} For $r=2$ \begin{align}&P_{2+1}=\dfrac{5-2}{2+1} P_2\\ \implies &P_3=1\cdot 10=10\end{align} For $r=3$ \begin{align}&P_{3+1}=\dfrac{5-3}{3+1} P_3\\ \implies
- Question 14 Exercise 4.2
- Question 14(i)===== Insert three arithmetic means between 6 and 41. GOOD ====Solution==== Let $A_1, A_2, A_3$ be three arithmetic means between 6 and 41. Then $6, A_1, A_2, A_3, 41$ are in A.P. We have $$a_1=6 \text{ and } a_6=41.$$ Now \begin{align}& a_5=11\\ \Rightarrow &a_1+4 d=41 \\ \R
- Question 12 & 13 Exercise 4.2
- f A.P and we have to find $a_{21}$. As, we have \begin{align} a_{21}&=a_1+20d\\ &=3500+20(750) \\ &=1... =====Question 13(i)===== Find the arithmetic mean between $12$ and $18$. GOOD ====Solution==== Here $a=12, b=18$.\\ Let say $A$ be arithmetic means. Then \\ \begin{align}A&=\dfrac{a+b}{2}\\&=\dfrac{12+18}{2}\\&=\dfrac{30}{2}=15.\end
- Question 1 Exercise 4.5
- of series.\\ We know that $$a_n=a_1 r^{n-1}$$,\\ \begin{align}3.2^9&=3(2)^{n-1} \text { or }(2)^{n-1}=... w }\quad S_n&=\dfrac{a_1(r^n-1)}{r-1},\end{align} becomes in the given case\\ \begin{align}S_{10}&=\dfrac{3[2^{10}-1]}{2-1} \\ \Rightarrow \quad S_{10}&=3... d $n$ and then the sum of series. We know that\\ \begin{align}a_n&=a_1 r^{n-1} \text {, }\\ \therefore
- Question 16 Exercise 4.2
- ====Question 16===== Insert five arithmetic means between $5$ and $8$ and show that their sum is five times the arithmetic mean between $5$ and $8$. GOOD ====Solution==== Let $A_1, A_2, A_3, A_4, A_5$ be five arithmetic means between $5$ and $8$. Then $5, A_1, A_2, A_3, A_4, A_5, 8$ are in A.P, where $$a
- Question 5 & 6 Exercise 4.3
- d,$$ $Condition-1$\\ Their sum is $20$ , thus\\ \begin{align}a-3 d+a-d+a+d+a+3 d&=20 \\ \Rightarrow 4... \\ The sum of their square is $120$, therefore\\ \begin{align}(a-3 d)^2+(a-d)^2+(a+d)^2+(a+2 d)^2&=120... d^2&=120 \\ \Rightarrow 20 d^2&=120-4 \cdot(5)^2 \because a=5 \\ \Rightarrow 20 d^2&=20 \\ \Rightarrow... ign} When $a=5$ and $d=1$ then the numbers are\\ \begin{align} a-3d&=5-3=2, \\ a-d&=5-1=4, \\ a+d&=5+1
- Question 3 and 4 Exercise 4.1
- sequence to pick the pattern of the sequence as: \begin{align} &(-1)^2 \cdot 2 \cdot 1, (-1)^3 \cdot 2... sequence to pick the pattern of the sequence as: \begin{align}(-1)^2,(-1)^3,(-1)^4,(-1)^5, \ldots, (-1... on==== Given $$a_1=3, a_{n+1}=5-a_n.$$ For $n=1$ \begin{align}a_{1+1}&=5-a_1\\ \Rightarrow a_2&=5-3=2\end{align} For $n=2$ \begin{align}a_{2+1}&=5-a_2\\ \Rightarrow a_3&=5-2=3\
- Question 11 Exercise 4.4
- that the prodect of $\mathrm{n}$ geometric means between $a$ and $b$ is equal to the $nth$ power for the single geometric mean between them. ====Solution==== Let $G_1, G_2, G_9, \ldots, G_n$ be the $n$ geometric means between $a$ and $b$,\\ then $a, G_1, G_2, G_3, \ldots, G_n, b$ is a geometric
- Question 1 Exercise 4.3
- h term; 20 terms. GOOD ====Solution==== Let $a_1$ be first term and $d$ be common difference of given A.P. Then \begin{align}&a_1=9 \\ &d=7-9=-2 \\ &n=20. \end{align} We know that \begin{align}&a_n=a_1+(n-1)d \\ \implies &a_20=9+(20-
- Question 9 & 10 Exercise 4.3
- estion 9===== Find the sum 'of all multiples of 9 between 300 and 700. ====Solution==== All the multiples of 9 between 300 and 700 are:\\ $$306,315,324,333, \ldots... ext { and } a_n=693 .$$\\ Let the number of terms be $n$. Then\\ \begin{align}a_n&=a_1+(n-1) d \text { becomes } \\ \Rightarrow a_1+(n-1) d&=693 \\ \Right
- Question 9 Exercise 4.4
- ===Question 9(i)===== Insert five geometric means between $3 \dfrac{5}{9}=\dfrac{32}{9}\quad$ and $\qu... ==Solution==== Let $G_1, G_2, G_3, G_4$ and $G_5$ be the five geometric means between $\dfrac{32}{9}$ and $\dfrac{81}{2}$,\\ then $\dfrac{32}{9}, G_1, G_2... ac{81}{2}$ and $a_1=\dfrac{32}{9}$.\\ Therefore, \begin{align}a_1 r^6&=\dfrac{81}{2} \\ \Rightarrow \d
- Question 2 Exercise 4.5
- n $S_n$\\ We know $a_n=a_1 r^{n-1}$, therefore\\ \begin{align}64&=(-2)^{n-1}\\ \Rightarrow(-2)^{n-1}&=... \ We know $$a_9=a_1 r^8$$\\ therefore we have\\ \begin{align}1&=a_1(\dfrac{1}{2})^{9-1}\\ &=a_1 \dfra... \dfrac{a_1[1-r^{\prime \prime}]}{1-r},\end{align} becomes in the given case\\ \begin{align}\Rightarrow S_9&=\dfrac{2^8[1-(\dfrac{1}{2})^9]}{1-\dfrac{1}{2}
- Question 4 Exercise 4.5
- e $$0 . \overline{8}=0.888888 \ldots$$\\ That can be written in the form\\ \begin{align}0 . \overline{8}&=0.8+0.08+0.008 \div 0.0008+ \ldots\\ \text { or ... ction $1 . \overline{63}$ ====Solution==== Since \begin{align}1 . \overline{63}&=1+0.63+0.0063+0.00006... 1$.\\ Therefore the sum exists and is given by\\ \begin{align}S_{\infty}&=\dfrac{a_1}{1-r}\\ &=\dfrac{
- Question 2 Exercise 4.3
- 3)=50.$$ Also $$S_n=\dfrac{n}{2}[a_1+a_n]$$ Thus \begin{align}S_{17}&=\dfrac{17}{2}(a_1+a_17) \\ &=\df... $n=21$ and we have to find $a_{21}$ and $d$. As \begin{align}&S_{21}=\dfrac{21}{2}(a_1+a_{21}) \\ \im... 40=60.\end{align} Also $a_{21}=a_1+20 d$, then\\ \begin{align}&20d=60-(-40)=100 \\ \implies &d=\dfrac{... S_n=\dfrac{n}{2}[2 a_1+(n-1) d].$$ Thus, we have \begin{align} & 225=\dfrac{n}{2}[2 \cdot(-7)+(n-1) \c
- Question 3 & 4 Exercise 4.3
- , d=5$ and $a_n=350$\\ To find $n$, we know that \begin{align}a_n&=a_1+(n-1) d\end{align} in the given case it becomes,\\ \begin{align} 350&=25+(n-1)(5) \\ \Rightarrow 5 n-5+25&=350 \\ \Rightarrow 5 n&=350-20=330 \\... sum } \\ S_n&=\dfrac{n}{2}(a_1+a_n), \text { that becomes } \\ S_{66}&=\dfrac{66}{2}(25+350) \\ \Right