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- Question 7 & 8 Exercise 4.3
- find the sum of $\mathrm{n}$ terms of each, and\\ then adding the $n$ terms sum of each one will give us... .\\ For $$1+7+13+\ldots$$ here $$a_1=1, d=7-1=6$$ then sum of the $n$ terms\\ \begin{align}S_n&=\dfrac{n... \\ Now for $3+9+15+\ldots$, with $a_1=3, d=9-3=6$ then sum of the $\mathrm{n}$ terms, let denote\\ \begi... n}S_n^{\prime}&=\dfrac{n}{2}[2 a_1+(n-1) d] \text{then}\\ \cdot S_n^{\prime}&=\dfrac{n}{2}[2.3+(n-1) 6]\
- Question 12 & 13 Exercise 4.2
- $a_1$ represents salary of worker at first year. Then $$a_1=3500.$$ Increase in salary in each year $=d... $a=12, b=18$.\\ Let say $A$ be arithmetic means. Then \\ \begin{align}A&=\dfrac{a+b}{2}\\&=\dfrac{12+18... }, b=\dfrac{1}{4}$,\\ Let $A$ be arithmetic mean. Then\\ \begin{align}A&=\dfrac{a+b}{2}\\&=\dfrac{\dfrac... re $a=-6, b=-216$.\\ Let $A$ be arithmetic mean. Then\\ $$A=\dfrac{a+b}{2}=\dfrac{-6-216}{2}=-111$$ GOO
- Question 1 and 2 Exercise 4.1
- n=(-1)^{n-1} 2^{n+1}$$ For first term, put $n=1$, then $$a_1=(-1)^{1-1} 2^{1+1}=2^2=4$$ For second term, put $n=2$, then $$a_2=(-1)^{2-1} 2^{2+1}=-2^3=-8$$ For third term, put $n=3$, then $$a_3=(-1)^{3-1} 2^{3-1}=2^4=16$$ For fourth term, put $n=4$, then $$a_4=(-1)^{4 \cdots 1} 2^{4+1}=-2^5=-32 \text {.
- Question 1 Exercise 4.3
- t term and $d$ be common difference of given A.P. Then \begin{align}&a_1=9 \\ &d=7-9=-2 \\ &n=20. \end{... _n$ represents the sum of first $n$ terms of A.P. Then \begin{align} S_n&=\dfrac{n}{2}[a_1+a_n], \\ \imp... t term and $d$ be common difference of given A.P. Then \begin{align}&a_1=3 \\ &d=\dfrac{8}{3}-3=-\dfrac... _n$ represents the sum of first $n$ terms of A.P. Then \begin{align} S_n&=\dfrac{n}{2}[a_1+a_n], \\ \imp
- Question 2 & 3 Exercise 4.4
- { or } r= \pm 3 .\end{align} Putting this in (i), then\\ $$a_1(9)=27 \quad \text{then} \quad a_1=3$$ $$\text{Hence}\quad a_1=3, r= \pm 3$$ =====Question 3==... =-\dfrac{1}{\sqrt{2}}.\end{align} Putting in (i), then\\ \begin{align}\Rightarrow a_1(-\dfrac{1}{\sqrt{2... xt { putting } n&=7, \text { and } a_1, r \text { then we get } \\ a_7&=a_1 r^6=(-2 \sqrt{2})(-\dfrac{1}
- Question 8 Exercise 4.4
- 71$ ====Solution==== Here $a=3.14$ and $b=2.71$\\ then $$G= \pm \sqrt{(3.14)(2.71)}= \pm 2.94$$ Thus $$G... $-216$ ====Solution==== Here $a=-6$ and $b=-216$ then\\ \begin{align}G&= \pm \sqrt{(-6)(-216)}= \pm \sq... $x-y$ ====Solution==== Here $a=x+y$ and $b=x-y$\\ then $$G= \pm \sqrt{(x+y)(x-y)}= \pm \sqrt{x^2-y^2}$$ ... sqrt{2}+3$ and \begin{align}b&=\sqrt{2}-3 \text { then } \\ G&= \pm \sqrt{(\sqrt{2}+3)(\sqrt{2}-3)} \\ \
- Question 8 Exercise 4.2
- , \dfrac{c+a-b}{b}, \dfrac{a+b-c}{c}$ are in A.P, then prove $\dfrac{1}{a}, \dfrac{1}{b}, \dfrac{1}{c}$ ... dfrac{a+b+c}{2} \\ \Rightarrow a+b+c&=2 S\\ \text{then} \Rightarrow a+b-c&=2(S-c) \text {, }\\ a+c-b&=2(S-b), \quad\text{and}\\ b+c-a&=2(S-a)\end{align} then (1), becomes\\ \begin{align}\dfrac{2(S-b)}{b}-\df
- Question 15 Exercise 4.2
- presents the arithmetic mean between $a$ and $b$, then $$ A=\dfrac{a+b}{2}. --- (1) $$ Also, we have gi... lies & b^n(a-b)=a^n(a-b)\end{align} If $a\neq b$, then we have \begin{align} &b^n =a^n \\ \implies &\... }\right)^0\\ \implies &n=0.\end{align} If $a=b$, then from (3), we have \begin{align}&\dfrac{a+a}{2}=\d
- Question 2 Exercise 4.3
- {21}=20+40=60.\end{align} Also $a_{21}=a_1+20 d$, then\\ \begin{align}&20d=60-(-40)=100 \\ \implies &d=\... uadratic equation with $a=4, b=-11$ and $c=-225$, then \begin{align}n&=\dfrac{-b \pm \sqrt{b^2-4 a c}}{2... $d$. We know that $$S_n=\dfrac{n}{2}[a_1+a_n],$$ then we have \begin{align}&S_{15}=\dfrac{15}{2}[a_1+a_
- Question 3 & 4 Exercise 4.3
- s suppose the three numbers are $a-d, a, a+d$\\. then by first condition their sum is equal to $36$\\ \... w d^2&=16= \pm 4\end{align} When $a=12$ and $d=4$ then the numbers are\\ \begin{align} a-d&=12-4=8, a=12... a+d&=12+4=16 .\end{align} When $a=12$ and $d=-4$ then the numbers are\\ \begin{align}a-d&=12-(-4)=16, a
- Question 5 & 6 Exercise 4.3
- t { or } d= \pm 1\end{align} When $a=5$ and $d=1$ then the numbers are\\ \begin{align} a-3d&=5-3=2, \\ a... \\ a+3 d&=5+3=8.\end{align} When $a=5$ and $d=-1$ then the numbers are\\ \begin{align}a-3 d&=5-3(-1)=8, ... x_7+x_{10}&=-6\end{align}\\ When $a=5$ and $d=-1$ then the numbers are\\ \begin{align}\therefore x_1+(x_
- Question 1 Exercise 4.5
- 6}{3}=2$ and $a_n=3.2^9$.\\ We first find $n$ and then the sum of series.\\ We know that $$a_n=a_1 r^{n-... nd $$a_n=\dfrac{1}{16}$$.\\ We first find $n$ and then the sum of series. We know that\\ \begin{align}a_... =2$$ and $$a_n=2^{10}$$.\\ We first find $n$ and then the sum of series. We know that\\ \begin{align}a_
- Question 2 Exercise 4.5
- d a_n=64$. ====Solution==== We first find $n$ and then $S_n$\\ We know $a_n=a_1 r^{n-1}$, therefore\\ \b... _7&=\dfrac{a_1[r^{\prime \prime}-1]}{r-1}\\ \text{then}\\ S_7&=\dfrac{1[(-2)^7-1]}{-2-1}\\ \Rightarrow S... =1, n=9$ ====Solution==== We first find $a_1$ and then $S_9$.\\ We know $$a_9=a_1 r^8$$\\ therefore we
- Question 5 & 6 Exercise 4.5
- == We know that $$S_n=\dfrac{a_1(r^n-1)}{r-1}$$\\ then\\ $$S_{10}=\dfrac{a_1(r^{10}-1)}{r-1} \quad \text... \ Replacing $n$ by $2 n$, and $3 n$ in the above, then we get\\ \begin{align}S_{2 n}&=\dfrac{a_1(r^{2 n}... ^n-1)}{r-1}]^2...(1)\end{align} Now taking R.H.S, then\\ \begin{align}(S_n-S_{2 n})^2&=[\dfrac{a_1(r^n-1
- Question 7 & 8 Exercise 4.5
- t the three number are $\dfrac{a}{r}, a . a r$ \\ then by the given conditions $$\dfrac{a}{r}+a+a r=38 \... 3}{2}\end{align} When $a=12$ and $r=\dfrac{2}{3}$ then\\ \begin{align}\dfrac{a}{r}&=\dfrac{12}{\dfrac{2}... {3}=8\end{align} When $a=12$ and $r=\dfrac{3}{2}$ then\\ \begin{align}\dfrac{a}{r}&=\dfrac{12}{\dfrac{3}