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- Question 2 Exercise 4.3
- on==== Given: $a_1=-40$ and $S_{21}=210$.\\ So we have $n=21$ and we have to find $a_{21}$ and $d$. As \begin{align}&S_{21}=\dfrac{21}{2}(a_1+a_{21}) \\ \impl... ===Solution==== Given: $a_1=-7, d=8, S_n=225$, we have to find $n$ and $a_n$. We know that $$S_n=\dfrac{n}{2}[2 a_1+(n-1) d].$$ Thus, we have \begin{align} & 225=\dfrac{n}{2}[2 \cdot(-7)+(n-1
- Question 15 Exercise 4.2
- $, then $$ A=\dfrac{a+b}{2}. --- (1) $$ Also, we have given $$ A=\dfrac{a^{n+1}+b^{n+1}}{a^n+b^n}. --- (2) $$ Comparing (1) and (2), we have \begin{align}&\dfrac{a+b}{2}=\dfrac{a^{n+1}+b^{n+... ^n(a-b)=a^n(a-b)\end{align} If $a\neq b$, then we have \begin{align} &b^n =a^n \\ \implies &\dfrac{b^... lies &n=0.\end{align} If $a=b$, then from (3), we have \begin{align}&\dfrac{a+a}{2}=\dfrac{a^{n+1}+a^{n+
- Question 1 Exercise 4.4
- 1 r^3, a_1 r^4, \ldots$, so for $a_1=5 ; r=3$, we have \begin{align}&5,5.3,5.3^2, 5.3^3, 5.3^4, \ldots\\... \ldots$,\\ so for $$a_1=8 ; r=-\dfrac{1}{2}$$ we have\\ \begin{align}&8,8(-\dfrac{1}{2}), 8(-\dfrac{1}{... for $$a_1=-\dfrac{9}{16} ; r=-\dfrac{2}{3}$$ we have\\ \begin{align}&-\dfrac{9}{16} ,-\dfrac{9}{16}(-\... \\ so for $$a_1=\dfrac{x}{y} r=-\dfrac{y}{x}$$ we have,\\ \begin{align}&\dfrac{x}{y}, \dfrac{x}{y} \cdot
- Question 12 & 13 Exercise 4.2
- year $=d=750$. The given problem is of A.P and we have to find $a_{21}$. As, we have \begin{align} a_{21}&=a_1+20d\\ &=3500+20(750) \\ &=18500. \end{align} H
- Question 17 Exercise 4.2
- ies n&=\dfrac{27}{d}-1 ---(i)\end{align} Also, we have given $A_3:A_7=7:13$, where $$A_3=a_4=a_1+3d=5+3d$$ and $$A_7=a_8=a_1+7d=5+7d.$$ Thus we have \begin{align}&\dfrac{A_3}{A_7}=\dfrac{7}{13} \\
- Question 10 Exercise 4.4
- $$(a+b)-2 \sqrt{a b}=36 \text {. }$$ From (i), we have $a=b+48$ pulting in (ii), then we have\\ \begin{align}(b+48+b)-2 \sqrt{b(b+48)}&=36 \\ \Rightarrow(2 b+
- Question 3 Exercise 4.5
- 2$ and $a_3=1$\\ ====Solution==== We first try to have find $a_1$ and $r$.\\ We know that $$a_n=a_1 r^{n... {2} \text {, }\end{align} putting this in (i), we have\\ \begin{align}\dfrac{a_1}{2}&=2\\ \Rightarrow a_
- Question 6 Exercise 4.1
- definition. GOOD ====Solution==== For $n=5$, we have Pascal sequence as follows: $$P_0=1, P_{r+1}=\dfr
- Question 7 Exercise 4.2
- \implies & a_1+4d=3 --- (1) \end{align} Also, we have given \begin{align} &a_6-a_4=\dfrac{2}{3} \\ \imp
- Question 10 Exercise 4.2
- ear=$d=-500$. The given problem is of A.P and we have to find population in 1970, that is, $a_{11}$. A
- Question 11 Exercise 4.2
- s the given problem is of A.P with $a_n=5400$, we have to find $n$, which represent the number of hours
- Question 14 Exercise 4.2
- nd 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}&
- Question 16 Exercise 4.2
- in A.P, where $$a_1=5 \text{ and } a_7=8.$$ As we have \begin{align}&a_7=a+6d\\ \implies &8=5+6d\\ \impl
- Question 3 & 4 Exercise 4.3
- dition, the sum of their cubes is $6336$,\\ so we have\\ \begin{align}(a-d)^3+a^3+(a+d)^3&=6336 \\ \Righ
- Question 5 & 6 Exercise 4.3
- arrow d&=-1.\end{align} Putting $d=-1$ in (i), we have\\ \begin{align}3 x_1+15(-1)&=-6 \\ \Rightarrow 3