Search
You can find the results of your search below.
Fulltext results:
- Question 6 Exercise 4.1
- }&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 &P_4=\dfrac{10}{2}=5.\end{align} For $r=4$ \begin{align}&P_{4+1}=\dfrac{5-4
- Question 3 and 4 Exercise 4.1
- ot 2 \cdot 3, (-1)^{4+1} \cdot 2 \cdot 4, \ldots \end{align} Hence the general term of the sequence is ... (-1)^3,(-1)^4,(-1)^5, \ldots, (-1)^{n+1}, \ldots \end{align} Hence the general term of the sequence is ... in{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\end{align} For $n=3$ \begin{align}a_{3+1}&=5-a_3\\ \R
- Question 5 & 6 Exercise 4.3
- d&=20 \\ \Rightarrow 4 a&=20\\ \Rightarrow a&=5 .\end{align} $Condition-2$\\ The sum of their square is... 2&=20 \\ \Rightarrow d^2&=1 \text { or } d= \pm 1\end{align} When $a=5$ and $d=1$ then the numbers are\... 1=4, \\ a+d&=5+1=6 \text { and } \\ a+3 d&=5+3=8.\end{align} When $a=5$ and $d=-1$ then the numbers are... text { and } \\ a+3 d&=5-3=2\\ & 2,4,6,8; 8,6,4,2\end{align} =====Question 6===== If $x_1, x_2, x_3, \
- Question 1 Exercise 4.5
- \text {. Now }\quad S_n&=\dfrac{a_1(r^n-1)}{r-1},\end{align} becomes in the given case\\ \begin{align}S... 1]}{2-1} \\ \Rightarrow \quad S_{10}&=3(2^{10}-1)\end{align}\\ is the required sum. =====Question 1(ii... ad n=8.\\ S_n&=\dfrac{a_1(r^n-1)}{r-1} \text {, }\end{align} becomes in the given case\\ \begin{align}S... Rightarrow S_3&=\dfrac{2^8-1}{16}=\dfrac{255}{16}\end{align} is the required sum. =====Question 1(iii)
- Question 4 Exercise 4.5
- ^2(0.8)+\ldots \ldots \ldots \ldots .(\mathrm{i})\end{align}\\ It is geometric series with $$a_1=0.8, \... (0.63)-(0.01)^2 0.63+\ldots \ldots \text { (i) }\end{align} The serics in braces is infinite gcometric... \ldots \ldots \ldots \ldots . . . \text { (ii) }\end{align} Putting (ii) in (i), we get\\ $$1.63=1+\df... }&=2+[0.15+(0.01) 0.15+ .(0.01)^2 0.15+\ldots ..]\end{align}\\ The sequence in bracket is infinite geom
- Question 14 Exercise 4.2
- \Rightarrow &d=\dfrac{41-6}{4}\\ &=\dfrac{35}{4}.\end{align} Now \begin{align} A_1&=a+d=6+\dfrac{35}{4} \\ &=\dfrac{59}{4}=14\dfrac{3}{4},\end{align} \begin{align} A_2&=a+2 d\\ &=6+2 \cdot \dfrac{35}{4} \\ &=\dfrac{47}{2}=23\dfrac{1}{2}\end{align} and \begin{align}A_3&=a+3 d=6+3 \cdot \dfrac{35}{4}\\ &=\dfrac{129}{4}=32\dfrac{1}{4}.\end{align} Hence three arithmetic means between 6 and
- Question 2 Exercise 4.3
- ac{17}{2}(a_1+a_17) \\ &=\dfrac{17}{2}(2+50)=442.\end{align} Hence $a_{17}=50$ and $S_{17}=442$. GOOD ... \times 210}{21}=20 \\ \implies &a_{21}=20+40=60.\end{align} Also $a_{21}=a_1+20 d$, then\\ \begin{alig... 0d=60-(-40)=100 \\ \implies &d=\dfrac{100}{20}=5 \end{align} Hence $a_{21}=60$, $d=5$ and $n=21$. GOOD ... & 8 n^2-22 n-450=0 \\ \implies & 4 n^2-11 n-225=0\end{align} This is quadratic equation with $a=4, b=-1
- Question 5 Exercise 4.1
- ) \\ \implies \sum_{j=1}^6(2 j-3)&=-1+1+3+5+7+9 .\end{align} =====Question 5(ii)===== Write each of th... \\ \implies \sum_{k=1}^5(-1)^k 2^k& =-1+2-4+8-16.\end{align} =====Question 5(iii)===== Write each of t... }+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\ldots \end{align} or we can simply write \begin{align}\sum_{... rac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+\ldots \end{align} =====Question 5(iv)===== Write each of th
- Question 3 & 4 Exercise 4.3
- d $n$, we know that \begin{align}a_n&=a_1+(n-1) d\end{align} in the given case it becomes,\\ \begin{ali... 2}(25+350) \\ \Rightarrow S_{66}&=33(375)=12375 .\end{align} =====Question 4===== The sum of three num... )&=36 \\ \Rightarrow 3 a&=36 \\ \Rightarrow a&=12\end{align} Now by the second condition, the sum of th... d^2&=6336-5184=1152 \\ \Rightarrow d^2&=16= \pm 4\end{align} When $a=12$ and $d=4$ then the numbers are
- Question 7 & 8 Exercise 4.3
- & (5+11+17+\ldots) \ldots \ldots \ldots . . .(1)\end{align} Now the series each one in parenthesis is ... \Rightarrow S_n&=\dfrac{n(6 n-4)}{2}\\ &=n(3 n-2)\end{align}\\ Now for $3+9+15+\ldots$, with $a_1=3, d=... \Rightarrow S_n^{\prime}&=\dfrac{6 n^2}{2}=3 n^2\end{align}\\ and for $5+11+17+\ldots$ with $a_1=5$ an... \prime}&=\dfrac{n(6 n+4)}{2}=n(3 n+2) \text {. }\end{align} Putting (i),(ii) and (iii) in (1), we get\
- Question 2 & 3 Exercise 4.4
- \begin{align}a_3&=a_1 r^2=27\\ a_5&=a_1 r^4=243.\end{align} Dividing (ii) by (i), we get\\ \begin{alig... }=9 \\ \Rightarrow r^2&=9 \text { or } r= \pm 3 .\end{align} Putting this in (i), then\\ $$a_1(9)=27 \... {the third term is}=a_3&=a_1 r^2=-\sqrt{2}...(ii)\end{align} Dividing (ii) by (i), we get\\ \begin{alig... {\sqrt{2}}\\ \Rightarrow r&=-\dfrac{1}{\sqrt{2}}.\end{align} Putting in (i), then\\ \begin{align}\Right
- Question 2 Exercise 4.5
- row S_7&=\dfrac{-128-1}{-3}\\ s_7&=\dfrac{129}{3}\end{align} =====Question 2(ii)===== Some of the comp... 2^8\\ S_9&=\dfrac{a_1[1-r^{\prime \prime}]}{1-r},\end{align} becomes in the given case\\ \begin{align}\... row \quad S_9&=2^9-1\\ \Rightarrow \quad S_9&=511\end{align} =====Question 2(iii)===== Some of the com... r-1} \\ \text { or } S_n&=\dfrac{a_n r-a_1}{r-1}.\end{align}\\ becomes in the given case\\ \begin{align
- Question 7 Exercise 4.2
- mplies & 2a_1+8d=6\\ \implies & a_1+4d=3 --- (1) \end{align} Also, we have given \begin{align} &a_6-a_4... es & 2d=\dfrac{2}{3}\\ \implies & d=\dfrac{1}{3} \end{align} Using the value of $d$ in (1), we get \beg... )=3\\ \implies &a_1=3-\dfrac{4}{3}=\dfrac{5}{3}. \end{align} Also \begin{align} a_2&=a_1+d\\ &=\dfrac{5}{3}+\dfrac{1}{3}=2\end{align} and \begin{align} a_3&=a_1+2d\\ &=\dfrac{
- Question 12 & 13 Exercise 4.2
- ked. If he received a raise of dollars 750 at the end of each year for 20 years, what was his salary du... gn} a_{21}&=a_1+20d\\ &=3500+20(750) \\ &=18500. \end{align} Hence, the salary of the man during his 21... {a+b}{2}\\&=\dfrac{12+18}{2}\\&=\dfrac{30}{2}=15.\end{align} Hence 15 is A.M between 12 and 18. GOOD =... 2}\\&=\dfrac{\dfrac{4+3}{12}}{2}\\&=\dfrac{7}{24}\end{align} GOOD =====Question 13(iii)===== Find the
- Question 1 Exercise 4.3
- Then \begin{align}&a_1=9 \\ &d=7-9=-2 \\ &n=20. \end{align} We know that \begin{align}&a_n=a_1+(n-1)d \\ \implies &a_20=9+(20-1)(-2)=-29. \end{align} Assume $S_n$ represents the sum of first $... es S_{20}&=\dfrac{20}{2}[9-29] \\ &=10(-20)=-200 \end{align} Hence 20th term is -29 and sum of first 20... =3 \\ &d=\dfrac{8}{3}-3=-\dfrac{1}{3} \\ &n=11. \end{align} We know that $$a_n=a_1+(n-1) d,$$ This giv