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- Question 20, 21 and 22, Exercise 4.3
- .\\ As \begin{align} &S_n=\frac{n}{2}[a_1+a_n]\\ \implies & 876=\frac{n}{2}[7+139]\\ \implies & 1752=146n\\ \implies & n=\frac{1752}{146}=12. \end{align} Also we have \begin{align} &a_n=a_1+(n-1)d\\ \implies & 139=7+(12-1)d\\ \implies & 139-7=11d\\ \implies
- Question 17, 18 and 19, Exercise 4.3
- $.\\ We have \begin{align} & a_n=a_1+(n-1)d \\ \implies & 96=6+(n-1)(6) \\ \implies & 96=6+6n-6 \\ \implies & 6n=96 \\ \implies & n = 24. \end{align} Now \begin{align} S_n&=\frac{n}{2}[a_1+a_n] \\ \implies S_{24}&
- Question 20 and 21, Exercise 4.4
- 1}. $$ This gives \begin{align*} &a_5=a_1 r^4 \\ \implies & 48=3r^4 \\ \implies & r^4 = 16 \\ \implies & r^4 = 2^4 \\ \implies & r = 2. \end{align*} Thus \begin{align*} & a_2=a_1 r= (3)(2) = 6 \\ & a_3=a_1 r^2
- Question 5 and 6, Exercise 4.2
- +(n-1)d$$ Given \begin{align*} & a_{17} = -40 \\ \implies &a_1 + 16d = -40 \quad \cdots (1) \end{align*} Also \begin{align*} &a_{28}=-73\\ \implies &a_1 + 27d = -73 \quad \cdots (2) \end{align*} No... s_{+}40 \\ \hline & 11d &= -33\\ \end{array}&\\ \implies \boxed{d = -3} \quad \\ \end{align*} Putting the ... d$ in (1) \begin{align*} & a_1 +16(-3) = -40 \\ \implies & a_1 = -40+48 \\ \implies & \boxed{a_1=8} \end{a
- Question 15 and 16, Exercise 4.3
- =?$. We have \begin{align} & a_n=a_1+(n-1)d \\ \implies & 15=91+(n-1)(-4) \\ \implies & 15=91-4n+4 \\ \implies & 4n=95-15 \\ \implies & 4n = 80\\ \implies & n = 20. \end{align} Now \begin{align} S_n&=\frac{n}{2}[a_1+
- Question 14 and 15, Exercise 4.2
- \begin{align*} &\text{A.M.} = \frac{a + b}{2} \\ \implies & 10 = \frac{b + 20}{2} \\ \implies & 20 = b + 20 \\ \implies & b = 20 - 20 \\ \implies & b = 0 \end{align*} Hence $b = 0$. GOOD <callout type="warning" icon="true"> T
- Question 22 and 23, Exercise 4.4
- \\ This gives\\ \begin{align*} a_6 &= a_1 r^5 \\ \implies \frac{1}{4} &= 8 \cdot r^5 \\ \implies r^5 &= \frac{1}{4 \cdot 8} \\ \implies r^5 &= \frac{1}{32} \\ \implies r &= \left(\frac{1}{32}\right)^{\frac{1}{5}} \\ \implies r &= \frac{1}{2}.
- Question 11, 12 and 13, Exercise 4.5
- in{align*} & 244=\frac{a_1(1-(-3)^5)}{1-(-3)} \\ \implies & 244=\frac{a_1(1+243)}{4} \\ \implies & 976=244a_1\\ \implies & a_1=4.\\ \end{align*} Hence $a_1=4$. =====Question 12===== Find $a_{1}$ for ... 4)}{-1} \\ &= \frac{a_1(-63)}{-1} \\ &= 63a_1 \\ \implies a_1 &= \frac{32}{63}\\ \implies a_1&= 0.51 \end{a
- Question 7 and 8, Exercise 4.2
- 1)d.$$ This gives \begin{align*} &70=-6+(n-1)4\\ \implies &70=-6+4n-4\\ \implies &70=4n-10\\ \implies &4n=80\\ \implies & n=20 \end{align*} Hence $a_{20}=70$. GOOD =====Question 8===== Which term of the sequ
- Question 24 and 25, Exercise 4.4
- }.$$ This gives \begin{align*} a_5 &= a_1 r^4 \\ \implies 80 &= 5 \cdot r^4 \\ \implies r^4 &= \frac{80}{5} \\ \implies r^4 &= 16 \\ \implies r &= 2. \end{align*} Thus, we can find the missing terms: \begin{align*} a_2 &= a_1
- Question 7 and 8, Exercise 4.8
- +(3k-2)B \ldots (2) \end{align*} Put $3k-2=0$ $\implies k=\dfrac{2}{3}$ in (2), we have \begin{align*} &1 = \left(3\times\frac{2}{3}+1 \right)A+0 \\ \implies &A = \frac{1}{3}. \end{align*} Now put $3k+1=0$ $\implies k=-\dfrac{1}{3}$ in (2), we have \begin{align*} &... 0+\left(3\left(-\frac{1}{3}\right)-2 \right)B \\ \implies &B = -\frac{1}{3}. \end{align*} Using value of $
- Question 9 and 10, Exercise 4.3
- n \begin{align} S_n&=\frac{n}{2}[2a_1+(n-1)d] \\ \implies S_{50}&=\frac{50}{2}[2(1)+(50-1)(2)]\\ &=25\times... $.\\ We have \begin{align} & a_n=a_1+(n-1)d \\ \implies & 520=16+(n-1)(4) \\ \implies & 520=16+4n-4 \\ \implies & 4n=520-16+4 \\ \implies & 4n = 508\\ \implies & n = 127. \end{align} Let $S_n
- Question 9 and 10, Exercise 4.5
- gin{align*} &\frac{a_6}{a_3}=\frac{3/32}{3/4} \\ \implies & \frac{a_1 r^5}{a_1 r^2}=\frac{4}{32}\\ \implies & r^3=\frac{1}{8} \\ \implies & r^3=\left(\frac{1}{2} \right)^3\\ \implies & r=\frac{1}{2}. \end{align*} Now \begin{align*} & a_3 = a_1
- Question 12, Exercise 4.6
- +(n-1)d.$$ Thus \begin{align*}\\ &11 =3+(6-1)d\\ \implies &5d = 11-3\\ \implies &5d=8\\ \implies & d=\frac{8}{5}\end{align*} Now \begin{align*} &\frac{1}{H_1}=a_1+d=3+\frac{8}{5}=\frac{23}{5}\\ \implies &H_1=\frac{5}{23} \end{align*} \begin{align*} &\f
- Question 11 and 12, Exercise 4.8
- k=0$ in (2), we have \begin{align*} &1=2A + 0 \\ \implies & A = \frac{1}{2}. \end{align*} Put $k+2=0 \implies k=-2$ in (2), we have \begin{align*} &1=0-2B\\ \implies &B = -\frac{1}{2}. \end{align*} Using the values o... +(3k-2)B \ldots (2) \end{align*} Put $3k-2=0$ $\implies k=\dfrac{2}{3}$ in (2), we have \begin{align*} &1