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- Question 6 Exercise 4.1
- 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 &P_4=\dfrac{10}{2}=5.\end{align} For $r=4$ \begin
- Question 2 Exercise 4.3
- begin{align}&S_{21}=\dfrac{21}{2}(a_1+a_{21}) \\ \implies &210=\dfrac{21}{2}(-40+a_{21}) \\ \implies &-40+a_{21}=\dfrac{2 \times 210}{21}=20 \\ \implies &a_{21}=20+40=60.\end{align} Also $a_{21}=a_1+20 d$, then\\ \begin{align}&20d=60-(-40)=100 \\ \implies &d=\dfrac{100}{20}=5 \end{align} Hence $a_{21}=60
- Question 17 Exercise 4.2
- gn}a_{n+2}&=a_1+(n+2-1) d \\ & =a_1+(n+1) d \\ \implies 32&=5+(n+1)d \\ \implies (n+1)d&=32-5\\ \implies n+1&=\dfrac{27}{d}\\ \implies n&=\dfrac{27}{d}-1 ---(i)\end{align} Also, we have given $A_3:A_7=7:13$, wh
- Question 15 Exercise 4.2
- 2}=\dfrac{a^{n+1}+b^{n+1}}{a^n+b^n}, --- (3) \\ \implies &(a^n+b^n)(a+b)=2(a^{n+1}+b^{n+1}) \\ \implies &a^{n+1}+a b^n+a^n b+b^{n+1}=2 a^{n+1}+2 b^{n+1} \\ \implies &a b^n+b^{n+1}-2 b^{n+1}=2 a^{n+1}-a^{n+1}-a^n b \\ \implies &a b^n-b^{n+1}=a^{n+1}-a^n b \\ \implies & b^n(a
- Question 7 Exercise 4.2
- ce of A.P. As given \begin{align} &a_6+a_4=6 \\ \implies & a_1+5d+a_1+3d=6\\ \implies & 2a_1+8d=6\\ \implies & a_1+4d=3 --- (1) \end{align} Also, we have given \begin{align} &a_6-a_4=\dfrac{2}{3} \\ \implies & a_1+5d-a_1-3d=\dfrac{2}{3}\\ \implies & 2d=\dfr
- Question 1 and 2 Exercise 4.2
- $ This gives \begin{align} &a_{21}=8+(21-1) d \\ \implies &108=8+20 d\\ \implies &20 d=108-8=100 \\ \implies &d=\dfrac{100}{20}\\ \implies &d=5. \end{align} Now \begin{align}&a_7=a_1+(n-1) d \\ \implies &a_7=8+(7-1
- Question 5 and 6 Exercise 4.2
- \begin{align}& (6 k-2)-(2 k+7)=(8 k-4)-(6 k-2)\\ \implies & 6 k-2 k-2-7=8 k-6 k-4+2 \\ \implies & 4 k-9=2 k-2 \\ \implies & 4 k-2 k=-2+9 \\ \implies & 2 k=7\\ \implies & k=\dfrac{7}{2}.\end{align} Now the terms are: \begin{align
- Question 11 Exercise 4.2
- top. We know \begin{align} &a_n=a_1+(n-1)d \\ \implies &5400=1000+(n-1)100\\ \implies &5400=900+100n \\ \implies &100n=5400-900\\ \implies &100n=4500\\ \implies &n=45.\end{align}\\ Hence Ahmad and Akram will take 45
- Question 3 and 4 Exercise 4.1
- or $n=1$ \begin{align}&a_{1+1}=\dfrac{a_1}{1} \\ \implies &a_2=\dfrac{3}{1}=3.\end{align} For $n=2$ \begin{align}&a_{2+1}=\dfrac{a_2}{2} \\ \implies &a_3=\dfrac{3}{2}.\end{align} For $n=3$ \begin{align}&a_{3+1}=\dfrac{a_3}{3} \\ \implies &a_4=\dfrac{\dfrac{3}{2}}{3}=\dfrac{1}{2}.\end{al... or $n=4$ \begin{align}&a_{4+1}=\dfrac{a_4}{4} \\ \implies &a_5=\dfrac{\dfrac{1}{2}}{4}=\dfrac{1}{8}.\end{al
- Question 3 and 4 Exercise 4.2
- -1) d$$ This gives \begin{align}&78=6+(n-1) 3 \\ \implies &3(n-1)=78-6 \\ \implies &n-1=\dfrac{72}{3} \\ \implies &n=24+1=25.\end{align} Thus, the number of terms in given progression are $
- Question 16 Exercise 4.2
- d } a_7=8.$$ As we have \begin{align}&a_7=a+6d\\ \implies &8=5+6d\\ \implies &6d=8-5\\ \implies &d=\dfrac{3}{6}=\dfrac{1}{2}. \end{align} Now \begin{align} A_1&=a+d=5+\dfrac{1}{2}=\df
- Question 1 Exercise 4.3
- gn} We know that \begin{align}&a_n=a_1+(n-1)d \\ \implies &a_20=9+(20-1)(-2)=-29. \end{align} Assume $S_n$ ... hen \begin{align} S_n&=\dfrac{n}{2}[a_1+a_n], \\ \implies S_{20}&=\dfrac{20}{2}[9-29] \\ &=10(-20)=-200 \en... hen \begin{align} S_n&=\dfrac{n}{2}[a_1+a_n], \\ \implies S_{11}&=\dfrac{11}{2}\left[3-\dfrac{1}{3}\right]
- Question 5 Exercise 4.1
- 3)+(2.2-3)+(2.3-3)+(2.4-3)\\&+(2.5-3)+(2.6-3) \\ \implies \sum_{j=1}^6(2 j-3)&=-1+1+3+5+7+9 .\end{align} =... &+(-1)^3 2^{3-1}+(-1)^4 2^4 1 +(-1)^5 2^{5-1} \\ \implies \sum_{k=1}^5(-1)^k 2^k& =-1+2-4+8-16.\end{align}
- Question 6 & 7 Exercise 4.4
- ac{a_{n+1}}{a_n}$ is independent of $n$, \\ which implies that sequence in (ii) is also geometric sequence