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- Question 13 & 14 Exercise 4.5
- from which it is dropped. If it is dropped from $10 \mathrm{ft}$, how far does it travel from the mom... s eighth bounce? ====Solution==== \begin{align}S&=10+[10(\dfrac{1}{2})+10(\dfrac{1}{2})]+ \\ & {[10(\dfrac{1}{2})^2+10(\dfrac{1}{2})^2]+\ldots} \\ & +[10(\dfr
- Question 11 Exercise 4.2
- =====Question 11===== Ahmad and Akram can climb 1000 feet in the first hour and 100 feet in each succeeding hour. When will they reach the top of a 5400... limb by Ahmad and Akram in first hour. Then $$a_1=1000.$$ Distance climb in each succeeding hour $= d=100$. As the given problem is of A.P with $a_n=5400
- Question 5 & 6 Exercise 4.5
- tan. =====Question 5===== Find $r$ such that $S_{10}=244 S_5$. ====Solution==== We know that $$S_n=\dfrac{a_1(r^n-1)}{r-1}$$\\ then\\ $$S_{10}=\dfrac{a_1(r^{10}-1)}{r-1} \quad \text{and}\quad S_5=\dfrac{a_1(r^5-1)}{r-1}$$\\ Putting both the $S_{10}$ and $S_S$ in the given equation, we get\\ \begi
- Question 1 Exercise 4.5
- )^{n-1}&=2^9 \\ \Rightarrow n-1&=9 \text { or } n=10 \\ \text {. Now }\quad S_n&=\dfrac{a_1(r^n-1)}{r... lign} becomes in the given case\\ \begin{align}S_{10}&=\dfrac{3[2^{10}-1]}{2-1} \\ \Rightarrow \quad S_{10}&=3(2^{10}-1)\end{align}\\ is the required sum. =====Question 1
- Question 6 Exercise 4.1
- 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}
- Question 1 and 2 Exercise 4.2
- rm of an arithmetic sequence is 8 and the 21st is 108. Find the 7th term. GOOD ====Solution==== Since $a_1=8$ and $a_{21}=108$. We know that $$a_n=a_1+(n-1) d.$$ 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 \
- Question 10 Exercise 4.2
- ====== Question 10 Exercise 4.2 ====== Solutions of Question 10 of Exercise 4.2 of Unit 04: Sequence and Series. This is... KPTB or KPTBB) Peshawar, Pakistan. =====Question 10===== The population of a town is decreasing by $5... , $a_{11}$. As we know \begin{align} a_{11}&=a_1+10d \\ &=20135+10(-500)\\ &=15135. \end{align} Hence
- Question 9 & 10 Exercise 4.3
- ====== Question 9 & 10 Exercise 4.3 ====== Solutions of Question 9 & 10 of Exercise 4.3 of Unit 04: Sequence and Series. Thi... 0$ and $700$ is equal to $21,978$. =====Question 10===== The sum of Rs. $1000$ is distributed among four people so that each person after the first recei
- Question 4 & 5 Exercise 4.4
- c{1}{2})^{n-1}&=\dfrac{1}{64 \times 16}=\dfrac{1}{1024} \\ \Rightarrow(\dfrac{1}{2})^{n-1}&=\dfrac{1}{2^{10}}=(\dfrac{1}{2})^{10} \\ \Rightarrow n-1&=10 \text { or } n=11\end{align} Hence the total number of terms are $11$ in sequ
- Question 6 & 7 Exercise 4.4
- Peshawar, Pakistan. =====Question 6===== If $a_{10}=l, a_{13}=m$ and $a_{16}=n;\quad$ show that $\ln... hat $a_n=a_1 r^{n-1}$ therefore\\ \begin{align}a_{10}&=a_1 r^9=l \\ a_{13}&=a_1 r^{12}=m\\ \text{and} ... ltiplying (i) and (iii), we get\\ \begin{align}a_{10} \cdot a_{16}&=\ln =(a_1 r^9)(a_1 r^{15})\\ \Righ
- Question 10 Exercise 4.4
- ====== Question 10 Exercise 4.4 ====== Solutions of Question 10 of Exercise 4.4 of Unit 04: Sequence and Series. This is... KPTB or KPTBB) Peshawar, Pakistan. =====Question 10===== Find two numbers if the difference between t
- Question 9 & 10 Exercise 4.5
- ====== Question 9 & 10 Exercise 4.5 ====== Solutions of Question 9 & 10 of Exercise 4.5 of Unit 04: Sequence and Series. Thi... } But $r$ can not be 1 thus $r=3$. =====Question 10==== How many terms of the series: $1+\sqrt{3}+3+\
- Question 1 and 2 Exercise 4.1
- or forth term, put $n=4,$ $$a_4=\dfrac{4(4+1)}{2}=10$$ Hence first four terms of the sequence are $1,3,6, 10$. =====Question 2(ii)==== Find first four terms
- Question 1 Exercise 4.3
- _n], \\ \implies S_{20}&=\dfrac{20}{2}[9-29] \\ &=10(-20)=-200 \end{align} Hence 20th term is -29 and ... w that $$a_n=a_1+(n-1) d,$$ This gives $$a_{11}=3+10\left(-\dfrac{1}{3}\right)=-\dfrac{1}{3}$$ Assume
- Question 2 Exercise 4.3
- . Find the one that are missing $a_1=-40, S_{21}=210$. GOOD ====Solution==== Given: $a_1=-40$ and $S_{21}=210$.\\ So we have $n=21$ and we have to find $a_{21}... n}&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}