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- Question 5 & 6 Exercise 4.3
- $Condition-2$\\ The sum of their square is $120$, therefore\\ \begin{align}(a-3 d)^2+(a-d)^2+(a+d)^2+(a+2... $ and $d=-1$ then the numbers are\\ \begin{align}\therefore x_1+(x_1+6 d)+(x_1+9 d)&=-6 \\ \Rightarrow 3 ... text { and } \\ x_{22}&=x_1+21 d=3+21(-1)=-18 \\ \therefore \quad x_3+x_8+x_{22}&=1-4-18=-21\end{align}
- Question 9 & 10 Exercise 4.3
- arrow 9 n&=396 \\ \Rightarrow n&=44.\end{align} $\therefore$ Required sum is:\\ \begin{align} S_{44}&=\df... = The total money for distribution $S_4=1000$, \\ therefore we have $n=4$\\ \begin{align}\text{Let the fi... \ \begin{align}S_n&=\dfrac{n}{2}[2 a+(n-1) d] \\ \therefore S_4&=\dfrac{4}{2}[2 a+3(-20)] . \\ \Rightarro
- Question 13 & 14 Exercise 4.3
- in the third row and so forth. How many seats are there in the theater? ====Solution==== \begin{align}\te... ver this is including $1$ and $50$ as terms,\\ so therefore there would need to be $16$ terms between $1$ and $50$. ====Go To==== <text align="left"><btn type=
- Question 1 Exercise 4.5
- hat\\ \begin{align}a_n&=a_1 r^{n-1} \text {, }\\ \therefore \dfrac{1}{16}&=8(\dfrac{1}{2})^{n-1} or\\ (\d... We know that\\ \begin{align}a_n&=a_1 r^{n-1}, \\ \therefore 2^{10}&=2^4(2)^{n-1} \text { or } 2^{n-1}=\df... ign}r&=\dfrac{-1}{\dfrac{8}{5}}=-\dfrac{5}{8} \\ \therefore S_{\infty}&=\dfrac{a_1}{1-r}=\dfrac{\dfrac{8}
- Question 11 & 12 Exercise 4.5
- ==== The general term of G.P $a_n=a_1 r^{n-1}$.\\ Therefore, $a_p=a_1 r^{p-1}=a \quad a_q=a_1 r^{q-1}=b$ ... frac{a_1}{1-r}$, but we are given $S_{\infty}=6$. Therefore,\\ $$\dfrac{a_1}{1-r}=6 \text { or } 6 a_1=1-... four times the sum of all the terms following it, therefore\\ \begin{align}a_1&=4(a_1 r+a_1 r^2+a_1 r^3+\
- Question 4 & 5 Exercise 4.4
- akistan. =====Question 4===== How many terms are there in a geometric sequence in which the first and th... a three terms geometric sequence,\\ \begin{align}\therefore \dfrac{a_2}{a_1}&=\dfrac{a_3}{a_2} \\ \Righta
- Question 9 Exercise 4.4
- th $a_7=\dfrac{81}{2}$ and $a_1=\dfrac{32}{9}$.\\ Therefore, \begin{align}a_1 r^6&=\dfrac{81}{2} \\ \Righ... =-\dfrac{7}{64} \text { and }\\ a_1&=14 . \text { Therefore, } \\ a_1 r^7&=-\dfrac{7}{64} \\ \Rightarrow
- Question 10 Exercise 4.4
- ition-$1$\\ The difference between them is $48$\\ Therefore, $$\quad a-b=48....(i)$$. The geometric mean ... dition-$2$\\ Their A.M exceeds their G.M by 18.\\ Therefore $A \cdot M=G \cdot M+18$ or $A \cdot M-G \cdo
- Question 2 Exercise 4.5
- d $n$ and then $S_n$\\ We know $a_n=a_1 r^{n-1}$, therefore\\ \begin{align}64&=(-2)^{n-1}\\ \Rightarrow(-... a_1$ and then $S_9$.\\ We know $$a_9=a_1 r^8$$\\ therefore we have\\ \begin{align}1&=a_1(\dfrac{1}{2})^{
- Question 5 and 6 Exercise 4.2
- , in second term the power of $b$ is 1 and so on, therefore $$a_n=\log (a b^{n-1}).$$ We show that the g
- Question 8 Exercise 4.2
- dfrac{a+b-c}{c}$ are in A.P, thus\\ \begin{align}\therefore \dfrac{c+a-b}{b}-\dfrac{b+c-a}{a}&=\dfrac{a+b
- Question 17 Exercise 4.2
- KPTBB) Peshawar, Pakistan. =====Question 17===== There are $n$ arithmetic means between 5 and 32 such th
- Question 7 & 8 Exercise 4.3
- \begin{align}S_n&=\dfrac{n}{2}[2 a_1+(n-1) d] \\ \therefore S_n&=\dfrac{n}{2}[2 \cdot 1+(n-1) \cdot 2] \\
- Question 11 & 12 Exercise 4.3
- \begin{align}S_n&=\dfrac{n}{2}[2 a_1+(n-1) d] \\ \therefore S_6&=\dfrac{6}{2}(2.16+5.32) \\ \Rightarrow S
- Question 6 & 7 Exercise 4.4
- $ ====Solution==== We know that $a_n=a_1 r^{n-1}$ therefore\\ \begin{align}a_{10}&=a_1 r^9=l \\ a_{13}&=a