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- Question 1 and 2 Exercise 4.1
- Question 3 and 4 Exercise 4.1
- Question 5 Exercise 4.1
- Question 6 Exercise 4.1
- Question 1 and 2 Exercise 4.2
- Question 3 and 4 Exercise 4.2
- Question 5 and 6 Exercise 4.2
- Question 7 Exercise 4.2
- Question 8 Exercise 4.2
- Question 9 Exercise 4.2
- Question 10 Exercise 4.2
- Question 11 Exercise 4.2
- Question 12 & 13 Exercise 4.2
- Question 14 Exercise 4.2
- Question 15 Exercise 4.2
- Question 16 Exercise 4.2
- Question 17 Exercise 4.2
- Question 1 Exercise 4.3
- Question 2 Exercise 4.3
- Question 3 & 4 Exercise 4.3
- Question 5 & 6 Exercise 4.3
- Question 7 & 8 Exercise 4.3
- Question 9 & 10 Exercise 4.3
- Question 11 & 12 Exercise 4.3
- Question 13 & 14 Exercise 4.3
- Question 1 Exercise 4.4
- Question 2 & 3 Exercise 4.4
- Question 4 & 5 Exercise 4.4
- Question 6 & 7 Exercise 4.4
- Question 8 Exercise 4.4
- Question 9 Exercise 4.4
- Question 10 Exercise 4.4
- Question 11 Exercise 4.4
- Question 12 Exercise 4.4
- Question 1 Exercise 4.5
- Question 2 Exercise 4.5
- Question 3 Exercise 4.5
- Question 4 Exercise 4.5
- Question 5 & 6 Exercise 4.5
- Question 7 & 8 Exercise 4.5
- Question 9 & 10 Exercise 4.5
- Question 11 & 12 Exercise 4.5
- Question 13 & 14 Exercise 4.5
- Question 15 & 16 Exercise 4.5
Fulltext results:
- Question 2 Exercise 4.3
- \ \implies & 8 n^2-22 n-450=0 \\ \implies & 4 n^2-11 n-225=0\end{align} This is quadratic equation with $a=4, b=-11$ and $c=-225$, then \begin{align}n&=\dfrac{-b \pm \sqrt{b^2-4 a c}}{2 a} \\ n&=\dfrac{11 \pm \sqrt{(-11)^2-4(4)(-225)}}{2.4} \\ \implies n&=\dfrac{11 \pm \sqrt{121+3600}}{8} \\ \implies n&=\
- Question 16 Exercise 4.2
- Now \begin{align} A_1&=a+d=5+\dfrac{1}{2}=\dfrac{11}{2},\\ A_2&=a+2 d=5+2 \cdot \dfrac{1}{2}=6,\\ A_3... rac{1}{2}=\dfrac{15}{2}.\end{align} Hence $\dfrac{11}{2},6,\dfrac{13}{2},7, \dfrac{15}{2}$ are five A.... Now \begin{align}A_1&+A_2+A_3+A_4+A_5 \\ &=\dfrac{11}{2}+6+\dfrac{13}{2}+7+\dfrac{15}{2}\\ &=\dfrac{11}{2}+\dfrac{12}{2}+\dfrac{13}{2}+\dfrac{14}{2}+\dfra
- Question 1 Exercise 4.3
- dfrac{8}{3}, \dfrac{7}{3}, 2, \ldots$; 11th term; 11 terms. GOOD ====Solution==== Let $a_1$ be first t... }&a_1=3 \\ &d=\dfrac{8}{3}-3=-\dfrac{1}{3} \\ &n=11. \end{align} We know that $$a_n=a_1+(n-1) d,$$ This gives $$a_{11}=3+10\left(-\dfrac{1}{3}\right)=-\dfrac{1}{3}$$ A... align} S_n&=\dfrac{n}{2}[a_1+a_n], \\ \implies S_{11}&=\dfrac{11}{2}\left[3-\dfrac{1}{3}\right] \\ &=\
- Question 5 & 6 Exercise 4.3
- +x_7+x_{10}&=-6 \text { and } \\ x_3+x_8+x_{12}&=-11, \text { find } \\ x_3+x_8+x_{22} .\end{align} ==... ldots . . .(i) \\ \text { Also } x_3+x_8+x_{12}&=-11 \\ \Rightarrow(x_1+2 d)+(x_1+7 d)+(x_1+11 d)& =-11 \\ \Rightarrow 3 x_1+20 d&=-11 \ldots \ldots \ldots \ldots . .(ii) \end{align} Subtracting (
- Question 7 & 8 Exercise 4.3
- Pakistan. =====Question 7===== Find: $1+3-5+7+9-11+13+15-$ $17+\ldots$ upto $3 n$ terms. ====Solutio... n{align}&(1+7+13+\ldots)+(3+9+15+\ldots)- \\ & (5+11+17+\ldots) \ldots \ldots \ldots . . .(1)\end{alig... }&=\dfrac{6 n^2}{2}=3 n^2\end{align}\\ and for $5+11+17+\ldots$ with $a_1=5$ and $d=11-5=6$ then sum of the $n$ terms,\\ let say denote by $S_n^{\prime \p
- Question 10 Exercise 4.2
- we have to find population in 1970, that is, $a_{11}$. As we know \begin{align} a_{11}&=a_1+10d \\ &=20135+10(-500)\\ &=15135. \end{align} Hence the pop... === <text align="left"><btn type="primary">[[math-11-kpk:sol:unit04:ex4-2-p6 |< Question 9]]</btn></te... t> <text align="right"><btn type="success">[[math-11-kpk:sol:unit04:ex4-2-p8|Question 11 >]]</btn></te
- Question 11 Exercise 4.2
- ====== Question 11 Exercise 4.2 ====== Solutions of Question 11 of Exercise 4.2 of Unit 04: Sequence and Series. This is... KPTB or KPTBB) Peshawar, Pakistan. =====Question 11===== Ahmad and Akram can climb 1000 feet in the f... === <text align="left"><btn type="primary">[[math-11-kpk:sol:unit04:ex4-2-p7 |< Question 10]]</btn></t
- Question 11 & 12 Exercise 4.3
- ====== Question 11 & 12 Exercise 4.3 ====== Solutions of Question 11 & 12 of Exercise 4.3 of Unit 04: Sequence and Serie... KPTB or KPTBB) Peshawar, Pakistan. =====Question 11===== The distance which an object dropped from a ... == <text align="left"><btn type="primary">[[math-11-kpk:sol:unit04:ex4-3-p6 |< Question 9 & 10 ]]</bt
- Question 11 Exercise 4.4
- ====== Question 11 Exercise 4.4 ====== Solutions of Question 11 of Exercise 4.4 of Unit 04: Sequence and Series. This is... KPTB or KPTBB) Peshawar, Pakistan. =====Question 11===== Prove that the prodect of $\mathrm{n}$ geome... === <text align="left"><btn type="primary">[[math-11-kpk:sol:unit04:ex4-4-p7 |< Question 10 ]]</btn></
- Question 11 & 12 Exercise 4.5
- ====== Question 11 & 12 Exercise 4.5 ====== Solutions of Question 11 & 12 of Exercise 4.5 of Unit 04: Sequence and Serie... KPTB or KPTBB) Peshawar, Pakistan. =====Question 11===== If $p^{t h}, q^{t h}$ and $r^{t h}$ terms of... === <text align="left"><btn type="primary">[[math-11-kpk:sol:unit04:ex4-5-p7 |< Question 9 & 10 ]]</bt
- Question 12 & 13 Exercise 4.2
- ean. Then\\ $$A=\dfrac{a+b}{2}=\dfrac{-6-216}{2}=-111$$ GOOD =====Question 13(iv)===== Find the arith... === <text align="left"><btn type="primary">[[math-11-kpk:sol:unit04:ex4-2-p8 |< Question 11 ]]</btn></text> <text align="right"><btn type="success">[[math-11-kpk:sol:unit04:ex4-2-p10|Question 14 >]]</btn></t
- Question 14 Exercise 4.2
- 1=6 \text{ and } a_6=41.$$ Now \begin{align}& a_5=11\\ \Rightarrow &a_1+4 d=41 \\ \Rightarrow &6+4 d=4... === <text align="left"><btn type="primary">[[math-11-kpk:sol:unit04:ex4-2-p9 |< Question 12 & 13 ]]</b... t> <text align="right"><btn type="success">[[math-11-kpk:sol:unit04:ex4-2-p11|Question 15 >]]</btn></text>
- Question 4 & 5 Exercise 4.4
- 1}{2})^{10} \\ \Rightarrow n-1&=10 \text { or } n=11\end{align} Hence the total number of terms are $11$ in sequence.\\ =====Question 5===== Find $x$ so ... === <text align="left"><btn type="primary">[[math-11-kpk:sol:unit04:ex4-4-p2 |< Question 2 & 3 ]]</btn... t> <text align="right"><btn type="success">[[math-11-kpk:sol:unit04:ex4-4-p4|Question 6 & 7 >]]</btn><
- Question 9 & 10 Exercise 4.3
- === <text align="left"><btn type="primary">[[math-11-kpk:sol:unit04:ex4-3-p5 |< Question 7 & 8 ]]</btn... t> <text align="right"><btn type="success">[[math-11-kpk:sol:unit04:ex4-3-p7|Question 11 & 12 >]]</btn></text>
- Question 10 Exercise 4.4
- === <text align="left"><btn type="primary">[[math-11-kpk:sol:unit04:ex4-4-p6 |< Question 9 ]]</btn></t... t> <text align="right"><btn type="success">[[math-11-kpk:sol:unit04:ex4-4-p8|Question 11 >]]</btn></text>