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- Question 10, Exercise 1.2
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Fulltext results:
- Question 10, Exercise 1.2 @math-11-nbf:sol:unit01
- ====== Question 10, Exercise 1.2 ====== Solutions of Question 10 of Exercise 1.2 of Unit 01: Complex Numbers. This is un... extbook Board, Islamabad, Pakistan. ====Question 10(i)==== For $z_{1}=-3+2 i$, verify: $$\left|z_{1}\... ht| = \sqrt{13}.$$ As required. GOOD ====Question 10(ii)==== For $z_{1}=-3+2 i$ and $z_{2}=1-3 i$ veri
- Question 29 and 30, Exercise 4.7 @math-11-nbf:sol:unit04
- nd sum to infinity of the series: $$1+4 x+7 x^{2}+10 x^{3}+\ldots$$ ** Solution. ** The given arithm... as:\\ \[ 1 \times 1 + 4 \times x + 7 \times x^2 + 10 \times x^3 + \ldots \] The numbers \(1, 4, 7, 10, \ldots\) are in AP with \(a = 1\) and \(d = 4 - 1... Find sum to infinity of the series: $$3+\frac{6}{10}+\frac{9}{100}+\frac{12}{1000}+\ldots$$ ** Solut
- Unit 04: Sequences and Seeries
- * [[math-11-nbf:sol:unit04:ex4-1-p5|Question 9 & 10]] * [[math-11-nbf:sol:unit04:ex4-1-p6|Question ... ion 17 & 18]] * [[math-11-nbf:sol:unit04:ex4-1-p10|Question 19 & 20]] * [[math-11-nbf:sol:unit04:e... * [[math-11-nbf:sol:unit04:ex4-2-p6|Question 9 & 10]] * [[math-11-nbf:sol:unit04:ex4-2-p7|Question ... ion 14 & 15]] * [[math-11-nbf:sol:unit04:ex4-2-p10|Question 16 & 17]] </panel> <panel type="default
- Question 4, Exercise 2.6 @math-11-nbf:sol:unit02
- {1}-4 x_{2}+3 x_{3}=7$\\ $4 x_{1}+2 x_{2}-5 x_{3}=10$\\ ** Solution. ** \begin{align*} 2x_1 - x_2 -... 3x_1 - 4x_2 + 3x_3 &= 7, \\ 4x_1 + 2x_2 - 5x_3 &= 10, \end{align*} The associative augment matrix: \b... & : & 2 \\ 3 & -4 & 3 & : & 7 \\ 4 & 2 & -5 & : & 10 \end{bmatrix}\\ &\sim \text{R}\begin{bmatrix} 1 &... \\ 3 & -4 & 3 & \vert & 7 \\ 4 & 2 & -5 & \vert & 10 \end{bmatrix}\quad \dfrac{1}{2}\\ &\sim \text{R}\
- Question 6, Exercise 2.6 @math-11-nbf:sol:unit02
- \right|\\ &=5(4 - 6) - 3(8 - 3) + 1(4 - 1)\\ &= -10 - 15 + 3\\ &=-22 \end{align*} This system is cons... 3 & 1 \\ 2 & 4 \end{array} \right| = -(12 - 2) = -10\\ A_{22} &= (-1)^{2+2} \left| \begin{array}{cc} 5... array}{cc} 5 & 3 \\ 1 & 2 \end{array} \right| = -(10 - 3) = -7\\ A_{31} &= (-1)^{3+1} \left| \begin{ar... nd{bmatrix}\\ A &=\begin{bmatrix} -2 & -5 & 3 \\ -10 & 19 & -7 \\ 8 & -18 & -1 \end{bmatrix}\\ \text{a
- Question 5 and 6, Exercise 4.1 @math-11-nbf:sol:unit04
- given, find the first 4 terms; the 10th term, $a_{10}$ and the 15 th term: $a_{15}$.$a_{n}=n^{2}-2 n$ ... 16 - 8 = 8\\ \end{align*} Now \begin{align*} a_{10} &= (10)^2 - 2(10) = 100 - 20 = 80\\ a_{15} &= (15)^2 - 2(15) = 225 - 30 = 195 \end{align*} So, $a_1 = -1$,$
- Question 10, Exercise 8.1 @math-11-nbf:sol:unit08
- ====== Question 10, Exercise 8.1 ====== Solutions of Question 10 of Exercise 8.1 of Unit 08: Fundamental of Trigonometry... ook Board, Islamabad, Pakistan. ===== Question 10(i)===== Verify: $\sin \left(\dfrac{\pi}{2}-\alph... s\alpha = R.H.S \end{align*} GOOD ===== Question 10(ii)===== Verify: $\cos (\pi-\alpha)=-\cos \alpha
- Question 1 and 2, Exercise 4.1 @math-11-nbf:sol:unit04
- given, find the first 4 terms; the 10th term, $a_{10}$ and the 15 th term: $a_{15}$. $$a_{n}=3 n+1$$ ... 3(2) + 1 = 6 + 1 = 7\\ a_3 &= 3(3) + 1 = 9 + 1 = 10\\ a_4 &= 3(4) + 1 = 12 + 1 = 13\\ \end{align*} Now \begin{align*} a_{10} &= 3(10) + 1 = 30 + 1 = 31\\ a_{15} &= 3(15) + 1 = 45 + 1 = 46. \end{align*} Hence $a_1=4$, $a_2=7$,
- Question 1 and 2, Exercise 4.8 @math-11-nbf:sol:unit04
- n}-T_{n-1}\right)-T_{n}. \\ \implies 0=&3+(4+6+8+10+\ldots \text { up to } (n-1) \text { terms })-T_... end{align*} Then \begin{align*} T_{n} & =3+(4+6+8+10+\ldots \text { up to }(n-1) \text { terms }) \\ ... {6}(2n^2+2n+n+1+3n+3+6) \\ & =\frac{n}{6}(2n^2+6n+10) \\ & =\frac{n}{3}(n^2+3n+5) \end{align*} Hence ... d of difference, find the sum of the series: $1+4+10+22+\ldots$ to $n$ term. ** Solution. ** Let $$
- Question 10(vi-x), Review Exercise @math-11-nbf:sol:unit09
- ====== Question 10(vi-x), Review Exercise ====== Solutions of Question 10(vi-x) of Review Exercise of Unit 09: Trigonometri... tbook Board, Islamabad, Pakistan. =====Question 10(vi)===== ** Solution. ** =====Question 10(vii)===== ** Solution. ** =====Question 10(viii
- Question 9 and 10, Exercise 4.1 @math-11-nbf:sol:unit04
- ====== Question 9 and 10, Exercise 4.1 ====== Solutions of Question 9 and 10 of Exercise 4.1 of Unit 04: Sequence and Series. ... given, find the first 4 terms; the 10th term, $a_{10}$ and the 15 th term, $a_{15}$: $a_{n}=(-1)^{n}(n... ** Solution. ** Do yourself. =====Question 10===== The $n$th term of the sequence is given, fin
- Question 5 and 6, Exercise 4.3 @math-11-nbf:sol:unit04
- mplies S_{20}&=\frac{20}{2}[2(50)+(20-1)(-4)]\\ &=10\times [100-76]\\ &=240. \end{align} Hence $S_{20}=240$. GOOD =====Question 6===== Find the sum of series. $-3+(-7)+(-11)+\cdots +a_{10}$ ** Solution. ** Given series is arithmetic series with $a_1=-3$, $d=-7-(-3)=-4$, $n=10$.\\ Let $S_n$ represents sum of arithmetic series
- Question 7 and 8, Exercise 4.7 @math-11-nbf:sol:unit04
- - 4 + 3) + (9 - 6 + 3)\\ &+ (16 - 8 + 3) + (25 - 10 + 3) \\ &= 3 + 2 + 3 + 6 + 11 + 18 \\ &= 43 \end{... ===Question 8===== Evaluate the sum: $\sum_{k=1}^{10} \frac{1}{k(k+1)}$ ** Solution. ** \begin{align*} \sum_{k=1}^{10} \frac{1}{k(k+1)} &= \frac{1}{1(2)} + \frac{1}{2(... )} + \frac{1}{7(8)} + \frac{1}{8(9)} + \frac{1}{9(10)} + \frac{1}{10(11)} \\ &= \frac{1}{2} + \frac{1}
- Question 5 and 6, Exercise 4.8 @math-11-nbf:sol:unit04
- of difference, find the sum of the series: $3+4+6+10+18+34+66+\dots$ to $n$ term. ** Solution. ** Let $$ S_{n}=3+4+6+10+18+\ldots +T_{n} $$ Also $$ S_{n}=3+4+6+10+\ldots +T_{n-1}+T_{n}. $$ Subtracting the second expressio... ssion, we have \begin{align*} S_{n}-S_{n}& =3+4+6+10+18+\ldots +T_{n} \\ & -\left(3+4+6+10+\ldots +T_
- Question 3 and 4, Exercise 4.1 @math-11-nbf:sol:unit04
- given, find the first 4 terms; the 10th term, $a_{10}$ and the 15 th term: $a_{n}=\frac{n}{n+1}$ ** S... frac{4}{5}\\ \end{align*} Now \begin{align*} a_{10} &= \frac{10}{10+1} = \frac{10}{11}\\ a_{15} &= \frac{15}{15+1} = \frac{15}{16} \end{align*} So, $a_1 = \frac{1}{