1. Convert $\frac{3-5i}{2+7i}$ into the form $a+bi$. Use the formula for division of complex numbers: $$\frac{z_1}{z_2} = \frac{z_1 \cdot \overline{z_2}}{|z_2|^2}$$ where $\overline{z_2}$ is the conjugate of $z_2$. Calculate conjugate: $\overline{2+7i} = 2 - 7i$. Multiply numerator and denominator: $$ (3-5i)(2-7i) = 3\times2 - 3\times7i - 5i\times2 + 5i\times7i = 6 - 21i - 10i - 35i^2 $$ Since $i^2 = -1$, this becomes: $$ 6 - 31i + 35 = 41 - 31i $$ Calculate denominator: $$ |2+7i|^2 = 2^2 + 7^2 = 4 + 49 = 53 $$ So, $$ \frac{3-5i}{2+7i} = \frac{41 - 31i}{53} = \frac{41}{53} - \frac{31}{53}i $$ 2. Find length of median through vertex $A$ of triangle $ABC$ with $A(0,0,0)$, $B(3,1,2)$, $C(4,2,3)$. Median from $A$ goes to midpoint $M$ of $BC$: $$ M = \left( \frac{3+4}{2}, \frac{1+2}{2}, \frac{2+3}{2} \right) = (3.5, 1.5, 2.5) $$ Length of median $AM$: $$ |AM| = \sqrt{(3.5-0)^2 + (1.5-0)^2 + (2.5-0)^2} = \sqrt{12.25 + 2.25 + 6.25} = \sqrt{20.75} $$ 3. Find rank of matrix $$ A = \begin{bmatrix} 4 & 5 & 2 \\ 1 & 3 & 4 \\ 2 & 1 & 5 \end{bmatrix} $$ Perform row operations to find rank: - $R_2 \to R_2 - \frac{1}{4}R_1 = (0, 3 - \frac{5}{4}, 4 - \frac{2}{4}) = (0, \frac{7}{4}, \frac{14}{4})$ - $R_3 \to R_3 - \frac{1}{2}R_1 = (0, 1 - \frac{5}{2}, 5 - 1) = (0, -\frac{3}{2}, 4)$ Next, eliminate below $R_2$: - $R_3 \to R_3 + \frac{6}{7} R_2 = (0, 0, 4 + \frac{6}{7} \times \frac{14}{4}) = (0, 0, 4 + 3) = (0, 0, 7)$ Since all three rows have pivots, rank is 3. 4. Find projection of $\mathbf{a} = 2\hat{i} + 3\hat{j} + 7\hat{k}$ on $\mathbf{b} = 2\hat{i} + 7\hat{j} + \hat{k}$. Projection formula: $$ \text{proj}_{\mathbf{b}} \mathbf{a} = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{b}|^2} \mathbf{b} $$ Calculate dot product: $$ \mathbf{a} \cdot \mathbf{b} = 2\times2 + 3\times7 + 7\times1 = 4 + 21 + 7 = 32 $$ Calculate magnitude squared: $$ |\mathbf{b}|^2 = 2^2 + 7^2 + 1^2 = 4 + 49 + 1 = 54 $$ Projection vector: $$ \frac{32}{54} (2\hat{i} + 7\hat{j} + \hat{k}) = \frac{16}{27} (2\hat{i} + 7\hat{j} + \hat{k}) = \left( \frac{32}{27}, \frac{112}{27}, \frac{16}{27} \right) $$ 5. Show determinant $$ \begin{vmatrix} b+c & c+b & a+b \\ a & b & c \\ 1 & 1 & 1 \end{vmatrix} = 0 $$ Note $b+c = c+b$, so first two columns are identical, determinant of matrix with two identical columns is zero. 6. Arithmetic series: first term $a=5$, last term $l=45$, sum $S=500$. Sum formula: $$ S = \frac{n}{2} (a + l) $$ Calculate $n$: $$ 500 = \frac{n}{2} (5 + 45) = 25n \Rightarrow n = 20 $$ Common difference $d$: $$ l = a + (n-1)d \Rightarrow 45 = 5 + 19d \Rightarrow d = \frac{40}{19} $$ 7. Find sum $$ \sum_{k=1}^4 \frac{2^k + 1}{3k - 1} $$ Calculate each term: - $k=1$: $\frac{2^1 + 1}{3(1) - 1} = \frac{3}{2}$ - $k=2$: $\frac{4 + 1}{6 - 1} = \frac{5}{5} = 1$ - $k=3$: $\frac{8 + 1}{9 - 1} = \frac{9}{8}$ - $k=4$: $\frac{16 + 1}{12 - 1} = \frac{17}{11}$ Sum: $$ \frac{3}{2} + 1 + \frac{9}{8} + \frac{17}{11} = \frac{33}{22} + \frac{22}{22} + \frac{24.75}{22} + \frac{34}{22} = \frac{113.75}{22} \approx 5.17 $$ 8. Number of arrangements of letters in "SHOPKEEPER". Count letters: S(1), H(1), O(1), P(2), K(1), E(3), R(1) Total letters = 10 Number of arrangements: $$ \frac{10!}{2! \times 3!} = \frac{3628800}{2 \times 6} = \frac{3628800}{12} = 302400 $$