1. **Find the 20th term of an A.P. with first term 5 and common difference 4.** The nth term of an arithmetic progression (A.P.) is given by: $$a_n = a + (n-1)d$$ where $a$ is the first term and $d$ is the common difference. Given: $a=5$, $d=4$, $n=20$ Calculate: $$a_{20} = 5 + (20-1) \times 4 = 5 + 19 \times 4 = 5 + 76 = 81$$ --- 2. **Prove that for all natural numbers $n$,** $$1\cdot 2 + 2\cdot 3 + 3\cdot 4 + \cdots + n(n+1) = \frac{n(n+1)(n+2)}{3}$$ Proof by induction: Base case $n=1$: $$1\cdot 2 = 2$$ Right side: $$\frac{1 \times 2 \times 3}{3} = 2$$ True for $n=1$. Assume true for $n=k$: $$\sum_{i=1}^k i(i+1) = \frac{k(k+1)(k+2)}{3}$$ For $n=k+1$: $$\sum_{i=1}^{k+1} i(i+1) = \frac{k(k+1)(k+2)}{3} + (k+1)(k+2)$$ Factor: $$= \frac{k(k+1)(k+2)}{3} + \frac{3(k+1)(k+2)}{3} = \frac{(k+1)(k+2)(k+3)}{3}$$ Hence, true for $k+1$. By induction, the formula holds for all natural $n$. --- 3. **If $x^y = e^{x-y}$, show that** $$\frac{dy}{dx} = \frac{\log x}{(1+\log x)^2}$$ Take natural log on both sides: $$y \log x = x - y$$ Rearranged: $$y \log x + y = x$$ $$y(1 + \log x) = x$$ Differentiate both sides w.r.t $x$: $$\frac{dy}{dx}(1 + \log x) + y \frac{d}{dx}(1 + \log x) = 1$$ Since $\frac{d}{dx}(1 + \log x) = \frac{1}{x}$: $$\frac{dy}{dx}(1 + \log x) + y \frac{1}{x} = 1$$ Solve for $\frac{dy}{dx}$: $$\frac{dy}{dx} = \frac{1 - \frac{y}{x}}{1 + \log x}$$ Recall from earlier: $$y = \frac{x}{1 + \log x}$$ Substitute $y$: $$\frac{dy}{dx} = \frac{1 - \frac{x}{x(1 + \log x)}}{1 + \log x} = \frac{1 - \frac{1}{1 + \log x}}{1 + \log x} = \frac{\frac{(1 + \log x) - 1}{1 + \log x}}{1 + \log x} = \frac{\log x}{(1 + \log x)^2}$$ --- 4. **Prove: $\binom{1000}{98} = \binom{999}{97} + \frac{x}{901} \binom{901}{C}$, find $x$.** Assuming the problem means: $$\binom{1000}{98} = \binom{999}{97} + \frac{x}{901} \binom{901}{C}$$ Using Pascal's identity: $$\binom{n}{r} = \binom{n-1}{r} + \binom{n-1}{r-1}$$ So: $$\binom{1000}{98} = \binom{999}{98} + \binom{999}{97}$$ Given the equation, comparing terms suggests: $$\frac{x}{901} \binom{901}{C} = \binom{999}{98}$$ Since $\binom{999}{98} = \frac{999}{901} \binom{901}{C}$ (assuming $C=97$), then: $$\frac{x}{901} = \frac{999}{901} \Rightarrow x = 999$$ Hence, $x=999$. --- 5. **Find the number of distinct permutations of the letters of the word MATHEMATICS.** Count letters: - M: 2 - A: 2 - T: 2 - H: 1 - E: 1 - I: 1 - C: 1 - S: 1 Total letters = 11 Number of distinct permutations: $$\frac{11!}{2! \times 2! \times 2!} = \frac{39916800}{8} = 4989600$$ --- 6. **Find the rank of matrix $A = \begin{bmatrix}1 & 3 & 4 & 3 \\ 3 & 9 & 12 & 9 \\ -1 & -3 & -4 & 3 \end{bmatrix}$ by row-echelon form.** Start with $A$: $$\begin{bmatrix}1 & 3 & 4 & 3 \\ 3 & 9 & 12 & 9 \\ -1 & -3 & -4 & 3 \end{bmatrix}$$ Row2 = Row2 - 3*Row1: $$\begin{bmatrix}1 & 3 & 4 & 3 \\ 0 & 0 & 0 & 0 \\ -1 & -3 & -4 & 3 \end{bmatrix}$$ Row3 = Row3 + Row1: $$\begin{bmatrix}1 & 3 & 4 & 3 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 6 \end{bmatrix}$$ Nonzero rows: 2 Rank of $A$ is 2. --- 7. **Define Consumer and Producer Surplus.** - Consumer Surplus: The difference between what consumers are willing to pay for a good and what they actually pay. - Producer Surplus: The difference between the amount producers receive for a good and the minimum amount they are willing to accept. --- 8. **Two industries input-output relationship:** Matrix $A$: $$\begin{bmatrix}50 & 75 \\ 100 & 50 \end{bmatrix}$$ Final demand vector: $$d = \begin{bmatrix}75 \\ 50 \end{bmatrix}$$ Gross output vector: $$x = \begin{bmatrix}400 \\ 600 \end{bmatrix}$$ Final demand satisfied: $$d' = x - A x = \begin{bmatrix}400 \\ 600 \end{bmatrix} - \begin{bmatrix}50 & 75 \\ 100 & 50 \end{bmatrix} \begin{bmatrix}400 \\ 600 \end{bmatrix}$$ Calculate $A x$: $$\begin{bmatrix}50 \times 400 + 75 \times 600 \\ 100 \times 400 + 50 \times 600 \end{bmatrix} = \begin{bmatrix}20000 + 45000 \\ 40000 + 30000 \end{bmatrix} = \begin{bmatrix}65000 \\ 70000 \end{bmatrix}$$ Final demand satisfied: $$d' = \begin{bmatrix}400 \\ 600 \end{bmatrix} - \begin{bmatrix}65000 \\ 70000 \end{bmatrix} = \begin{bmatrix}-64600 \\ -69400 \end{bmatrix}$$ Negative values indicate inconsistency; likely the matrix $A$ entries are in different units or need normalization. --- 9. **Test Hawkins-Simon conditions:** Matrix $I - A$: $$I - A = \begin{bmatrix}1-0.5 & -0.75 \\ -1 & 1-0.5 \end{bmatrix} = \begin{bmatrix}0.5 & -0.75 \\ -1 & 0.5 \end{bmatrix}$$ Check principal minors: - First order minors: $0.5 > 0$, $0.5 > 0$ - Determinant: $$0.5 \times 0.5 - (-0.75) \times (-1) = 0.25 - 0.75 = -0.5 < 0$$ Since determinant is negative, Hawkins-Simon conditions are not satisfied. --- **Final answers:** 1. $81$ 2. Proven formula holds. 3. $\frac{dy}{dx} = \frac{\log x}{(1+\log x)^2}$ 4. $x=999$ 5. $4989600$ 6. Rank = 2 7. Definitions given. 8. Final demand vector calculation shows inconsistency. 9. Hawkins-Simon conditions not satisfied.