1. Problem on Series: Find the sum of the first 10 terms of the arithmetic series where the first term $a_1=3$ and the common difference $d=5$. Formula: The sum of the first $n$ terms of an arithmetic series is given by $$S_n=\frac{n}{2}(2a_1+(n-1)d)$$ Step 1: Substitute $n=10$, $a_1=3$, and $d=5$ into the formula. Step 2: Calculate $$S_{10}=\frac{10}{2}(2\times3+(10-1)\times5)$$ Step 3: Simplify inside the parentheses: $$2\times3=6$$ and $$(10-1)\times5=9\times5=45$$ Step 4: So, $$S_{10}=5(6+45)=5\times51=255$$ Answer: The sum of the first 10 terms is 255. 2. Problem on Factorisation: Factorise the quadratic expression $$x^2 - 5x + 6$$. Step 1: Find two numbers that multiply to 6 and add to -5. Step 2: These numbers are -2 and -3. Step 3: Write the factorised form: $$(x-2)(x-3)$$ Answer: The factorised form is $$(x-2)(x-3)$$. 3. Problem on Quadratic Equations: Solve the quadratic equation $$x^2 - 4x - 5 = 0$$. Step 1: Use the quadratic formula $$x=\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=1$, $b=-4$, and $c=-5$. Step 2: Calculate the discriminant: $$b^2 - 4ac = (-4)^2 - 4\times1\times(-5) = 16 + 20 = 36$$ Step 3: Substitute into the formula: $$x=\frac{-(-4) \pm \sqrt{36}}{2\times1} = \frac{4 \pm 6}{2}$$ Step 4: Calculate the two solutions: $$x=\frac{4+6}{2} = \frac{10}{2} = 5$$ $$x=\frac{4-6}{2} = \frac{-2}{2} = -1$$ Answer: The solutions are $x=5$ and $x=-1$. 4. Problem on Coordinate Geometry and Lines: Find the equation of the line passing through points $(2,3)$ and $(4,7)$. Step 1: Calculate the slope $$m=\frac{y_2 - y_1}{x_2 - x_1} = \frac{7-3}{4-2} = \frac{4}{2} = 2$$ Step 2: Use point-slope form $$y - y_1 = m(x - x_1)$$ with point $(2,3)$: $$y - 3 = 2(x - 2)$$ Step 3: Simplify: $$y - 3 = 2x - 4$$ $$y = 2x - 4 + 3$$ $$y = 2x - 1$$ Answer: The equation of the line is $$y = 2x - 1$$. 5. Problem on Systems of Linear Equations: Solve the system $$\begin{cases} 2x + y = 7 \\ 3x - y = 8 \end{cases}$$ Step 1: Add the two equations to eliminate $y$: $$ (2x + y) + (3x - y) = 7 + 8 $$ $$ 5x + \cancel{y} - \cancel{y} = 15 $$ Step 2: Simplify: $$5x = 15$$ Step 3: Solve for $x$: $$x = \frac{15}{5} = 3$$ Step 4: Substitute $x=3$ into the first equation: $$2(3) + y = 7$$ $$6 + y = 7$$ Step 5: Solve for $y$: $$y = 7 - 6 = 1$$ Answer: The solution is $x=3$, $y=1$.