1. **Problem Statement:** Find the A.M. (Arithmetic Means) between given numbers and solve the arithmetic progression (A.P.) problem. 2. **Arithmetic Mean formula:** The $n$ arithmetic means between $a$ and $b$ are given by terms of an arithmetic progression where the first term is $a$, the last term is $b$, and there are $n+2$ terms total. 3. The $k^{th}$ arithmetic mean $A_k$ is given by: $$A_k = a + \frac{k}{n+1}(b - a)$$ for $k=1,2,\ldots,n$. ### Part (i) Find 1 A.M. between 19 and -5 - Given $a=19$, $b=-5$, $n=1$ - $A_1 = 19 + \frac{1}{1+1}(-5 - 19) = 19 + \frac{1}{2}(-24) = 19 - 12 = 7$ ### Part (ii) Find 1 A.M. between -10 and 45 - $a=-10$, $b=45$, $n=1$ - $A_1 = -10 + \frac{1}{2}(45 + 10) = -10 + \frac{1}{2}(55) = -10 + 27.5 = 17.5$ ### Part (iii) Find 1 A.M. between $2 + \sqrt{3}$ and $2 - \sqrt{3}$ - $a = 2 + \sqrt{3}$, $b = 2 - \sqrt{3}$, $n=1$ - $A_1 = (2 + \sqrt{3}) + \frac{1}{2}[(2 - \sqrt{3}) - (2 + \sqrt{3})] = (2 + \sqrt{3}) + \frac{1}{2}(-2\sqrt{3}) = 2 + \sqrt{3} - \sqrt{3} = 2$ ### Part (iv) Find 1 A.M. between $x + b$ and $x - b$ - $a = x + b$, $b = x - b$, $n=1$ - $A_1 = (x + b) + \frac{1}{2}[(x - b) - (x + b)] = x + b + \frac{1}{2}(-2b) = x + b - b = x$ ### Insert 2 A.M.'s between -5 and 40 - $a = -5$, $b = 40$, $n=2$ - $A_1 = -5 + \frac{1}{3}(40 + 5) = -5 + \frac{1}{3}(45) = -5 + 15 =10$ - $A_2 = -5 + \frac{2}{3}(45) = -5 + 30 = 25$ ### Insert 4 A.M.'s between $\frac{\sqrt{2}}{2}$ and $\frac{3\sqrt{2}}{2}$ - $a= \frac{\sqrt{2}}{2}$, $b = \frac{3\sqrt{2}}{2}$, $n=4$ - Difference $d = \frac{b - a}{n+1} = \frac{\frac{3\sqrt{2}}{2} - \frac{\sqrt{2}}{2}}{5} = \frac{\sqrt{2}}{2.5} = \frac{\sqrt{2}}{5/2} = \frac{2\sqrt{2}}{5}$ - The arithmetic means: - $A_1 = a + d = \frac{\sqrt{2}}{2} + \frac{2\sqrt{2}}{5} = \frac{5\sqrt{2}}{10} + \frac{4\sqrt{2}}{10} = \frac{9\sqrt{2}}{10}$ - $A_2 = a + 2d = \frac{\sqrt{2}}{2} + 2 \times \frac{2\sqrt{2}}{5} = \frac{ \sqrt{2}}{2} + \frac{4\sqrt{2}}{5} = \frac{5\sqrt{2}}{10} + \frac{8\sqrt{2}}{10} = \frac{13\sqrt{2}}{10}$ - $A_3 = a + 3d = \frac{\sqrt{2}}{2} + \frac{6\sqrt{2}}{5} = \frac{5\sqrt{2}}{10} + \frac{12\sqrt{2}}{10} = \frac{17\sqrt{2}}{10}$ - $A_4 = a + 4d = \frac{\sqrt{2}}{2} + \frac{8\sqrt{2}}{5} = \frac{5\sqrt{2}}{10} + \frac{16\sqrt{2}}{10} = \frac{21\sqrt{2}}{10}$ ### Insert 5 A.M.'s between 10 and 25 - $a=10$, $b=25$, $n=5$ - Common difference $d = \frac{25 - 10}{6} = \frac{15}{6} = 2.5$ - Arithmetic means: - $A_1 = 10 + 1 \times 2.5 = 12.5$ - $A_2 = 10 + 2 \times 2.5 = 15$ - $A_3 = 10 + 3 \times 2.5 = 17.5$ - $A_4 = 10 + 4 \times 2.5 = 20$ - $A_5 = 10 + 5 \times 2.5 = 22.5$ ### Find $x$ if $x + 1$, $4x + 1$, $8x - 1$ are consecutive terms of an A.P. - Condition for A.P. consecutive terms: $$2 \times \text{middle term} = \text{first term} + \text{third term}$$ - Substitute: $$2(4x + 1) = (x + 1) + (8x - 1)$$ $$8x + 2 = 9x$$ - Solve for $x$: $$9x - 8x = 2$$ $$x = 2$$ **Final answers:** (i) 7 (ii) 17.5 (iii) 2 (iv) $x$ Two A.M.'s between -5 and 40: 10, 25 Four A.M.'s between $\frac{\sqrt{2}}{2}$ and $\frac{3\sqrt{2}}{2}$: $\frac{9\sqrt{2}}{10}$, $\frac{13\sqrt{2}}{10}$, $\frac{17\sqrt{2}}{10}$, $\frac{21\sqrt{2}}{10}$ Five A.M.'s between 10 and 25: 12.5, 15, 17.5, 20, 22.5 Value of $x$ in A.P.: 2