The problem has two parts: (a) Simplify the expression $\frac{x^2 - 8x + 16}{x^2 - 7x + 12}$. (b) Given that $\frac{1}{2}$, $\frac{1}{x}$, and $\frac{1}{3}$ are successive terms of an arithmetic progression (A.P.), show that $\frac{2-x}{x-3} = \frac{2}{3}$.

1. Problem Description

The problem has two parts:

(a) Simplify the expression $\frac{x^2 - 8x + 16}{x^2 - 7x + 12}$.

(b) Given that $\frac{1}{2}$, $\frac{1}{x}$, and $\frac{1}{3}$ are successive terms of an arithmetic progression (A.P.), show that $\frac{2-x}{x-3} = \frac{2}{3}$.

2. Solution Steps

(a) Simplify $\frac{x^2 - 8x + 16}{x^2 - 7x + 12}$.

First, we factor the numerator and the denominator.

$x^2 - 8x + 16 = (x - 4)(x - 4) = (x - 4)^2$

$x^2 - 7x + 12 = (x - 3)(x - 4)$

Therefore,

$\frac{x^2 - 8x + 16}{x^2 - 7x + 12} = \frac{(x - 4)^2}{(x - 3)(x - 4)} = \frac{x - 4}{x - 3}$ for $x \neq 4$.

(b) Given that $\frac{1}{2}$, $\frac{1}{x}$, and $\frac{1}{3}$ are successive terms of an arithmetic progression. In an A.P., the difference between consecutive terms is constant.

Thus,

$\frac{1}{x} - \frac{1}{2} = \frac{1}{3} - \frac{1}{x}$

$\frac{2}{x} = \frac{1}{2} + \frac{1}{3} = \frac{3 + 2}{6} = \frac{5}{6}$

$x = \frac{2 \cdot 6}{5} = \frac{12}{5}$

Now, we need to show that $\frac{2 - x}{x - 3} = \frac{2}{3}$. Substituting $x = \frac{12}{5}$, we get:

$\frac{2 - \frac{12}{5}}{\frac{12}{5} - 3} = \frac{\frac{10 - 12}{5}}{\frac{12 - 15}{5}} = \frac{\frac{-2}{5}}{\frac{-3}{5}} = \frac{-2}{-3} = \frac{2}{3}$.

Therefore, $\frac{2-x}{x-3} = \frac{2}{3}$ is true when $\frac{1}{2}$, $\frac{1}{x}$ and $\frac{1}{3}$ are in A.P.

3. Final Answer

(a) $\frac{x - 4}{x - 3}$

(b) Shown that $\frac{2-x}{x-3} = \frac{2}{3}$.