3 a) Simplify (2x+14)/(2x2-98). b) Given that the third and sixth terms of a geometric progression are 3/4 and 3/32 respectively, find the i) first term and the common ratio, ii) nth term, iii) sum to infinity of the progression.

The numerator is $2x+14$. We can factor out 2: $2(x+7)$.
The denominator is $2x^2-98$. We can factor out 2: $2(x^2-49)$.

The expression $x^2-49$ is in the form $a^2-b^2 = (a-b)(a+b)$, where $a=x$ and $b=7$.
So, $x^2-49 = (x-7)(x+7)$.

Cancel out the common factors of 2 and $(x+7)$.

The simplified expression is $\boxed{\frac{1}{x-7}}$.

For $T_3$: $ar^{3-1} = ar^2 = \frac{3}{4}$ (Equation 1)
For $T_6$: $ar^{6-1} = ar^5 = \frac{3}{32}$ (Equation 2)

The first term is $\boxed{a=3}$ and the common ratio is $\boxed{r=\frac{1}{2}}$.

The $n^{th}$ term is $\boxed{3\left(\frac{1}{2}\right)^{n-1}}$.

This formula is valid when $|r| < 1$. Since $r=\frac{1}{2}$, $|r|=\frac{1}{2} < 1$, so the sum to infinity exists.

The sum to infinity of the progression is $\boxed{6}$.