An exponential sequence GP has a positive common ratio. If the sum of the infinity sequence is 25 and the sum of the first two terms is 16, find the (a) 5th term (b) The sum of the first four terms.

Step 1: Set up equations for the given information.
The sum to infinity of a GP is given by $S_{\infty} = \frac{a}{1-r}$, where $a$ is the first term and $r$ is the common ratio.
The sum of the first $n$ terms of a GP is given by $S_n = \frac{a(1-r^n)}{1-r}$.
We are given:

$$S_{\infty} = 25 \implies \frac{a}{1-r} = 25 \quad (*)$$

The sum of the first two terms is $S_2 = a + ar = a(1+r)$.
We are given:

$$S_2 = 16 \implies a(1+r) = 16 \quad (**)$$

Step 2: Solve for $a$ and $r$.
From equation $(*)$, we can express $a$ in terms of $r$:

$$a = 25(1-r)$$

Substitute this expression for $a$ into equation $(**)$:

$$25(1-r)(1+r) = 16$$ $$25(1-r^2) = 16$$ $$1-r^2 = \frac{16}{25}$$ $$r^2 = 1 - \frac{16}{25}$$ $$r^2 = \frac{25-16}{25}$$ $$r^2 = \frac{9}{25}$$ $$r = \pm\sqrt{\frac{9}{25}}$$ $$r = \pm\frac{3}{5}$$

Since the problem states that the common ratio is positive, we take $r = \frac{3}{5}$.
Now substitute $r = \frac{3}{5}$ back into the expression for $a$:

$$a = 25\left(1-\frac{3}{5}\right)$$ $$a = 25\left(\frac{5-3}{5}\right)$$ $$a = 25\left(\frac{2}{5}\right)$$ $$a = 5 \times 2$$ $$a = 10$$

So, the first term is $a=10$ and the common ratio is $r=\frac{3}{5}$.