The sum of the first n terms of a series is Sn = 5(2ⁿ − 1). Find the general term (an) and show that the sequence is a G.P.

The general term $a_n$ is given by $a_n = S_n - S_{n-1}$ for $n > 1$, and $a_1 = S_1$.

First, find $a_1$:

$$S_1 = 5(2^1 - 1) = 5(2-1) = 5(1) = 5$$

So, $a_1 = 5$.

Next, find $S_{n-1}$:

$$S_{n-1} = 5(2^{n-1} - 1)$$

Now, find $a_n$ for $n > 1$:

$$a_n = S_n - S_{n-1}$$ $$a_n = 5(2^n - 1) - 5(2^{n-1} - 1)$$ $$a_n = 5 \left[ (2^n - 1) - (2^{n-1} - 1) \right]$$ $$a_n = 5 \left[ 2^n - 1 - 2^{n-1} + 1 \right]$$ $$a_n = 5 \left[ 2^n - 2^{n-1} \right]$$ $$a_n = 5 \left[ 2 \cdot 2^{n-1} - 2^{n-1} \right]$$ $$a_n = 5 \left[ (2-1) \cdot 2^{n-1} \right]$$ $$a_n = 5 \cdot 2^{n-1}$$

This formula also holds for $n=1$ ($a_1 = 5 \cdot 2^0 = 5$).

The general term is $\boxed{a_n = 5 \cdot 2^{n-1}}$.

A sequence is a G.P. if the ratio of consecutive terms is constant. This constant ratio is the common ratio, $r$.
We need to show that $\frac{a_n}{a_{n-1}}$ is a constant for all $n > 1$.

Using the general term $a_n = 5 \cdot 2^{n-1}$:

$$a_{n-1} = 5 \cdot 2^{(n-1)-1} = 5 \cdot 2^{n-2}$$

Now, calculate the ratio:

$$\frac{a_n}{a_{n-1}} = \frac{5 \cdot 2^{n-1}}{5 \cdot 2^{n-2}}$$ $$= \frac{2^{n-1}}{2^{n-2}}$$ $$= 2^{(n-1) - (n-2)}$$ $$= 2^{n-1-n+2}$$ $$= 2^1$$ $$= 2$$

Since the ratio of consecutive terms is a constant value of 2, the sequence is a Geometric Progression.