3a) Find the range of value of $x$ for which the series $\sum_{r=1}^{\infty} \left(\frac{3}{x+2}\right)^r$ converges. Step 1: Identify the common ratio of the geometric series. A geometric series $\sum_{r=1}^{\infty} ar^{r-1}$ converges if and only if the common ratio $r$ satisfies $|r| < 1$. For the given series $\sum_{r=1}^{\infty} \left(\frac{3}{x+2}\right)^r$, the common ratio is $r = \frac{3}{x+2}$. Step 2: Set up the inequality for convergence. $$\left|\frac{3}{x+2}\right| < 1$$ Step 3: Solve the inequality. This inequality can be rewritten as $3 < |x+2|$. This implies two separate inequalities: $$x+2 > 3 \quad \text{or} \quad x+2 < -3$$ Step 4: Solve each inequality for $x$. For the first inequality: $$x+2 > 3$$ $$x > 3-2$$ $$x > 1$$ For the second inequality: $$x+2 < -3$$ $$x < -3-2$$ $$x < -5$$ The range of values for $x$ for which the series converges is $x < -5$ or $x > 1$. The range of values for $x$ is $\boxed{x < -5 \text{ or } x > 1}$. 3b) The sum of the first $n$ terms of a series is $S_n = 5(2^n - 1)$. Find the general term ($a_n$) and show that the sequence is a G.P. Step 1: Find the general term $a_n$. 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 $\boxed{a_n = 5 \cdot 2^{n-1}}$. Step 2: Show that the sequence is a Geometric Progression (G.P.). 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.