1. The problem asks to find the sum of the first $n$ terms of several geometric progressions (G.P.). The sum of the first $n$ terms of a G.P. is given by $$S_n = a \frac{1-r^n}{1-r}$$ where $a$ is the first term and $r$ is the common ratio. --- a) Series: $\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots + \frac{1}{1024}$ - First term $a = \frac{1}{2}$ - Common ratio $r = \frac{1}{4} \div \frac{1}{2} = \frac{1}{2}$ - Last term $\frac{1}{1024} = \left(\frac{1}{2}\right)^n$ implies $n = 10$ - Sum: $$S_{10} = \frac{1}{2} \frac{1-\left(\frac{1}{2}\right)^{10}}{1-\frac{1}{2}} = \frac{1}{2} \frac{1-\frac{1}{1024}}{\frac{1}{2}} = 1 - \frac{1}{1024} = \frac{1023}{1024}$$ b) Series: $4 + 8 + 22 + 112 + \cdots$ (find sum to 6 terms) - Verify if G.P.: $8/4=2$, $22/8=2.75$, not constant, so not G.P. - Since series is not geometric, sum of G.P. cannot be found here. c) Series: $10 - 24 + 60 + \cdots$ to 7 terms - Check ratio: $-24/10 = -2.4$, $60/-24 = -2.5$, not constant - Not G.P.; sum cannot be found. d) Series: $1 + 0.2 + 0.04 + \cdots$ to 5 terms - $a=1$, $r=0.2$, $n=5$ - Sum: $$S_5 = 1 \times \frac{1-0.2^5}{1-0.2} = \frac{1 - 0.00032}{0.8} = \frac{0.99968}{0.8} = 1.2496$$ --- e) Series: $54 - 18 + 6 - \cdots$ to the term $\frac{2}{9}$ - $a=54$, $r = -18/54 = -\frac{1}{3}$ - Find $n$: last term $t_n = a r^{n-1} = \frac{2}{9}$ - $$54 \left(-\frac{1}{3}\right)^{n-1} = \frac{2}{9}$$ - $$\left(-\frac{1}{3}\right)^{n-1} = \frac{2}{9} \times \frac{1}{54} = \frac{2}{486} = \frac{1}{243}$$ - Since $\left(-\frac{1}{3}\right)^5 = -\frac{1}{243}$, and we want positive $\frac{1}{243}$, exponent even, so $(n-1)=5$ or $(n-1)=10$... Checking positive power gives $n-1=5$ or 10? - By absolute value, $|r|=\frac{1}{3}$, so $\left(\frac{1}{3}\right)^{n-1} = \frac{1}{243} = \left(\frac{1}{3}\right)^5$ so $n-1=5 \Rightarrow n=6$ - Sum for $n=6$: $$S_6 = 54 \frac{1-\left(-\frac{1}{3}\right)^6}{1 - (-\frac{1}{3})} = 54 \frac{1 - \frac{1}{729}}{1 + \frac{1}{3}} = 54 \frac{\frac{728}{729}}{\frac{4}{3}} = 54 \times \frac{728}{729} \times \frac{3}{4} = 54 \times \frac{728 \times 3}{729 \times 4}$$ - Simplify: $\frac{54}{729} = \frac{2}{27}$ - So $$S_6 = \frac{2}{27} \times \frac{728 \times 3}{4} = \frac{2 \times 728 \times 3}{27 \times 4} = \frac{4368}{108} = 40.4444$$ approx. f) Series: $45 + 30 + 20 + \cdots$ to 6 terms - $a=45$, $r=30/45=\frac{2}{3}$, $n=6$ - Sum: $$S_6 = 45 \frac{1 - (\frac{2}{3})^6}{1 - \frac{2}{3}} = 45 \frac{1 - \frac{64}{729}}{\frac{1}{3}} = 45 \times (1 - 0.0878) \times 3 = 45 \times 0.9122 \times 3 = 123.63$$ approx. --- **Final answers:** a) $\frac{1023}{1024}$ d) 1.2496 e) approx. 40.4444 f) approx. 123.63 Parts b) and c) are not geometric progressions, so no sums computed.