'n Meetkundige reeks het 'n konstante verhouding van $rac{1}{2}$ en 'n som tot oneindigheid van 6 - NSC Mathematics - Question 3 - 2018 - Paper 1

The general formula for the $n^{th}$ term of a geometric series is: $T_n = ar^{n-1}$ For $n = 8$, substituting the known values: $T_8 = 3 \left( \frac{1}{2} \right)^{8-1} = 3 \left( \frac{1}{2} \right)^{7} = 3 \times \frac{1}{128} = \frac{3}{128}$ Thus, the 8th term of the series is $T_8 = \frac{3}{128}$.

We know: $\sum_{k=1}^{n} 3(2)^{k} = 5.8125$ This is a geometric series with first term $a = 3(2)^{1}$ and common ratio $r = 2$. The sum of the first $n$ terms is given by: $S_n = \frac{a(1 - r^{n})}{1 - r}$ Substituting $a$ and $r$ results in: $S_n = \frac{3(2)(1 - 2^n)}{1 - 2} = 3(2)(1 - 2^n) = 6(1 - 2^n)$ Setting this equal to 5.8125 and solving for $n$ gives: $6(1 - 2^n) = 5.8125 \implies 1 - 2^n = \frac{5.8125}{6} = 0.96875$ Thus: $2^n = 1 - 0.96875 = 0.03125 = \frac{1}{32} \implies n = 5$

The expression can be expanded as follows: $\sum_{k=1}^{20} 24(2)^{k} = 24 \sum_{k=1}^{20} (2)^{k} = 24(2 + 2^2 + \ldots + 2^{20})$ Utilizing the sum formula for a geometric series, we find: $\sum_{k=1}^{20} (2)^{k} = \frac{a(1 - r^{n})}{1 - r}$ Where $a = 2$, $r = 2$, and $n = 20$: $\sum_{k=1}^{20} (2)^{k} = 2 \frac{1 - 2^{20}}{1 - 2} = 2(2^{20} - 1) = 2^{21} - 2$ Thus: $\sum_{k=1}^{20} 24(2)^{k} = 24(2^{21} - 2) = 24 \cdot 2(2^{20} - 1) = 48(2^{20} - 1)$ Since $p = 6 - p = 4$, we express: $4p = 48(2^{20} - 1)$