Given the geometric series: $x + 90 + 81 + ...$ 2.1 Calculate the value of $x$: 2.2 Show that the sum of the first $n$ terms is $S_n = 1 000(1-(0.9)^n)$ - NSC Mathematics - Question 2 - 2021 - Paper 1

The sum of the first $n$ terms of a geometric series can be calculated using the formula:

$S_n = \frac{a(1 - r^n)}{1 - r}$ where:

• $a$ is the first term ( (100)), and
• $r$ is the common ratio ($\frac{90}{100} = 0.9$).

Substituting these values into the formula gives:

$S_n = \frac{100(1 - (0.9)^n)}{1 - 0.9} = \frac{100(1 - (0.9)^n)}{0.1} = 1000(1 - (0.9)^n)$

Therefore, we have shown that $S_n = 1000(1 - (0.9)^n)$.

To calculate the sum to infinity of a geometric series, we can use the formula:

$S = \frac{a}{1 - r}$

In this case:

• The first term $a = 100$,
• The common ratio $r = 0.9$.

Thus, substituting these into the formula:

$S = \frac{100}{1 - 0.9} = \frac{100}{0.1} = 1000$

Therefore, the sum to infinity is $1000$.