AHL 1.11—Sum of infinite geometric sequences

For an infinite geometric series ${\sum }_{n=1}^{\infty }{u}_1{r}^{n-1}$ to have a finite sum, the terms $r^n$ must approach zero as $n\to \infty$.

The expression $r^n\to 0$ as $n\to \infty$ if and only if $|r|<1$. This is the precise mathematical condition needed.

The condition $|r|<1$ accounts for both positive and negative values of $r$ (e.g. $r=-0.5$ also gives convergence). No other listed condition is equivalent.

The correct answer is $|r|<1$.

We need the condition that guarantees a finite sum to infinity for a geometric series.

$r>0$ does not guarantee convergence. For example, $r=2>0$ gives a divergent series with terms growing without bound.

$r<1$ is insufficient — it allows values like $r=-5$, where $|r|=5>1$ and the series diverges.

The first term $u_1$ has no bearing on convergence. A series with $u_1=0.001$ and $r=3$ still diverges.

$|r|<1$ is the correct and complete convergence condition, covering both positive and negative ratios.

The first term is $u_1=18$. The common ratio is $r=\frac{6}{18}=\frac{1}{3}$.

$|r|=\frac{1}{3}<1$, so the series converges and $S_{\infty }$ exists.

$S_{\infty }=\frac{u_1}{1-r}=\frac{18}{1-\frac{1}{3}}=\frac{18}{\frac{2}{3}}=18\times \frac{3}{2}=27$

The sum to infinity is $27$.

$u_1=18$, $r=\frac{1}{3}$. We need $S_{\infty }=\frac{18}{1-\frac{1}{3}}$.

$24$ would result from incorrectly summing just the first few terms or using the wrong denominator, e.g. $\frac{18}{\frac{3}{4}}$. This is not consistent with $r=\frac{1}{3}$.

$20$ does not arise from the formula $\frac{18}{\frac{2}{3}}$; it reflects an arithmetic error.

$36=18\times 2$ could arise from using $1-r=\frac{1}{2}$ instead of $\frac{2}{3}$, which is incorrect for $r=\frac{1}{3}$.

$\frac{18}{\frac{2}{3}}=27$, so the answer is $27$.