Geometric sequences and seriesEdexcel International A Level Maths: Revision notes

In a geometric sequence each term is the previous term multiplied by a constant common ratio $r$. With first term $a$: $u_n=ar^{n-1}.$ Example: $a=6$, $r=\frac12$: $u_5=6\left(\frac12\right)^4=\frac38$. Given two terms, divide to eliminate $a$: $ar=24$ and $ar^3=6$ give $r^2=\frac14$, so $r=\frac12$ if $r>0$, and then $a=48$.

For $r\ne1$: $S_n=\frac{a(1-r^n)}{1-r}=\frac{a(r^n-1)}{r-1}.$ Use the first form when $|r|<1$ and the second when $r>1$ to keep the numbers positive. Example: $a=6$, $r=\frac12$, $n=8$: $S_8=\frac{6(1-\frac1{256})}{\frac12}=\frac{765}{64}$. For $a=5$, $r=1.2$, $n=15$: $S_{15}=\frac{5(1.2^{15}-1)}{0.2}\approx360$.

If $|r|<1$ the terms shrink towards $0$ and the series converges: $S_\infty=\frac{a}{1-r}.$ If $|r|\ge1$ there is no sum to infinity. Example: $a=6$, $r=\frac12$ gives $S_\infty=12$. The gap between the sum to infinity and a partial sum is $S_\infty-S_n=\frac{ar^n}{1-r}$. When $a$ and $r$ are unknown, use $ar=$ given term and $\frac{a}{1-r}=S_\infty$ to form an equation in $r$; check every solution satisfies $|r|<1$.

When $n$ is an exponent, take logs. To find the first term above $1000$ when $a=5$ and $r=1.2$: $5(1.2)^{n-1}>1000\Rightarrow(n-1)\log1.2>\log200\Rightarrow n-1>29.06$, so $n=31$. If dividing by a logarithm, check the direction of the inequality: when $\log r<0$ (that is, $0<r<1$) the inequality flips. For $96\left(\frac12\right)^n<0.001$: $n\log\frac12<\log\frac{0.001}{96}$, and dividing by the negative $\log\frac12$ gives $n>\frac{\log96\,000}{\log2}=16.55$, so $n=17$.

You must know this proof. $S_n=a+ar+\dots+ar^{n-1}$ $rS_n=ar+ar^2+\dots+ar^n$ Subtract: $S_n-rS_n=a-ar^n$, so $(1-r)S_n=a(1-r^n)$ and $S_n=\frac{a(1-r^n)}{1-r}$ for $r\ne1$. The middle terms cancel, which is why multiplying by $r$ is the key step.