Geometric sequences and series

A geometric sequence has a common ratio $r=\frac{T_2}{T_1}=\frac{T_3}{T_2}$.

$T_n=a{r}^{n-1}$

Example. $3;6;12;\dots$ has $a=3$ and $r=2$, so $T_{10}=3\times 2^9=1536$.

Two terms given. If $T_2=12$ and $T_5=96$: $\frac{a{r}^4}{ar}=r^3=8$, so $r=2$ and $a=6$.

Which term? Set $T_n$ equal to the value. Match powers or use logarithms: $2\cdot 3^{n-1}=39366\Rightarrow 3^{n-1}=19683=3^9\Rightarrow n=10$.

$S_n=\frac{a(r^n-1)}{r-1},r\ne 1$

Proof (examinable).

$S_n=a+ar+a{r}^2+\dots +a{r}^{n-1}$

$r{S}_n=ar+a{r}^2+\dots +a{r}^{n-1}+a{r}^n$

Subtracting, all middle terms cancel:

$r{S}_n-S_n=a{r}^n-a\Rightarrow S_n(r-1)=a(r^n-1)\Rightarrow S_n=\frac{a(r^n-1)}{r-1}$

Example. With $a=6$ and $r=2$: $S_8=\frac{6(2^8-1)}{1}=1530$.

Unknown $n$. How many terms of $2+6+18+\dots$ must be added for the sum to exceed 1 000 000?

$3^n-1>1000000\Rightarrow n>\frac{\log 1000001}{\log 3}\approx 12,58$

So 13 terms are needed.

For a geometric sum written in sigma notation, write out the first few terms to find $a$, $r$ and $n$.

Example. $\sum _{k=1}^{6}2(3{)}^k$

• $k=1$: $6$; $k=2$: $18$, so $a=6$ and $r=3$
• $n=6-1+1=6$

$S_6=\frac{6(3^6-1)}{3-1}=2184$

Do not take $a$ from the number in front of the bracket. Here the multiplier is 2, but the first term is 6.