When the terms of a geometric sequence are added together, a geometric series is formed.

Sequence Versus Series

$2, 4, 8, 16, 32, 64$ is a finite geometric sequence.

$2+4+8+16+32+64$ is a finite geometric series.

Turning a Sequence into a Series

The sum of $n$ terms, $S_n$, of a geometric sequence equals:

$S_n=\dfrac {a(r^{n-1})}{r-1}$ if $r<-1$ or $r>1$ (eg. $r=2, \frac {-3}{2},-2$)

$S_n=\dfrac {a(1-r^n)}{1-r}$ if $-1 (eg. $r=0.2, \frac {1}{8},-0.25$)

Example 1

Find the sum of the first ten terms of the sequence to five decimal places.

$\dfrac{1}{3}, \dfrac{1}{9}.\dfrac {1}{27}\dfrac {1}{81}, ...$

$a=\frac 13$
$n=10$
$r=\frac {1}{9}\div \frac{1}{3}=\frac 13$

As $-1, use the second equation.

$S_n=\dfrac {a(1-r^n)}{1-r}$
$S_{10}=\dfrac {\frac 13(1-\frac 13^{10})}{1-\frac 13}$
$S_{10}=0.4999915$

The question says that we need to answer to five decimal places so:

The sum of the first ten terms is 0.49999.

The sum can be a positive or a negative number.

Example 2

The second term of a geometric is 22 and the fifth term is -176. Find the sum of the first eight terms of the sequence correct to one decimal place.

We need to find the value of $a$ and $r$.

So far we have this information:

[1] $t_2=ar^1=22$

[2] $t_5=ar^4=-176$

To find $r$, we need to solve these two equations simultaneously.

$\frac {ar^4}{ar^1}=\frac {-176}{22}$
$r^3=-8$
$r=-2$

Now that we have the value of $r$, we can substitute it into either of the equations to find the value of $a$.

[1] $ar^1=22$
$a(-2)^1=22$
$-2a=22$
$a=-11$

We know know each of the values so we can substitute these into the equation. As $r<-1$, use the first equation.

$S_n=\dfrac {a(r^{n-1})}{r-1}$
$S_8=\dfrac {-11(-2^{8-1})}{(-2)-1}$
$S_8=\dfrac {-11(-128)}{-3}$
$S_8=-469.33333$