The Sum of a Finite Geometric Sequence

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<r<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
r=\frac {1}{9}\div \frac{1}{3}=\frac 13

As -1<r<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}

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}

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

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}

The question asked for the answer correct to one decimal place.

The sum of the first eight terms of the geometric series is -469.3.

See also

Geometric Sequences
Finding the Terms of a Geometric Sequence
Sum of an Infinite Geometric Sequence