How to Find the Sum of a Geometric Sequence
A geometric sequence is a sequence of numbers where each term is found by multiplying the previous term by a set amount. We call this set amount the 'common ratio'.
For example; 2, 4, 8, 16, 32, 64, … is a geometric sequence that starts with two and has a common ratio of two.
6, 30, 150, 750, … is a geometric sequence starting with six and having a common ratio of five.
You can also have fractional multipliers such as in the sequence 48, 24, 12, 6, 3, … which has a common ratio of 1/2.
Sometimes, we want to find the sum of the first however many terms of a geometric sequence. If there aren't many terms to count, this is nice and easy. However, if you want to quickly add the first 50 terms, for example, adding them manually would take a long time. We want a shortcut. Using some algebra and a clever trick, we can create a formula to quickly find the sum no matter how many terms you are counting.
To create this formula, we must first see that any geometric sequence can be written in the form a, ar, ar^2, ar^3, … where a is the first term and r is the common ratio. Notice that because we start with a, and the ratio, r, is only involved from the second term onwards, the nth term = ar^n−1. For example, the 6th term = ar^5, the 100th term = ar^99, and so on.
We, therefore, have that the sum of the first n terms, Sn, is given by the following formula:
Sn = a + ar + ar^2 + ar^3 + ... + ar^n−2 + ar^n−1
If we multiply both sides by r, we get:
rSn = ar + ar^2 + ar^3 + ... + ar^n−1 + ar^n
If we subtract the second equation from the first equation we can see from the diagram below that we will get ar − ar, ar^2 − ar^2, and so on. In fact, most of the terms on the right will cancel out, leaving us with just a − ar^n.
For example; 2, 4, 8, 16, 32, 64, … is a geometric sequence that starts with two and has a common ratio of two.
6, 30, 150, 750, … is a geometric sequence starting with six and having a common ratio of five.
You can also have fractional multipliers such as in the sequence 48, 24, 12, 6, 3, … which has a common ratio of 1/2.
Sometimes, we want to find the sum of the first however many terms of a geometric sequence. If there aren't many terms to count, this is nice and easy. However, if you want to quickly add the first 50 terms, for example, adding them manually would take a long time. We want a shortcut. Using some algebra and a clever trick, we can create a formula to quickly find the sum no matter how many terms you are counting.
To create this formula, we must first see that any geometric sequence can be written in the form a, ar, ar^2, ar^3, … where a is the first term and r is the common ratio. Notice that because we start with a, and the ratio, r, is only involved from the second term onwards, the nth term = ar^n−1. For example, the 6th term = ar^5, the 100th term = ar^99, and so on.
We, therefore, have that the sum of the first n terms, Sn, is given by the following formula:
Sn = a + ar + ar^2 + ar^3 + ... + ar^n−2 + ar^n−1
If we multiply both sides by r, we get:
rSn = ar + ar^2 + ar^3 + ... + ar^n−1 + ar^n
If we subtract the second equation from the first equation we can see from the diagram below that we will get ar − ar, ar^2 − ar^2, and so on. In fact, most of the terms on the right will cancel out, leaving us with just a − ar^n.
By cancelling terms, we are now left with:
Sn − rSn = a − ar^n
Factorising both sides gives us:
(1 − r)Sn = a(1 − r^n)
Dividing both sides by (1 − r) gives us the final formula of:
Sn = a(1 − r^n)/(1 − r)
where a is the first term of the sequence and r is the common ratio.
Sn − rSn = a − ar^n
Factorising both sides gives us:
(1 − r)Sn = a(1 − r^n)
Dividing both sides by (1 − r) gives us the final formula of:
Sn = a(1 − r^n)/(1 − r)
where a is the first term of the sequence and r is the common ratio.
Using the Formula
Take the sequence 2, 6, 18, 54, 162, … . We can see quickly that a = 2. To find the common ratio simply divide any term by the previous term so r = 6 ÷ 2 = 3.
If we wanted to find the sum of the first ten terms using our formula we would get:
Sn = a(1 − r^n) / (1 − r)
S10 = 2(1 − 3^10) / (1 − 3)
= 2 × -59048 / -2
= 59 048
If we wanted to find the sum of the first ten terms using our formula we would get:
Sn = a(1 − r^n) / (1 − r)
S10 = 2(1 − 3^10) / (1 − 3)
= 2 × -59048 / -2
= 59 048
Summing a Geometric Sequence to Infinity
For any geometric sequence with a common ratio between -1 and 1, we can see that the terms will get smaller in absolute size as the sequence progresses (if you multiply a number by a number between -1 and 1, the magnitude will decrease).
As the terms get smaller and smaller, there comes a point where adding them makes a negligible difference to the total and the sum just ends up tending towards a particular value but never quite reaching it or surpassing it. We call this limit the 'sum to infinity', and we can adapt our formula to find out what this is.
We have the formula Sn = a(1 − r^n) / (1 − r)
If -1 < r < 1, then as n → ∞, r^n → 0. Therefore, as we approach infinity, the r^n on the top row of our fraction disappears, and so we get:
S∞ = a(1 − 0) / (1 − r)
= a / (1 − r)
As the terms get smaller and smaller, there comes a point where adding them makes a negligible difference to the total and the sum just ends up tending towards a particular value but never quite reaching it or surpassing it. We call this limit the 'sum to infinity', and we can adapt our formula to find out what this is.
We have the formula Sn = a(1 − r^n) / (1 − r)
If -1 < r < 1, then as n → ∞, r^n → 0. Therefore, as we approach infinity, the r^n on the top row of our fraction disappears, and so we get:
S∞ = a(1 − 0) / (1 − r)
= a / (1 − r)
Using the Formula
Take the sequence 1, 1/2, 1/4, 1/8, 1/16, … which has a = 1 and r = 1/2. As -1 < r < 1, we can find the sum to infinity of this sequence.
S∞= a / (1 − r)
=1 / (1 − 1/2)
= 2
So if we do the sum 1 + 1/2 + 1/4 + 1/8 + 1/16 + … our answer tends towards 2.
S∞= a / (1 − r)
=1 / (1 − 1/2)
= 2
So if we do the sum 1 + 1/2 + 1/4 + 1/8 + 1/16 + … our answer tends towards 2.