The geometric sequence calculator lets you calculate various important values for an geometric sequence. You can calculate the first term, nthterm, common ratio, sum of n terms, number of terms, or position of a term in the geometric sequence.
The calculator will not only give you the answer but also a step-by-step solution.
Here's a quick overview of what a geometric sequence is.
Geometric Sequence
A geometric sequence is a sequence of numbers such that the ratio between any two consecutive terms is constant.
Here's an example of an geometric sequence.
2,6,18,54,162,486,...
The constant ratio is known as the common ratio and is usually denoted by r. For the sequence above, the common ratio would be 6/2=3.
The nth term of an geometric sequence is denoted by an.
General Term
For a geometric sequence with a first term a1 and common ratio r, the nth term is given by —
an=a1⋅rn−1
Sum of Terms
The sum of n terms (denoted by Sn) of a geometric sequence would be
Sn=1−ra(1−rn),r=1
A Special Case
Consider a geometric sequence with a common difference is between −1 and 1. So ∣r∣<1.
Now, as n becomes larger, rn becomes smaller. As n approaches infinity rn approaches 0.
Putting it all together, for an
infinite geometric sequence with ∣r∣<0
, the sum of terms S is
S=1−ra(1−rn)=1−ra(1−0)=1−ra
Examples
Let's look at a couple of simple examples.
Example
Given the first term a1=4 and the common ratio r=2, find the 7th term of the geometric sequence.
Solution
We know the nth term of a geometric progression is given by —
an=a1⋅rn−1
Substituting the values of a1, r, and n, we have
a7=4⋅27−1=256
So the 7th term of the geometric sequence is 256.
Example
The 5th term of a geometric sequence is 162 and the 8th is 6. Find the sum of this infinite geometric sequence.
Solution
The first thing to note is that the sum of an infinite geometric sequence exists only if the common ratio is smaller than 1 in absolute value. So, let's calculate r and see whether this condition is satisfied.
Here's how we can calculate r. We know
an=a1⋅rn−1
The question gives us the values of the 5th and the 8th terms. We can use them to get two equations.
a5162=a1⋅25−1=a1⋅r4(1)
a86=a1⋅28−1=a1⋅r7(2)
Dividing the second equation by the first, we have
1626271=a1⋅r4a1⋅r7=r3
Taking the cube root on both sides,
r=31
Great, since r<1, the sum of the infinite sequence exists. And as we saw earlier, it is given by the formula —
S=1−ra1
Now, to be able to use this formula, we need to find a1 too. For that, we can use the equation (1).
162162a1=a1⋅r4=a1⋅(31)4=13122
Finally, substituting the values of a1 and r in the formula for sum, we get
S=1−ra1=1−3113122=19683
Done.