Geometric Sequences

Learn how to solve Geometric Sequence problems using the following step-by-step guide with detailed solutions.

Step by step guide to solve Geometric Sequence Problems

• It is a sequence of numbers where each term after the first is found by multiplying the previous item by the common ratio, a fixed, non-zero number. For example, the sequence $$2, 4, 8, 16, 32$$, … is a geometric sequence with a common ratio of $$2$$.
• To find any term in a geometric sequence use this formula: $$\color{blue}{x_{n}=ar^{(n – 1)}}$$
• $$a =$$ the first term, $$r =$$ the common ratio, $$n =$$ number of items

Example 1:

Given the first term and the common ratio of a geometric sequence find the first five terms of the sequence. $$a_1=3,r=-2$$

Solution:

Use geometric sequence formula: $$\color{blue}{x_{n}=ar^{(n – 1)}}$$ $$→x_{n}=0.8 .(-5)^{n-1}$$
If $$n=1$$ then: $$x_{1}=3 .(-2)^{1-1}=3 (1)=3$$, First Five Terms: $$3,-6,12,-24,48$$

Example 2:

Given two terms in a geometric sequence find the 8th term. $$a_{3}=10$$ and $$a_{5}=40$$

Solution:

Use geometric sequence formula: $$\color{blue}{x_{n}=ar^{(n – 1)}}$$ $$→a_{3}=ar^{(3 – 1)}=ar^2=10$$
$$x_{n}=ar^{(n – 1)}→a_5=ar^{(5 – 1)}=ar^4=40$$
Now divide $$a_{5}$$ by $$a_{3}$$. Then: $$\frac{a_{5}}{a_{3}} =\frac{ar^4}{ar^2 }=\frac{40}{10}$$, Now simplify: $$\frac{ar^4}{ar^2 }=\frac{40}{10}→r^2=4→r=2$$
We can find a now: $$ar^2=12→a(2^2 )=10→a=2.5$$
Use the formula to find the 8th term: $$x_{n}=ar^{(n – 1)}→a_8=(2.5) (2)^8=2.5(256)=640$$

Example 3:

Given the first term and the common ratio of a geometric sequence find the first five terms of the sequence. $$a_{1}=0.8,r=-5$$

Solution:

Use geometric sequence formula: $$\color{blue}{x_{n}=ar^{(n – 1)}}$$ $$→x_{n}=0.8 .(-5)^{n-1}$$
If $$n=1$$ then: $$x_{1}=0.8 .(-5)^{1-1}=0.8 (1)=0.8$$, First Five Terms: $$0.8,-4,20,-100,500$$

Example 4:

Given two terms in a geometric sequence find the 8th term. $$a_3=12$$ and $$a_5=48$$

Solution:

Use geometric sequence formula: $$\color{blue}{x_{n}=ar^{(n – 1)}}$$ $$→a_3=ar^{(3 – 1)}=ar^2=12$$
$$\color{blue}{x_{n}=ar^{(n – 1)}}$$ $$→a_5=ar^{(5 – 1)}=ar^4=48$$
Now divide $$a_{5}$$ by $$a_{3}$$. Then: $$\frac{a_{5}}{a_{3} }=\frac{ar^4}{ar^2}=\frac{48}{12}$$, Now simplify: $$\frac{ar^4}{ar^2}=\frac{48}{12}→r^2=4→r=2$$
We can find a now: $$ar^2=12→a(2^2 )=12→a=3$$
Use the formula to find the $$8^{th}$$ term: $$\color{blue}{x_{n}=ar^{(n – 1)}}$$ $$→a_{8}=(3) (2)^8=3(256)=768$$

Exercises

Determine if the sequence is geometric. If it is, find the common ratio.

• $$\color{blue}{1, – 5, 25, – 125, …}$$
• $$\color{blue}{– 2, – 4, – 8, – 16, …}$$
• $$\color{blue}{4, 16, 36, 64, …}$$
• $$\color{blue}{– 3, – 15, – 75, – 375, …}$$

• $$\color{blue}{r=-5}$$
• $$\color{blue}{r=2}$$
• $$\color{blue}{not \ geometric}$$
• $$\color{blue}{r=5}$$