## Related Topics

- How to Solve Finite Geometric Series
- How to Solve Infinite Geometric Series
- How to Solve Arithmetic Sequences

## 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**

### Geometric Sequences – 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: \(3,-6,12,-24,48\)

### Geometric Sequences – 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: \(x_{n}=ar^{(n – 1)}→a_8=(2.5) (2)^8=2.5(256)=640\)

### Geometric Sequences – 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: \(0.8,-4,20,-100,500\)

### Geometric Sequences – 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 for Solving Geometric Sequences

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

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

### Download Geometric Sequences Worksheet

- \(\color{blue}{r=-5}\)
- \(\color{blue}{r=2}\)
- \(\color{blue}{not \ geometric}\)
- \(\color{blue}{r=5}\)

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