Arithmetic and Geometric Sequences

Arithmetic Sequences

An arithmetic sequence is a type of number sequence that has a constant difference between consecutive terms, known as the common difference.

To identify an arithmetic sequence, examine the differences between each consecutive term. If this difference remains constant, then the sequence is arithmetic. For example, the sequence 4, 6, 8, 10, 12 is an arithmetic sequence with a common difference of 2, as each term is 2 greater than the previous term.

To find the Nᵗʰ term of an arithmetic sequence, we can use the formula:

Nᵗʰ term = a + (n - 1)d

where ‘a‘ is the first term, ‘n‘ is the term number, and ‘d‘ is the common difference.

Let’s look at an example in more detail:

Example

Consider the arithmetic sequence 2, 6, 10, 14, 18. In this case, the first term (a) is 2, and the common difference (d) is 4, as each term is 4 greater than the previous term.

To find the Nᵗʰ term of this sequence, we can apply the formula:

Nᵗʰ term = 2 + (n - 1)4

= 2 + 4n - 4 = 4n - 2

Now that we have the Nth term formula, we can find any term in the sequence without having to list all the previous terms. For example, if we want to find the 10th term, we can simply substitute n = 10 into the formula:

10ᵗʰ term = 4(10) - 2

40 - 2

= 38

Let’s look at some more examples:

First Term (a)Term Rule (Nth term formula)First 5 Terms
2an = 2 + 3(n - 1)2, 5, 8, 11, 14
4an = 4 + 7(n - 1)4, 11, 18, 25, 32
10an = 10 + 2(n - 1)10, 12, 14, 16, 18
-3an = -3 + 5(n - 1)-3, 2, 7, 12, 17
6an = 6 - 4(n - 1)6, 2, -2, -6, -10

  • an is the nᵗʰ term

Geometric Sequences

Geometric sequences are a type of number sequence that have a constant ratio between consecutive terms, known as the common ratio.

To identify a geometric sequence, look at the ratios between each consecutive term. If this ratio remains constant, then the sequence is geometric. For example, the sequence 1, 3, 9, 27, 81 is a geometric sequence with a common ratio of 3, as each term is 3 times greater than the previous term.

To find the Nth term of a geometric sequence, we can use the formula:

Nᵗʰ term = ar^{(n-1)}

where ‘a‘ is the first term, ‘r‘ is the common ratio, and ‘n‘ is the term number.

Let’s look at an example in more detail:

Example

Consider the geometric sequence 5, 15, 45, 135, 405. In this case, the first term (a) is 5, and the common ratio (r) is 3, as each term is 3 times greater than the previous term.

To find the Nᵗʰ term of this sequence, we can apply the formula:

Nᵗʰ term = 5 \times 3^{(n-1)}

With the Nᵗʰ term formula, we can find any term in the sequence without having to list all the previous terms. For example, if we want to find the sixth term, we can simply substitute n = 6 into the formula:

6ᵗʰ term = 5 \times 3^{(6-1)}

= 5 \times 3^5

= 5 \times 243

= 1215

First Term (a)Term Rule (Nth term formula)First 5 Terms
1an = 1 \times 2^(n - 1)1, 2, 4, 8, 16
3an = 3 \times 3^(n - 1)3, 9, 27, 81, 243
2an = 2 \times 0.5^(n - 1)2, 1, 0.5, 0.25, 0.125
5an = 5 \times 4^(n - 1)5, 20, 80, 320, 1280
7an = 7 \times (-2)^(n - 1)7, -14, 28, -56, 112

  • an is the nᵗʰ term

Examples

Example 1:

The first three terms of an arithmetic sequence are 5, 9, and 13. Find the 8th term in the sequence.

Step 1: Identify the first term (aa) and the common difference (dd) of the arithmetic sequence. a = 5a = 5 (first term) d = 9 - 5 = 4d = 9 - 5 = 4 (common difference)

Step 2: Use the Nᵗʰ term formula for an arithmetic sequence.

Nᵗʰ term = a + (n - 1)da + (n - 1)d

Step 3: Find the 8ᵗʰ term (when n = 8).

8ᵗʰ term = 5 + (8 - 1) \times 4 = 5 + 7 \times 45 + (8 - 1) \times 4 = 5 + 7 \times 4

= 5 + 28 = 33= 5 + 28 = 33

The 8ᵗʰ term in the arithmetic sequence is 33.

Example 2:

The first three terms of a geometric sequence are 6, 18, and 54. Find the 5th term in the sequence.

Step 1: Identify the first term (aa) and the common ratio (rr) of the geometric sequence. a = 6a = 6 (first term) r = \frac{18}{6} = 3r = \frac{18}{6} = 3 (common ratio)

Step 2: Use the Nᵗʰ term formula for a geometric sequence. Nᵗʰ term = a \times r^{(n-1)}a \times r^{(n-1)}

Step 3: Find the 5ᵗʰ term (when n = 5).

5ᵗʰ term = 6 \times 3^{(5-1)}= 6 \times 3^{(5-1)}

= 6 \times 3^4= 6 \times 3^4

= 6 \times 81 = 486= 6 \times 81 = 486

The 5ᵗʰ term in the geometric sequence is 486.

Example 3:

A person saves £5 in the first week and increases their savings by £2 each week. Find the total savings after 10 weeks.

Step 1: Recognise that this forms an arithmetic sequence with a first term (aa) of £5 and a common difference (dd) of £2.

Step 2: Find the Nᵗʰ term formula for this arithmetic sequence.

Nth term = a + (n - 1)d= a + (n - 1)d

In our example:

Nth term = 5 + (n - 1)2= 5 + (n - 1)2

Step 3: Find the savings in the 10ᵗʰ week (when n = 10).

Nᵗʰ term = 5 + (10 - 1)2= 5 + (10 - 1)2

= 5 + 9 \times 2= 5 + 9 \times 2

= 5 + 18 == 5 + 18 = £2323

So, the savings in the 10ᵗʰ week is £23.

Step 4: Calculate the total savings after 10 weeks using the arithmetic series formula.

The arithmetic series formula is:

Sum = \frac{n \times (a + l)}{2}= \frac{n \times (a + l)}{2}

where ‘n’ is the number of terms (10 weeks in this case), ‘a’ is the first term (£5), and ‘l’ is the last term (the savings in the 10ᵗʰ week, £23).

Sum = \frac{10 \times (5 + 23)}{2}= \frac{10 \times (5 + 23)}{2}

= \frac{10 \times 28}{2}= \frac{10 \times 28}{2}

= \frac{280}{2} == \frac{280}{2} = £140140

The total savings after 10 weeks is £140.

Example 4:

A population of bacteria starts with 100 individuals and doubles every hour. Find the population size after 4 hours.

Step 1: Identify the initial population (a)(a) and the common ratio (r)(r):

a = 100a = 100 (initial population) r = 2r = 2 (the population doubles every hour)

Step 2: Use the Nᵗʰ term formula for a geometric sequence:

Nᵗʰ term = ar^{(n-1)}ar^{(n-1)}

In this case, we want to find the population after 4 hours, so n = 4.

Step 3: Plug in the values into the formula:

Population after 4 hours = 100 \times 2^{(4-1)}100 \times 2^{(4-1)}

Step 4: Calculate the result:

Population after 4 hours = 100 \times 2^3100 \times 2^3

= 100 \times 8 = 800= 100 \times 8 = 800

After 4 hours, the population size will be 800 bacteria.

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