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If you have the sequence 2, 8, 14, 20, 26, then each term is 6 more than the previous term. This is an example of an **arithmetic progression (AP)** and the constant value that defines the difference between any two consecutive terms is called the **common difference**.

If an arithmetic difference has a first term **a** and a common difference of **d**, then we can write

**a, (a + d), (a + 2d),... {a + (n-1) d}**

where the **n ^{th} term = a + (n−1)d**

**Sum of Arithmetic series**

The sum of an arithmetic series of **n** terms is found by making n/2 pairs each with the value of the sum of the first and last term. *(Try this with the sum of the first 10 integers, by making 5 pairs of 11.)*

**This gives us the formula:**

where *a* = first term and *l* = last term.

As the last term is the n^{th} term = a + (n − 1)d we can rewrite this as:

(Use the first formula if you know the first and last terms; use the second if you know the first term and the common difference.)

If you have a sequence such as: **81, 27, 9, 3, 1, 1/3, 1/9,...** then each term is one third of the term before.

This can be written as **81, 81(1/3), 81(1/3) ^{2}, 81(1/3)^{3}, 81(1/3)^{4},...**

It is an example of a **Geometric Progression (GP)** where the each term is a multiple of the previous one. The multiplying factor is called the **common ratio**.

So a GP with a first term **a** and a common ratio **r** with **n** terms, can be stated as

a, ar, ar^{2}, ar^{3}, ar^{4}...ar^{n-1} , **where the n ^{th} term = ar^{n-1}**

*Example:*

In the sequence, 400, 200, 100, 50,... find the 8^{th} term.

a = 400, r = 0.5 and so the 8^{th} term = 400 × 0.5^{7} = 3.125

**Note:** To find which term has a certain value you will need to use **logarithms**.

*Example:*

**In the sequence, 2, 6, 18, 54 ... which is the first term to exceed 1,000,000?**

a = 2, r = 3.

2 × 3^{n-1} > 1,000,000

3^{n-1} > 500000

(n − 1) log 3 > log 500000

n > 12.94

Therefore:

n = 13

*Example:*

**In the earlier sequence, 400, 200, 100, 50 ... which is the first term that is less than 1?**

400 × 0.5^{(n-1]} < 1

0.5^{(n-1)} < 0.0025

(n-1) log 0.5 < log 0.0025

Therefore:

n > 9, or n = 10

**Note:** The inequality sign changed because we divided by a negative (log 0.5 < 0)

**Sum of Geometric series**

The sum of the terms can be written in two ways.

where a = first term, r = common ratio and r ≠ 1.

(use this formula when r < 1).

*Example:*

Evaluate,

(**Note:** there are 9 terms.)

The first term is when n = 2

(i.e 2.36^{2} = 5.5696)

Using the formula for the sum of a geometric progression gives:

which is approximately 9300 (to 3 s.f.).

**Convergence**

**The sum of an infinite series exists if:**

-1 < r < 1 or | r | < 1

This is because each successive term is getting smaller and so the series will tend towards a certain limit. This limit is found using the second of our two formulae:

If | r | < 1 then as n → ∞, r^{n} → 0

and so:

*Example:*

the series 1/3 + (1/3)^{2} + (1/3)^{3} + (1/3)^{4} + ... converges and its sum is 1 as n approaches ∞.

(A sequence such as n^{3} has the first 6 terms as 1 + 8 + 27 + 64 + 125 + 216. As n approaches infinity, the sum also increases. Therefore, it is not convergent. This series is **divergent**.

Every AP has a sum that approaches infinity as n increases, so every AP is divergent.)

*Example*

Find 1 - 1/2 + 1/4 - 1/8 + ...

1 - 1/2 + 1/4 - 1/8 + ... = 1 + (-1/2) + (-1/2)^{2} + (-1/2)^{3} + ...

This is a geometric progression where r = -½, so | r | < 1.

Therefore this series converges to:

Two final pieces of information that may be useful:

**Arithmetic mean**

The **arithmetic mean** of two numbers m and n is given by:

**Arithmetic mean** = ½(m+n)

This is the way of finding a missing term in between two known terms.

*Example:*

The 4^{th} term of an AP is 14, the 6^{th} term is 22. The 5^{th} term will be the **Arithmetic Mean** of these two values.

i.e. (14 + 22)/2 = 18 (here d = 4 and a = 2).

**Geometric mean**

The **geometric mean** of two numbers m and n is given by:

**Geometric mean** = √(mn)

This represents the value between two others in a GP.

*Example:*

The 7^{th} term of a GP is 6, the 9^{th} is 1.5. The 8^{th} term is:

√(6×1.5) = √9 = 3

Here r = 0.5 and a = 384.