# 1. Arithmetic Progressions

by M. Bourne

We want a sequence of numbers. Let's start with a number: a_1.

Now add a number d, (for "difference").

We get a_1 + d and the first 2 terms in our sequence are:

a_1,
a_1 + d

For the next term, let's add another d to that last term and we have a_1 + 2d.

Our sequence is now:

a_1,
a_1 + d,
a_1 + 2d

We continue this process for as long as we can stay awake. The resulting set of numbers is called an arithmetic progression (AP) or arithmetic sequence.

### Example 1

Let's start with a_1 = 4 and then add d=3 each time to get each new number in the sequence. We get:

4, 7, 10, 13, ...

Continues below

## General Term of an Arithmetic Progression

The nth term, a_n of an AP is:

a_n=a_1+(n-1)d

## Sum of an Arithmetic Progression

The sum to n terms of an AP is:

S_n=n/2(a_1+a_n)\ or\   S_n=n/2[2a_1+(n-1)d]

Proofs of sum formulas

### First Formula

The sum of the first term is, well, simply the first term:

S_1 = a_1

The sum of the first 2 terms is:

S_2 = a_1 + (a_1 + d) = 2a_1 + d

Of course, S_2 also equals:

S_2 = a_1 + a_2

We can write this answer as 2/2[a_1 + a_2].

(You'll see why we do this soon).

Putting the above 2 results together, we get:

S_2 = 2a_1 + d = 2/2[a_1 + a_2]

Now, the sum of the first 3 terms is:

S_3 = a_1 + ( a_1 + d) +  (a_1 + 2d)  = 3a_1 + 3d

And we examine some more relationships:

a_3 = a_1 + 2d

(This follows from the definition of an arithmetic progression.)

and

a_1 + a_3 = a_1 + (a_1 + 2d)  = 2a_1 + 2d

And now, look at these two results:

S_3 = 3a_1 + 3d

a_1 + a_3 = 2a_1 + 2d

Multiplying the last line throughout by 3/2 gives

3/2(a_1 + a_3) = 3/2(2a_1 + 2d)  = 3a_1 + 3d = S_3

So we can write the answer for the sum of 3 terms as

S_3 = 3a_1 + 3d  = 3/2[a_1 + a_3]

The sum of the first 4 terms is:

S_4 = a_1 + (a_1 + d) +  (a_1 + 2d) +  (a_1 + 3d)  = 4a_1 + 6d

Now, a_4 = a_1 + 3d and

a_1 + a_4 = a_1 + (a_1 + 3d)  = 2a_1 + 3d

Using a similar idea to the step above, we can write the answer for the sum of 4 terms as

S_4 = 4a_1 + 6d  = 4/2[a_1 + a_4]

In general, the sum to n terms is:

S_n = n/2[a_1 + a_n]

### Second Formula

From above, the sum of the first 3 terms is:

S_3 = a_1 + ( a_1 + d) +  ( a_1 + 2d)  = 3a_1 + 3d

We can write the answer as 3/2[2a_1 + (3-1)d]

We'll see in a minute why we want to do this rather odd step.

The sum of the first 4 terms is:

S_4 = 4a_1 + 6d

We can write the answer as 4/2[2a_1 + (4 − 1)d]

The sum of the first 5 terms is:

S_5 = 5a_1 + 10d which can be written 5/2[2a_1 + (5 − 1)d]

Are you getting the idea? In general, the sum to n terms is

S_n = n/2[2a_1 + (n − 1)d]

### Example 2

Using the second formula, find the sum of the first 10 terms for the series that we met above: 4, 7, 10, 13, ...

a_1 = 4

d = 3

n = 10

S_n = n/2[2a_1 + (n − 1)d]  = 10/2[2 × 4 + 9 × 3]  = 175

Can you see why we didn't use the first formula? We don't know the final term (but actually, this is no big deal - it is easy to find.)

### Example 3

Find the sum of the first 1000 odd numbers.

In this case, we have a_1 = 1, d = 2 and n = 1000.

So, using the formula

S_n=n/2[2a_1+(n-1)d]

the sum 1 + 3 + 5 + ... for 1000 steps is given by:

S_1000 =1000/2[2(1)+(1000-1)(2)] =1\ 000\ 000

Easy to understand math videos:
MathTutorDVD.com

### Example 4

A clock strikes the number of times of the hour. How many strikes does it make in one day?

1+2+3+...+12=12/2(1+12)
=78\ "times"
For the next 12 hours, there will be another 78 times, so in one day, the clock will strike 156 times.