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 + 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 + 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.
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, ...`
General Term of an Arithmetic Progression
The nth term, `a_n` of an AP is:
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
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.)
`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]`
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]`
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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.)
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
the sum `1 + 3 + 5 + ...` for 1000 steps is given by:
`S_1000` `=1000/2[2(1)+(1000-1)(2)]` `=1\ 000\ 000`
A clock strikes the number of times of the hour. How many strikes does it make in one day?
For the first 12 hours of the day, the clock will strike
For the next `12` hours, there will be another `78` times, so in one day, the clock will strike `156` times.