We start with the simplest kind of sequence: one that changes by the same amount at every step.
Definition 11.1.1. An arithmetic sequence (or arithmetic progression) is a sequence where the difference between any two consecutive terms is the same, or, more formally: an AP is a sequence such that for all .
Problem 11.1.2. An arithmetic progression has as its 1st term and as its 6th term.
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Common difference:
so CD and the th term is ; the 717th term is .
Or, starting from :, and with this is . So .
Problem 11.1.3. The th term of an AP with first term and common difference .
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Multiplied out: in general, is the th term finder.
Problem 11.1.4. Prove that if are in AP, then for all , are also in arithmetic progression.
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Say are in AP . Now see , but , so . are in AP.
The same goes for a whole AP. In an AP, consecutive terms differ by the common difference: , and for all . So note that for any with :
so consecutive terms differ by for all .
Problem 11.1.5. An AP has first and last terms of and , with .
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If there are terms,
So there are terms.
With the other form of the th term, , term is . To find (i.e. the max ):
So is the 19th term in the sequence.
Problem 11.1.6. The first four terms of an AP are, in order, ,,,. What’s the 5th term?
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The th term is .
The common difference:
Also, from the last two terms,
The third and fourth terms give
So the terms are :
and .
So , with
From :
(Solving the same equation for instead: , so ,,.)
Now put into (multiplied through by , this reads ):
Dividing out the root :
leaves , so and
The 5th term is the 4th term plus :
Another way, setting the consecutive differences equal:
The fifth term is :
Now , and :
so
With :, and the fifth term is
Example 11.1.7 (Roots in arithmetic progression). Solve , given that its roots are in AP.
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Candidates for rational roots: numerator from and denominator from , like ,,,,, … Testing :
.
Writing the roots as , their sum is
which is one root.
Or call the roots . Then
So, from the product of the roots,
and the roots are
Problem 11.1.8. Let be in AP. Prove that:
(a) , and
(b) ,, and are in AP.
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(a) in AP . ✓
(b) Since , and we wish to show , and are in AP, we must show that
that is,
By ,
So, proof: , as desired. Thus ,, are in AP.
Exercises
Exercise 11.1.9. For , an (infinite) AP has the value . For , the value is . What is the smallest value of such that the th term is negative?
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The th term is , or equally , counting from the next term. Here
Counting from the 4th term:
so the first such is .
Check:
Exercise 11.1.10. In an AP, and .
(a)
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, so the th term is ?
(a) .
Exercise 11.1.11. Show that if is the average of and , then are in AP.
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Given: , therefore they are in AP.
Exercise 11.1.12.Show solution
Let and add to each term, so we have ,,:
Exercise 11.1.13.Show solution
Exercise 11.1.14. Find all such that has roots in in AP.
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Call the roots . Their sum is :
The sum of the products two at a time is :
The product of the roots is :
Again, with the roots (so ). The sum:
The products two at a time:
so
The product:
and with :
Also is a root:
And the products three at a time: