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Arithmetic Progressions (Cambridge (CIE) AS Maths: Pure 1): Flashcards

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FrontArithmetic Progressions
Define arithmetic progression.

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Define arithmetic progression.
An arithmetic progression is a progression in which the difference between consecutive terms is always the same.
That constant difference is called the common difference, written $d$ .
In an arithmetic progression with first term $a$ and common difference $d$ , what is the $n$ th term?
The $n$ th term is:
$u_{n} = a + (n - 1) d$
Here $a$ is the first term and $d$ is the common difference.
The 15th term of an arithmetic progression with first term $a$ and common difference $d$ is $- 19$ . Complete the equation this gives:
$a +______d = - 19$
The completed equation is:
$a + 14 d = - 19$
The first term is already $a$ , so by the time you reach the 15th term the common difference has been added only 14 times.
True or False?
In an arithmetic progression, the common difference must be positive.
False.
The common difference can be negative, and then every term is smaller than the one before: $2, 0.5, - 1, - 2.5, - 4, \dots$ has $d = - 1.5$ .
The first three terms of an arithmetic progression are $12$ , $k^{2}$ and $5 k$ . What equation does being arithmetic give you?
There is only one common difference, so the second term minus the first must equal the third minus the second:
$k^{2} - 12 = 5 k - k^{2}$
Rearranging gives the quadratic $2 k^{2} - 5 k - 12 = 0$ .
You know the 9th term and the 15th term of an arithmetic progression. How do you find the first term and the common difference?
Substitute each of them into the $n$ th term formula, which gives two simultaneous equations in $a$ and $d$ .
Subtracting one equation from the other removes $a$ and leaves $d$ , which you then substitute back to find $a$ .
In an arithmetic progression with first term $a$ and common difference $d$ , complete the formula for the sum of the first $n$ terms:
$S_{n} = \frac{n}{2} (______+ (n - 1) d)$
The completed formula is:
$S_{n} = \frac{n}{2} (2 a + (n - 1) d)$
Note that the first term inside the bracket is $2 a$ , not $a$ .
In the proof of the arithmetic series formula, why divide by 2 at the end?
Because the sum has been written out twice, once forwards and once in reverse, and the two copies added together.
That total is $2 S_{n}$ , so halving it recovers the sum of the first $n$ terms.
True or False?
The sum of the first $n$ terms of an arithmetic progression is $n$ times the average of its first and last terms.
True.
That is exactly the formula $S_{n} = \frac{n}{2} (a + l)$ , where $l$ is the last term, since $\frac{a + l}{2}$ is the average of the first and last terms.
Why does setting the sum of the first $n$ terms of an arithmetic progression equal to a given total produce a quadratic equation in $n$ ?
Because the sum formula multiplies $n$ by a bracket that itself contains $n$ , so expanding it produces a term in $n^{2}$ .
Setting that equal to a number and rearranging therefore leaves a quadratic to solve.
Solving a problem about an arithmetic series gives $n = 40$ or $n = - \frac{125}{3}$ . Which do you take, and why?
Take $n = 40$ .
Since $n$ counts how many terms are being added, it must be a positive whole number, which rules the other root out.

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