Arithmetic Series: Proofs and Conditions

Prove S_n = n/2[2a + (n−1)d]: forwards and backwards, add, divide by 2, every line shown

In the exam for Edexcel International AS Maths · 5 real past-paper questions below

Prove the sum formula

You can already use $S_{n} = \frac{n}{2} [2 a + (n - 1) d]$ . AS papers also ask you to prove it. It is usually worth 3 marks, and the method never changes.

Worked example

1
An arithmetic series has first term $a$ and common difference $d$ . Prove that $S_{n} = \frac{n}{2} [2 a + (n - 1) d]$ .
2
$S_{n} = a + (a + d) + \dots + [a + (n - 2) d] + [a + (n - 1) d]$ — Write the sum forwards. Show at least three terms, including the first and the last.
3
$S_{n} = [a + (n - 1) d] + [a + (n - 2) d] + \dots + (a + d) + a$ — Write the same sum backwards.
4
$2 S_{n} = [2 a + (n - 1) d] + [2 a + (n - 1) d] + \dots + [2 a + (n - 1) d]$ — Add the two lines, pair by pair. Every pair gives the same total.
5
$2 S_{n} = n [2 a + (n - 1) d]$ — There are $n$ pairs. Do not skip this line: the final mark needs it.
6
$S_{n} = \frac{n}{2} [2 a + (n - 1) d]$ — Divide both sides by 2.

Rule

How the marks go: 1 mark for the forwards sum (first and last terms in $a$ and $d$ ), 1 mark for reversing and adding, 1 mark for reaching the result with the $2 S_{n}$ line and no errors. Use + signs, not commas: a list is not a sum.

Memory trick

Write it, flip it, add it, halve it.

Key idea

A famous story says that at about age 8, Carl Friedrich Gauss was told to add the whole numbers from 1 to 100. He paired 1 + 100, 2 + 99, 3 + 98… Fifty pairs of 101 gave 5050 in seconds. Your proof is the same trick.

Common mistake

Writing the last term as $a + n d$ . The $n$ th term is $a + (n - 1) d$ , and the term before it is $a + (n - 2) d$ .

Key idea

Reversing works because one line goes up by $d$ while the other goes down by $d$ . So every pair has the same total.

Three consecutive terms

If three terms come one after another in an AP, the jump from the 1st to the 2nd equals the jump from the 2nd to the 3rd. That gives one equation.

Formula

u_{2} - u_{1} = u_{3} - u_{2} ⟺ 2 u_{2} = u_{1} + u_{3}

Memory trick

The middle term is the average of its neighbours: like the middle step of an even staircase.

The terms are often trig expressions. Use an identity so only one trig function is left.

Worked example

1
$2 cos^{2} θ$ , $2 sin θ$ and $1 - 2 sin θ$ are the first three terms of an AP, where $0^{\circ} < θ < 18 0^{\circ}$ . Find $θ$ .
2
$2 sin θ - 2 cos^{2} θ = 1 - 2 sin θ - 2 sin θ$ — Second minus first = third minus second.
3
$6 sin θ = 2 cos^{2} θ + 1$ — Collect the sine terms on the left.
4
$6 sin θ = 2 (1 - sin^{2} θ) + 1$ — Use $cos^{2} θ = 1 - sin^{2} θ$ so there is only sine.
5
$2 sin^{2} θ + 6 sin θ - 3 = 0$ — A quadratic in $sin θ$ .
6
$sin θ = \frac{- 3 + 15}{2} = 0.4365 \dots$ — Quadratic formula. The other root, $\frac{- 3 - 15}{2}$ , is less than −1, so reject it.
7
$θ = 25. 9^{\circ} or 154. 1^{\circ}$ — Sine is positive in the first and second quadrants.

Or the terms can be binomial coefficients.

Worked example

1
In the expansion of $(1 + k x)^{7}$ , where $k \neq = 0$ , the coefficients of $x$ , $x^{2}$ and $x^{3}$ form an AP. Find $k$ .
2
$7 k, 21 k^{2}, 35 k^{3}$ — The three coefficients: $(1 7) k$ , $(2 7) k^{2}$ , $(3 7) k^{3}$ .
3
$2 (21 k^{2}) = 7 k + 35 k^{3}$ — Middle term doubled = sum of the outer two.
4
$7 k (5 k^{2} - 6 k + 1) = 0$ — Move everything to one side and take out $7 k$ .
5
$7 k (5 k - 1) (k - 1) = 0$ — Factorise the quadratic.
6
$k = \frac{1}{5} or k = 1$ — Reject $k = 0$ : the question says $k \neq = 0$ .

Rule

When the answer is printed ("show that"), every line must be there, including the trig identity you used. Skipping to the given answer loses the last mark.

Common mistake

Subtracting in different directions, such as $u_{2} - u_{1} = u_{2} - u_{3}$ . Use the same order both times: later term minus earlier term.

Key idea

Plain algebra, trig or binomial: the equation is always the same. Only the algebra after it changes.

Largest sum of a falling AP

When $d$ is negative, the terms fall and later turn negative. The running total grows while you add positive terms, then shrinks once you add negative ones. So the sum is largest when you stop at the last positive term.

Steps

1
1. Find where the terms change sign: solve $u_{n} > 0$ .
2. Take the largest whole number $n$ that works.
3. Work out $S_{n}$ for that $n$ .

Worked example

1
An arithmetic series begins $200 + 194 + 188 + \dots$ . Find the greatest possible value of $S_{n}$ .
2
$a = 200, d = - 6$ — Name the values. $a$ is positive and $d$ is negative, so a largest sum exists.
3
$200 + (n - 1) (- 6) > 0$ — Terms stay positive while $u_{n} > 0$ .
4
$206 - 6 n > 0 \Rightarrow n < 34.33 \dots$ — Solve for $n$ .
5
$n = 34$ — Check: $u_{34} = 2$ (positive), $u_{35} = - 4$ (negative).
6
$S_{34} = \frac{34}{2} (200 + 2) = 3434$ — First and last terms known, so use $\frac{n}{2} (a + l)$ .

Rule

You may also write $S_{n} = 203 n - 3 n^{2}$ and find its turning point, $n = 33.8 \dots$ . Either way, $n$ must be a whole number, so test the whole numbers near it.

Common mistake

Putting $n = 34.33$ into $S_{n}$ . You cannot add a third of a term. $n$ counts terms, so it is a whole number.

Key idea

Largest sum of a falling AP = sum up to the last positive term.

Which root makes sense?

In context questions, setting $S_{n}$ equal to a total gives a quadratic in $n$ . A quadratic has two roots, but the story usually allows only one.

Worked example

1
A farm harvests 460 kg of olives in week 1. Each week after that it harvests 20 kg less. The farm stops at the end of week $N$ , when 4800 kg have been harvested in total. Show that $N^{2} - 47 N + 480 = 0$ , and find $N$ .
2
$\frac{N}{2} [2 (460) + (N - 1) (- 20)] = 4800$ — Sum formula with $a = 460$ , $d = - 20$ .
3
$\frac{N}{2} (940 - 20 N) = 4800$ — Simplify the bracket: $920 - 20 N + 20$ .
4
$470 N - 10 N^{2} = 4800$ — Multiply out. Keep this line: "show that" needs it.
5
$N^{2} - 47 N + 480 = 0$ — Divide by −10 and rearrange. This is the printed result.
6
$(N - 15) (N - 32) = 0$ — Factorise.
7
$N = 15 or N = 32$ — Two roots. Now decide which one makes sense.
8
$N = 15$ — Reject 32: week 32 would be $460 - 20 \times 31 = - 160$ kg. You cannot harvest a negative amount.

Reject a root when… it is not a positive whole number, it is beyond the time in the story, or it gives a negative term.

A negative term means… the model has broken down. Here the harvest reaches 0 kg in week 24, so the pattern cannot carry on past it.

Rule

"State, with a reason" is 1 mark. Give the value AND a reason from the story, such as "week 32 gives −160 kg, which is impossible". Just "it is negative" is not enough.

Common mistake

Keeping both roots, or choosing the larger one because it "looks safer". Always test each root against the story.

Key idea

The algebra gives possible answers. The story picks the real one.

Which method? Read the question

The wording tells you which of these four AS question types you are facing.

You see	You do
"Prove that the sum to n terms is…"	Write the sum forwards and backwards, add, divide by 2
"…are consecutive terms of an AP"	$u_{2} - u_{1} = u_{3} - u_{2}$ , then solve
"Greatest value of $S_{n}$ " with $d < 0$	Last positive term, then sum to there
A quadratic in $n$ from a story	Solve, then reject the root that does not fit, with a reason

Common mistake

Using a calculator's equation solver without writing the quadratic. If the question says solutions relying on calculator technology are not acceptable, an answer with no working scores nothing.

Key idea

Every one of these ends with a check: the pairs match, the terms have the same gap, the last term is positive, the root fits the story.

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