Arithmetic Sequences and Series

Q: Which sum formula should I use?

Use $\frac{n}{2}(a + l)$ when you know the last term, and $\frac{n}{2}[2a + (n - 1)d]$ when you know $d$.

An arithmetic sequence grows by the same amount each time. Grade 12 adds the sum of the terms, the proof of the sum formula, and sigma notation. Number patterns carry about 25 marks in Paper 1, so these are reliable marks to secure.

General term of an arithmetic sequence

An arithmetic sequence has a common difference $d = T_{2} - T_{1} = T_{3} - T_{2}$ .

$T_{n} = a + (n - 1) d$

Example. $7; 11; 15; \dots$ has $a = 7$ and $d = 4$ .

$T_{n} = 7 + 4 (n - 1) = 4 n + 3 T_{50} = 203$

Two terms given. If $T_{3} = 13$ and $T_{10} = 41$ : $a + 2 d = 13$ and $a + 9 d = 41$ . Subtracting gives $7 d = 28$ , so $d = 4$ and $a = 5$ .

Which term? Solve $T_{n} =$ value for $n$ . The answer must be a natural number.

Sum of an arithmetic series (with proof)

$S_{n} = \frac{n}{2} [2 a + (n - 1) d] or S_{n} = \frac{n}{2} (a + l)$

where $l$ is the last term.

Proof (examinable). Write the sum forwards and backwards:

$S_{n} = a + (a + d) + \dots + (a + (n - 1) d)$

$S_{n} = (a + (n - 1) d) + (a + (n - 2) d) + \dots + a$

Adding, each pair of terms gives $2 a + (n - 1) d$ , and there are $n$ pairs:

$2 S_{n} = n [2 a + (n - 1) d] \Rightarrow S_{n} = \frac{n}{2} [2 a + (n - 1) d]$

Example. How many terms of $5; 9; 13; \dots$ add up to 1 224?

$\frac{n}{2} [10 + 4 (n - 1)] = 1224 \Rightarrow 2 n^{2} + 3 n - 1224 = 0 \Rightarrow n = 24$

Sigma notation

$\sum_{k = 1}^{n} T_{k} = T_{1} + T_{2} + \dots + T_{n}$

The number of terms is (top value $-$ bottom value $+ 1$ ).

Example. $k = 1 \sum 20 (3 k - 2)$

First term: $k = 1$ gives 1. Last term: $k = 20$ gives 58.
Number of terms: 20.

$S_{20} = \frac{20}{2} (1 + 58) = 590$

Sum of terms in the middle. $T_{11} + \dots + T_{30} = S_{30} - S_{10}$ .

Key terms

Arithmetic sequence: A sequence with a constant difference between consecutive terms.
Common difference: $d = T_{n + 1} - T_{n}$ .
Series: The sum of the terms of a sequence.
Sigma notation: Shorthand for a sum, using $\sum$ .

Common mistakes

Using $T_{n} = a + n d$ instead of $a + (n - 1) d$ .
Counting the number of terms in sigma notation as top minus bottom, forgetting the $+ 1$ .
Using $S_{30} - S_{11}$ for the sum from $T_{11}$ to $T_{30}$ ; it should be $S_{30} - S_{10}$ .
Keeping a negative or fractional value of $n$ .

Practice questions

Reveal steps one at a time, then tick the marks you earned.

Question 1

3 marks total

knowledge

For the sequence $7; 11; 15; \dots$ , determine $T_{n}$ and $T_{50}$ .

Worked solution

Hidden - try the question first

Solution

$a = 7$ , $d = 4$
$T_{n} = 7 + 4 (n - 1) = 4 n + 3$
$T_{50} = 203$

Question 2

5 marks total

routine

Calculate $k = 1 \sum 20 (3 k - 2)$ .

Worked solution

Hidden - try the question first

Solution

$a = 7$
$T_{n} = 7 + 4 (n - 1) = 4 n + 3$
$n = 20$
$S_{20} = \frac{20}{2} (1 + 58)$
$= 590$

Question 3

7 marks total

complex

(a) Prove that the sum of the first $n$ terms of an arithmetic series is $S_{n} = \frac{n}{2} [2 a + (n - 1) d]$ . (4)

(b) How many terms of the series $5 + 9 + 13 + \dots$ must be added to give 1 224? (3)

Worked solution

Hidden - try the question first

Solution

(a) $S_{n} = a + (a + d) + \dots + [a + (n - 1) d]$
$S_{n} = [a + (n - 1) d] + [a + (n - 2) d] + \dots + a$
Adding: $2 S_{n} = [2 a + (n - 1) d] + \dots$ ( $n$ times)
$2 S_{n} = n [2 a + (n - 1) d]$ , so $S_{n} = \frac{n}{2} [2 a + (n - 1) d]$
(b) $\frac{n}{2} [10 + 4 (n - 1)] = 1224$
$2 n^{2} + 3 n - 1224 = 0 \Rightarrow (n - 24) (2 n + 51) = 0$
$n = 24$

Question 4

9 marks total

problem-solving

In an arithmetic sequence, $T_{3} = 13$ and $T_{10} = 41$ .

(a) Determine $a$ and $d$ . (3)

(b) Calculate $T_{11} + T_{12} + \dots + T_{30}$ . (4)

Worked solution

Hidden - try the question first

Solution

(a) $S_{n} = a + (a + d) + \dots + [a + (n - 1) d]$ and $a + 9 d = 41$
$7 d = 28 \Rightarrow d = 4$
$a = 5$
(b) $S_{30} = 15 [10 + 29 (4)] = 1890$
$S_{10} = 5 [10 + 9 (4)] = 230$
$T_{11} + \dots + T_{30} = S_{30} - S_{10}$
$= 1660$
(c) $5 + 4 (n - 1) > 300 \Rightarrow n > 74, 75$
$T_{75} = 301$

FAQs

Is the proof of the arithmetic series formula examinable?

Yes. Learn the method of writing the series forwards and backwards and adding.

How do I find the number of terms in sigma notation?

Subtract the bottom value from the top value and add 1.

Which sum formula should I use?

Use $\frac{n}{2} (a + l)$ when you know the last term, and $\frac{n}{2} [2 a + (n - 1) d]$ when you know $d$ .

General term of an arithmetic sequence

Sum of an arithmetic series (with proof)

Sigma notation

Key terms

Common mistakes

Practice questions

FAQs

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