Arithmetic sums | Year 11 | Ascent Library

01 / LEARN

Understand the idea.

Pairing forwards and backwards

A series adds sequence terms. Pair an arithmetic list with a reversed copy: each of the $n$ pairs totals $a + u_{n}$ . Since this counts the original sum twice, $2 S_{n} = n (a + u_{n})$ .

Thus the sum is the number of terms multiplied by the mean of the endpoints. Substituting $u_{n} = a + (n - 1) d$ gives the alternative formula using $a, d, n$ . Negative differences and negative terms retain their signs.

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

Counting terms before summing

Count the terms before using a sum formula. Find $n$ from the last-term equation if necessary. An inclusive list of consecutive integers from $p$ to $q$ contains $q - p + 1$ terms, including both endpoints.

An unknown term count may lead to a quadratic equation. Solve it completely, then reject roots that are not positive integers. Finally substitute the retained count into the sum; a count must also fit the stated context.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

A sum with known endpoints

Find $12 + 17 + 22 + \dots + 97$ .

Worked solution

1
Identify the first term and difference.
$a = 12, d = 17 - 12 = 5$
Each term increases by five.
2
Find the number of terms.
$\begin{aligned} 97 & = 12 + 5 (n - 1) \\ 85 & = 5 (n - 1) \\ n & = 18 \end{aligned}$
The last term occurs after seventeen increases.
3
Sum using the endpoints.
$S_{18} = \frac{18 (12 + 97)}{2} = 981$
Pairing gives an average term of one half of the endpoint sum.

Answer

$981$ .

EXAMPLE 02

Recovering the row count

Rows contain $8, 11, 14, \dots$ seats. The total is 215. Find the number of rows.

Worked solution

1
Express the total for an unknown row count.
$215 = \frac{n}{2} [2 (8) + (n - 1) 3]$
Seat counts form an arithmetic sequence with first term eight and difference three.
2
Rearrange the equation.
$\begin{aligned} 430 & = n (3 n + 13) \\ 3 n^{2} + 13 n - 430 & = 0 \end{aligned}$
Multiplying by two removes the denominator.
3
Factor and select a valid count.
$\begin{aligned} (n - 10) (3 n + 43) & = 0 \\ n & = 10 or n = - \frac{43}{3} \end{aligned}$
Only the positive integer root can count rows.
4
Check the total for ten rows.
$u_{10} = 8 + 9 (3) = 35, S_{10} = \frac{10 (8 + 35)}{2} = 215$
The first and last seat counts reproduce the required total.

Answer

$10$ .

03 / YOUR TURN

Try it yourself.

Check your understanding

Find the sum of the first 20 positive even integers.

Show answer & reasoning

$420$ .

The first twenty positive even integers run from $2$ to $40$ . Their mean is $21$ , so their sum is $20 (21) = 420$ .

Keep practising

6 worksheet questions · 13 marks

Open worksheet Worked solutions

Read the questions online

Find $S_{15}$ when $a = 3, d = 4$ .
2 marks
Find $100 + 94 + 88 + \dots + 4$ .
2 marks
Find the sum of the integers from 23 to 68 inclusive.
2 marks
A 12-term arithmetic series starts at 7 and sums to 282. Find its last term and difference.
2 marks
A runner begins at 2 km on day 1 and adds 0.5 km each day. Find the distance over 14 days.
2 marks
Find positive integer $n$ if $1 + 4 + \dots + (3 n - 2) = 145$ .
3 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Find $S_{15}$ when $a = 3, d = 4$ .
2 marks
$\frac{15 [6 + 14 (4)]}{2} = 465$ .
$465$ .
Find $100 + 94 + 88 + \dots + 4$ .
2 marks
$4 = 100 - 6 (n - 1)$ gives $n = 17$ ; $\frac{17 (104)}{2} = 884$ .
$884$ .
Find the sum of the integers from 23 to 68 inclusive.
2 marks
There are 46 terms; $\frac{46 (23 + 68)}{2} = 2093$ .
$2093$ .
A 12-term arithmetic series starts at 7 and sums to 282. Find its last term and difference.
2 marks
$282 = 6 (7 + l)$ gives $l = 40$ ; $\frac{40 - 7}{11} = 3$ .
Last term 40; $d = 3$ .
A runner begins at 2 km on day 1 and adds 0.5 km each day. Find the distance over 14 days.
2 marks
$S = \frac{14 [4 + 13 (0.5)]}{2} = 73.5$ .
$73.5 km$ .
Find positive integer $n$ if $1 + 4 + \dots + (3 n - 2) = 145$ .
3 marks
$\frac{n (3 n - 1)}{2} = 145$ ; $(n - 10) (3 n + 29) = 0$ .
$10$ .

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 1 Mathematics sequence and series content with introductory induction, regrouped into an original school-sized chapter.

Official curriculum source ↗

In this lesson

Understand the idea.

Pairing forwards and backwards

Counting terms before summing

See the method in action.

A sum with known endpoints

Identify the first term and difference.

Find the number of terms.

Sum using the endpoints.

Recovering the row count

Express the total for an unknown row count.

Rearrange the equation.

Factor and select a valid count.

Check the total for ten rows.

Try it yourself.

Check your understanding

Keep practising