Summing an arithmetic progression

By the end of this lesson

By the end you can find the sum of the first $n$ terms of an arithmetic progression with $S_n = \tfrac12 n\{2a + (n - 1)d\}$ or $S_n = \tfrac12 n(a + l)$ , and work backwards to find $n$ , $a$ or $d$ .

Key to the symbols

The main symbols used in this lesson:

Symbol	Meaning
$S_n$	the sum of the first $n$ terms, e.g. $S_{40}$ is the sum of the first 40
$a$	the first term ( $u_1$ )
$d$	the common difference (the amount added each time)
$l$	the last term in the sum
$u_n$	the $n$ th term
$\{\ \}$	curly brackets: ordinary brackets, used so you can see which bracket is which

The 5050 trick

The story goes that a teacher told young Carl Friedrich Gauss's class to add up $1 + 2 + \dots + 100$ , hoping for a quiet hour. Gauss had the answer in moments: 5050.

(The details are probably embellished. The trick is real.)

His idea: pair the first term with the last, the second with the second-last, and so on. Every pair makes $101$ , and there are 50 pairs.

Forwards plus backwards

With small numbers first. Take $S = 2 + 5 + 8 + 11 + 14$ . Write it forwards, then backwards underneath:

forwards: $2, 5, 8, 11, 14$
backwards: $14, 11, 8, 5, 2$
add each column: $16, 16, 16, 16, 16$

Five columns of 16 make 80. That's two copies of the sum, so $2S = 80$ and $S = 40$ .

Now in general, with first term $a$ , last term $l$ and $n$ terms. Write the sum twice, once forwards and once backwards, and add the columns:

$\begin{aligned} S &= a + (a + d) + \dots + (l - d) + l \\ S &= l + (l - d) + \dots + (a + d) + a \\ 2S &= (a + l) + (a + l) + \dots + (a + l) + (a + l) \end{aligned}$

That's $n$ copies of $(a + l)$ , so $2S = n(a + l)$ . Halve it: $S = \tfrac12 n(a + l)$ .

Every column has the same total because one row goes up by $d$ exactly as fast as the other goes down.

2 + 5 + 8 + 11 + 14

: the two copies fill a

5 \times 16

rectangle, so

2S = 80

and

S = 40

Definition

Sum of an arithmetic progression

The sum of the first $n$ terms of an AP with first term $a$ , common difference $d$ and last term $l$ is

$S_n = \tfrac12 n(a + l) = \tfrac12 n\{2a + (n - 1)d\}.$

Where the second form comes from. The last term is the $n$ th term, so $l = a + (n - 1)d$ . Substitute:

$\tfrac12 n(a + l) = \tfrac12 n\{a + a + (n - 1)d\} = \tfrac12 n\{2a + (n - 1)d\}$

Use $\tfrac12 n(a + l)$ when you know the last term.
Use $\tfrac12 n\{2a + (n - 1)d\}$ when you don't.

Quick check

CalculateQuestion 1 · Recall

CalculateQuestion 2 · Recall

Answer the 2 questions above to continue.