Summing an arithmetic series

The two formulae for the sum of an arithmetic series, S(n) = n/2 (a + ℓ) and S(n) = n/2 (2a + (n − 1)d), and solving problems including finding how many terms give a given sum.

The two formulae

Sum of an AP

$$S_n = \frac{n}{2}(a + \ell) \qquad\qquad S_n = \frac{n}{2}\big(2a + (n - 1)d\big)$$

Use the first when you know the last term $\ell$. Use the second when you know the common difference $d$. The second comes from the first by putting $\ell = a + (n - 1)d$.

Example 1●

Find the sum of the first $20$ terms of $5 + 9 + 13 + \cdots$

$a = 5$, $d = 4$, $n = 20$.

$S_{20} = 10\,(10 + 19 \times 4) = 10 \times 86 = 860$

Example 2: count the terms first●●

Find the sum of all the multiples of $7$ between $100$ and $300$.

The first is $105$ and the last is $294$. The number of terms is $\dfrac{294 - 105}{7} + 1 = 28$.

$S = \dfrac{28}{2}(105 + 294) = 14 \times 399 = 5586$

How many terms for a given sum?

Example 3●●●

How many terms of $3 + 7 + 11 + \cdots$ add up to $990$?

$S_n = \dfrac{n}{2}\big(6 + 4(n - 1)\big) = n(2n + 1)$.

$2n^2 + n - 990 = 0$, so $n = \dfrac{-1 \pm \sqrt{1 + 7920}}{4} = \dfrac{-1 \pm 89}{4}$.

$n = 22$ terms.

n must be a positive integer.

S as a function of n

The partial sums of an AP lie on a parabola through the origin. When a and d have opposite signs, the sums can rise and then fall.

Try this

Set $a = 10$ and $d = -2$. For which $n$ is the sum largest?

Flashcards

Sum of an AP, last term known

$S_n = \dfrac{n}{2}(a + \ell)$

Sum of an AP, $d$ known

$S_n = \dfrac{n}{2}\big(2a + (n - 1)d\big)$

$1 + 2 + cdots + n$

$\dfrac{n(n + 1)}{2}$

Practice

Find $1 + 2 + 3 + \cdots + 100$.
Find the sum of the first $20$ terms of $5 + 9 + 13 + \cdots$
Find the sum of the multiples of $7$ between $100$ and $300$.
How many terms of $3 + 7 + 11 + \cdots$ add up to $990$?
Find $2 + 4 + 6 + \cdots + 50$.
An AP has $a = 10$ and $d = -2$. Find $S_{12}$.
Find the sum of the first $30$ odd numbers.
Which formula needs the last term?
For an AP, $S_n$ as a function of $n$ is: