IB Applications and Interpretation SL · Arithmetic and geometric sequences · Section 2 of 5

Sum of the first n terms of an arithmetic progression

3 worked examples · By Álvaro López

Key idea

Adding the terms of an arithmetic progression one by one is slow and error-prone. With a formula you can find the total of $n$ consecutive terms in two operations. In the IB exam this is useful for savings problems, rows of seats, stacks of objects or payments that grow by a fixed amount, and also for finding how many terms are needed to reach a given amount.

Series and sigma notation

If, instead of listing the terms of a sequence, you add them, you obtain a series. When the sequence is an arithmetic progression, the series is called an arithmetic series.

For example, for the progression $2, 5, 8, 11, 14, \dots$ we have $S_{3} = 2 + 5 + 8 = 15$ and $S_{5} = 2 + 5 + 8 + 11 + 14 = 40$ .

Where the formula comes from

The idea is the one attributed to the young Gauss for adding $1 + 2 + \dots + 100$ : write the sum twice, the second time in reverse, and add column by column.

$S_{n} S_{n} = u_{1} + u_{2} + u_{3} + \dots + u_{n - 1} + u_{n} = u_{n} + u_{n - 1} + u_{n - 2} + \dots + u_{2} + u_{1}$

In each column, one term goes up and the other goes down by the same amount $d$ , so every column gives the same result: $u_{1} + u_{n}$ . There are $n$ columns, so $2 S_{n} = n (u_{1} + u_{n})$ . Put another way: the sum is the number of terms multiplied by the mean of the first and last.

For the progression above, $S_{5} = \frac{5}{2} (2 + 14) = 40$ , the same as adding by hand. And $1 + 2 + \dots + 10 = \frac{10}{2} (1 + 10) = 55$ . Both expressions appear in the IB formula booklet, but it is worth understanding where they come from so you do not confuse them.

Summing a block and finding n

Two situations come up often in problems.

Summing a middle section. To add from the term $u_{k}$ to $u_{n}$ , calculate the sum up to $n$ and subtract the sum of the terms that come before: $u_{k} + \dots + u_{n} = S_{n} - S_{k - 1}$ .

Finding how many terms are needed. If you are asked for the $n$ for which the sum equals a given amount, substitute into the formula and you get an equation in $n$ . Since $S_{n} = \frac{n}{2} (2 u_{1} + (n - 1) d)$ contains an $n^{2}$ term, the result is a quadratic equation. Only one of its solutions is a positive integer.

Worked examples

Example 1 Basic

Adding the terms of a progression

A progression starts like this: $7, 11, 15, \dots$

a) Calculate the term $u_{12}$ .

b) Calculate the sum of the first 12 terms, $S_{12}$ .

Solution

a) The first term is $u_{1} = 7$ and the difference is $d = 11 - 7 = 4$ . Then $u_{12} = 7 + (12 - 1) \cdot 4 = 7 + 44 = 51$

b) With the last term already known, $S_{12} = \frac{12}{2} (7 + 51) = 6 \cdot 58 = 348$ .

Check using the other formula: $S_{12} = 6 (2 \cdot 7 + 11 \cdot 4) = 6 \cdot 58 = 348$ .

Example 2 Medium

A savings plan

A person saves 20 euros in the first month and, each following month, deposits 15 euros more than the previous month.

a) Calculate how much they deposit in month 18.

b) Calculate how much they have saved in total at the end of month 18.

c) Find the month in which the total saved reaches exactly 5000 euros.

Solution

The monthly deposits form an arithmetic progression with $u_{1} = 20$ and $d = 15$ .

a) $u_{18} = 20 + 17 \cdot 15 = 20 + 255 = 275$ euros.

b) $S_{18} = \frac{18}{2} (20 + 275) = 9 \cdot 295 = 2655$ euros.

c) We need to solve $S_{n} = 5000$ with the second formula: $\frac{n}{2} (40 + 15 (n - 1)) = 5000 \Rightarrow n (15 n + 25) = 10000 \Rightarrow 3 n^{2} + 5 n - 2000 = 0$ Applying the quadratic formula, $Δ = 25 + 24000 = 24025 = 15 5^{2}$ , so $n = \frac{- 5 \pm 155}{6} \Rightarrow n = 25 or n = - \frac{80}{3}$ The number of months is positive, so $n = 25$ . Check: $u_{25} = 20 + 24 \cdot 15 = 380$ and $S_{25} = \frac{25}{2} (20 + 380) = 5000$ .

The total reaches 5000 euros exactly at the end of month 25.

Example 3 Hard

The rows of a stand

In a stand, the number of seats in each row increases by a constant amount from one row to the next. Row 3 has 11 seats and the first 10 rows have 210 seats in total.

a) Find the number of seats in the first row and the increase per row.

b) Calculate how many seats there are from row 6 to row 10, both inclusive.

c) The stand is to be extended with more rows following the same rule. Find the smallest number of rows that gives a total of at least 1000 seats.

Solution

a) Let $u_{1}$ and $d$ be the required quantities. From the data for row 3, $u_{3} = u_{1} + 2 d = 11$ . From the sum data, $S_{10} = \frac{10}{2} (2 u_{1} + 9 d) = 210 \Rightarrow 2 u_{1} + 9 d = 42$ From the first equation, $u_{1} = 11 - 2 d$ . Substituting: $22 - 4 d + 9 d = 42$ , so $5 d = 20$ and $d = 4$ . Then $u_{1} = 11 - 8 = 3$ .

The first row has 3 seats and each row has 4 more than the previous one.

b) Rows 6 to 10 are the total of 10 rows minus that of the first 5. As $u_{5} = 3 + 16 = 19$ , we have $S_{5} = \frac{5}{2} (3 + 19) = 55$ . Then $S_{10} - S_{5} = 210 - 55 = 155$

c) The total of $n$ rows is $S_{n} = \frac{n}{2} (6 + 4 (n - 1)) = n (2 n + 1) = 2 n^{2} + n$ . We look for the smallest integer $n$ with $2 n^{2} + n \geq 1000$ .

For $n = 22$ : $S_{22} = 22 \cdot 45 = 990 < 1000$ .

For $n = 23$ : $S_{23} = 23 \cdot 47 = 1081 \geq 1000$ .

As $S_{n}$ increases with $n$ , the smallest value is $n = 23$ rows.

In the exam: in a problem like this, marks are given for setting up the two equations from the data, solving the system, and in the last part justifying the integer value (by comparing $S_{22}$ and $S_{23}$ , or solving the equation with the GDC and rounding up, because there are no fractions of a row).

Common mistakes

Summary

An arithmetic series is the sum of the terms of an arithmetic progression; $S_{n} = u_{1} + \dots + u_{n} = \sum_{r = 1}^{n} u_{r}$ .
If you know the last term: $S_{n} = \frac{n}{2} (u_{1} + u_{n})$ , that is, $n$ times the mean of the first and last.
If you know $u_{1}$ , $d$ and $n$ : $S_{n} = \frac{n}{2} (2 u_{1} + (n - 1) d)$ .
A section from $u_{k}$ to $u_{n}$ is summed as $S_{n} - S_{k - 1}$ (there are $n - k + 1$ terms).
Finding $n$ from a sum gives a quadratic equation; only a positive integer solution is valid.
In Paper 2, the GDC (table, sequence sum or equation) speeds up the calculation, but write down what you have calculated.

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