The sum of n terms of an A.P.

Adding the terms of an A.P. one by one is slow when there are fifty or a hun­dred of them. There is a neat trick, said to have been found by the young Carl Friedrich Gauss when he was asked in class to add the num­bers from 1 to 100: write the sum for­wards and back­wards, and add the two. This les­son turns that trick into a for­mula for the sum of any num­ber of terms of an A.P.

Adding 1 to 100#

Write the sum twice, once in each order:

$$\begin{aligned} S &= 1 + 2 + 3 + \cdots + 99 + 100 \\ S &= 100 + 99 + 98 + \cdots + 2 + 1 \end{aligned}$$

Add the two lines col­umn by col­umn. Every col­umn adds up to 101, and there are 100 columns. So $2S = 100 \times 101$, and $S = 5050$.

The sum of n terms of an A.P.#

Do the same with a gen­eral A.P. whose first term is $a$, com­mon dif­fer­ence $d$ and $n$th term $l = a + (n - 1)d$:

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

Adding col­umn by col­umn, each pair gives $a + l$, and there are $n$ pairs. So $2S_n = n(a + l)$, which gives

$$\displaystyle S_n = \frac{n}{2}(a + l)$$

Replac­ing $l$ by $a + (n - 1)d$ gives the sec­ond form:

$$\displaystyle S_n = \frac{n}{2}\left[2a + (n - 1)d\right]$$

Use the first form when you know the last term, and the sec­ond when you know the com­mon dif­fer­ence.

Use­ful spe­cial sums#

Three sums occur so often that they are worth remem­ber­ing. Each fol­lows from the for­mula.

The first $n$ nat­ural num­bers ($a = 1$, $l = n$): $\displaystyle 1 + 2 + 3 + \cdots + n = \frac{n(n + 1)}{2}$.

The first $n$ odd num­bers ($a = 1$, $d = 2$): $\displaystyle 1 + 3 + 5 + \cdots + (2n - 1) = \frac{n}{2}\left[2 + 2(n - 1)\right] = n^2$.

The first $n$ even num­bers ($a = 2$, $d = 2$): $2 + 4 + 6 + \cdots + 2n = n(n + 1)$.

Sum and term are linked#

Remem­ber from the first les­son that $t_n = S_n - S_{n-1}$. For an A.P. this gives a quick test: the sum $S_n$ of an A.P. is always of the form $pn^2 + qn$ (no con­stant term), and then $t_n = S_n - S_{n-1}$.

Exam­ple 1#

Find the sum of the first 20 terms of the A.P. $2, 5, 8, \ldots$

$a = 2$, $d = 3$, $n = 20$.

$$\displaystyle S_{20} = \frac{20}{2}\left[2 \times 2 + 19 \times 3\right] = 10 \times (4 + 57) = 10 \times 61 = 610$$

Exam­ple 2#

Find the sum $7 + 11 + 15 + \cdots + 207$.

First count the terms: $7 + 4(n - 1) = 207$ gives $4(n - 1) = 200$, so $n = 51$. Now use the form with the last term:

$$\displaystyle S_{51} = \frac{51}{2}(7 + 207) = \frac{51}{2} \times 214 = 51 \times 107 = 5457$$

Exam­ple 3#

Find the sum of the first 15 terms of the A.P. $30, 26, 22, \ldots$

$a = 30$, $d = -4$.

$$\displaystyle S_{15} = \frac{15}{2}\left[60 + 14 \times (-4)\right] = \frac{15}{2} \times (60 - 56) = \frac{15}{2} \times 4 = 30$$

Exam­ple 4#

How many terms of the A.P. $30, 27, 24, \ldots$ must be taken so that their sum is 162?

$a = 30$, $d = -3$. Set $S_n = 162$:

$$\displaystyle \begin{aligned} \frac{n}{2}\left[60 + (n - 1)(-3)\right] &= 162 \\ n(63 - 3n) &= 324 \\ 3n^2 - 63n + 324 &= 0 \\ n^2 - 21n + 108 &= 0 \\ (n - 9)(n - 12) &= 0 \end{aligned}$$

So $n = 9$ or $n = 12$. Both answers are cor­rect. The first nine terms are $30, 27, \ldots, 6$, with sum $\displaystyle \frac{9}{2}(30 + 6) = 162$. The 10th, 11th and 12th terms are $3, 0, -3$, which add up to 0, so tak­ing 12 terms also gives 162.

Exam­ple 5#

The sum of the first $n$ terms of a sequence is $S_n = 3n^2 + 2n$. Show that it is an A.P. and find its gen­eral term.

$$t_n = S_n - S_{n-1} = 3n^2 + 2n - \left[3(n - 1)^2 + 2(n - 1)\right]$$

Expand­ing, $3(n - 1)^2 + 2(n - 1) = 3n^2 - 6n + 3 + 2n - 2 = 3n^2 - 4n + 1$. So

$$t_n = 3n^2 + 2n - 3n^2 + 4n - 1 = 6n - 1$$

This also gives $t_1 = 5 = S_1$. Since $t_n = 6n - 1$ is first degree in $n$, the sequence is an A.P.: $5, 11, 17, \ldots$ with $d = 6$.

Exam­ple 6#

Find the sum of all odd num­bers between 0 and 100.

These are $1, 3, 5, \ldots, 99$, the first 50 odd num­bers. Their sum is $50^2 = 2500$.

Exam­ple 7#

The first term of an A.P. is 5, the last term is 45 and the sum is 400. Find the num­ber of terms and the com­mon dif­fer­ence.

From $\displaystyle S_n = \frac{n}{2}(a + l)$: $\displaystyle 400 = \frac{n}{2}(5 + 45) = 25n$, so $n = 16$. Then $45 = 5 + 15d$ gives $\displaystyle d = \frac{40}{15} = \frac{8}{3}$.

Prac­tice#

1. Find the sum of the first 25 nat­ural num­bers.
2. Find the sum of the first 12 terms of $3, 8, 13, \ldots$
3. Find the sum $10 + 14 + 18 + \cdots + 102$.
4. Find the sum of the first 10 terms of $20, 17, 14, \ldots$
5. Find the sum of the first 40 even num­bers.
6. How many terms of $3, 6, 9, \ldots$ must be added to get 165?
7. How many terms of $17, 15, 13, \ldots$ give a sum of 72? Explain why there are two answers.
8. If $S_n = 2n^2 + 5n$, find $t_n$ and the com­mon dif­fer­ence.
9. The first term of an A.P. is 8, the last term is 62 and the sum is 350. Find the num­ber of terms and the com­mon dif­fer­ence.
10. Find the sum of all mul­ti­ples of 6 between 1 and 200.

Answers#