An arithmetic sequence is a sequence of numbers, such that the difference between any term and the previous term
is a constant number called the common difference (\(d\)):
\[{T}_{n}={a} + \left(n-1\right)d\]
When we sum a finite number of terms in an arithmetic sequence, we get a finite arithmetic series.
A simple arithmetic sequence is when \({a} = 1\) and \(d=1\), which is the sequence of positive integers:
\begin{align*}
{T}_{n}& = {a} + \left(n-1\right)d \\
& = 1 + \left(n-1\right)(1) \\
& = n \\
\therefore \left\{{T}_{n}\right\} & = 1; 2; 3; 4; 5; \ldots
\end{align*}
If we wish to sum this sequence from \(n=1\) to any positive integer, for example \(\text{100}\), we would write
\[\sum _{n=1}^{100}n=1+2+3+\cdots + 100\]
This gives the answer to the sum of the first \(\text{100}\) positive integers.
The mathematician, Karl Friedrich Gauss, discovered the following proof when he was only 8 years old. His teacher
had decided to give his class a problem which would distract them for the entire day by asking them to add all the
numbers from \(\text{1}\) to \(\text{100}\). Young Karl quickly realised how to do this and shocked the teacher
with the correct answer, \(\text{5 050}\). This is the method that he used:
General formula for a finite arithmetic series (EMCDY)
If we sum an arithmetic sequence, it takes a long time to work it out term-by-term. We therefore derive the
general formula for evaluating a finite arithmetic series. We start with the general formula for an arithmetic
sequence of \(n\) terms and sum it from the first term (\(a\)) to the last term in the sequence (\(l\)):
\begin{align*}
\sum _{n=1}^{l}{T}_{n} &= S_{n} \\
S_{n} & = a + (a + d) + (a + 2d) + \cdots + (l - 2d) + (l -d) + l \\
+ \quad \underline{S_{n}} & = \underline{l + (l - d) + (l -2d) + \cdots + (a + 2d) + (a + d) + a } \\
\therefore 2S_{n} & = (a + l ) + (a + l ) + (a + l ) + \cdots + (a + l ) + (a + l ) + (a + l ) \\
\therefore 2S_{n} & = n \times (a + l ) \\
\therefore S_{n} & = \frac{n}{2}(a + l )
\end{align*}
This general formula is useful if the last term in the series is known.
We substitute \(l = a + (n-1)d\) into the above formula and simplify:
\begin{align*}
S_{n} & = \frac{n}{2}(a + [a + (n-1)d ]) \\
\therefore S_{n} &= \frac{n}{2}[2a + (n-1)d ]
\end{align*}
The general formula for determining the sum of an arithmetic series is given by:
\[{S}_{n}= \frac{n}{2} \left[ 2{a} + \left(n-1\right) d \right]\]
or
\[{S}_{n}= \frac{n}{2} (a + l)\]
For example, we can calculate the sum \({S}_{20}\) for the arithmetic sequence \({T}_{n}=3+7 \left(n-1\right)\)
by summing all the individual terms:
\begin{align*}
{S}_{20} & = \sum _{n=1}^{20} \left[3+7 \left(n-1\right)\right] \\
& = 3+10+17+24+31+38+45+52 \\
& +59+66+73+80+87+94+101 \\
& +108+115+122+129+136 \\
& = 1390
\end{align*}
or, more sensibly, we could use the general formula for determining an arithmetic series by substituting
\({a}=3\), \(d=7\) and \(n=20\):
\begin{align*}
{S}_{n} & = \frac{n}{2}(2a + (n-1)d ) \\
{S}_{20} & = \frac{20}{2} \left[2(3) + 7 \left(20-1\right)\right] \\
& = 1390
\end{align*}
This example demonstrates how useful the general formula for determining an arithmetic series is, especially
when the series has a large number of terms.
Worked example 7: General formula for the sum of an arithmetic sequence
Find the sum of the first \(\text{30}\) terms of an arithmetic series with \(T_{n} = 7n - 5\) by using the
formula.
Use the general formula to generate terms of the sequence and write down the known variables
\begin{align*}
T_{n} &= 7n - 5 \\
\therefore T_{1} &= 7(1) - 5 \\
&= 2 \\
T_{2} &= 7(2) - 5 \\
&= 9 \\
T_{3} &= 7 (3) - 5 \\
&= 16
\end{align*}
This gives the sequence: \(2; 9; 16 \ldots\)
\[a = 2; \quad d = 7; \quad n = 30\]
Write down the general formula and substitute the known values
\begin{align*}
S_{n} &= \frac{n}{2}[2a + (n-1)d ] \\
S_{30} &= \frac{30}{2}[2(2) + (30-1)(7) ] \\
&= 15(4 + 203) \\
&= 15 (207) \\
&= 3105
\end{align*}
Write the final answer
\(S_{30} = 3105\)
Worked example 8: Sum of an arithmetic sequence if first and last terms are known
Find the sum of the series \(-5 -3 -1 + \cdots \cdots + 123\)
Identify the type of series and write down the known variables
\begin{align*}
d &= T_{2} - T_{1} \\
&= -3 - (-5) \\
&= 2 \\
d &= T_{3} - T_{2} \\
&= -1 - (-3) \\
&= 2
\end{align*}\[a = -5; \quad d = 2; \quad l = 123\]
Determine the value of \(n\)
\begin{align*}
T_{n} &= a + (n-1)d \\
\therefore 123 &= -5 + (n-1)(2) \\
&= -5 + 2n - 2 \\
\therefore 130&= 2n \\
\therefore n &= 65
\end{align*}
Use the general formula to find the sum of the series
\begin{align*}
S_{n} &= \frac{n}{2}(a + l ) \\
S_{65} &= \frac{65}{2}(-5 + 123) \\
&= \frac{65}{2}(118) \\
&= 3835
\end{align*}
Write the final answer
\(S_{65} = 3835\)
Worked example 9: Finding \(n\) given the sum of an arithmetic sequence
Given an arithmetic sequence with \(T_{2} = 7\) and \(d = 3\), determine how many terms must be added
together to give a sum of \(\text{2 146}\).
Write down the known variables
\begin{align*}
d &= T_{2} - T_{1} \\
\therefore 3 &= 7 - a \\
\therefore a &= 4
\end{align*}\[a = 4 ; \quad d = 3; \quad S_{n} = 2146\]
Use the general formula to determine the value of \(n\)
\begin{align*}
S_{n} &= \frac{n}{2}(2a + (n-1)d ) \\
2146 &= \frac{n}{2}(2(4) + (n-1)(3) ) \\
4292 &= n(8 + 3n - 3) \\
\therefore 0 &= 3n^2 + 5n -4292 \\
&= (3n + 116)(n - 37 ) \\
\therefore n = -\frac{116}{3} &\text{ or } n = 37
\end{align*}
but \(n\) must be a positive integer, therefore \(n = 37\).
We could have solved for \(n\) using the quadratic formula but factorising by inspection is usually the
quickest method.
Write the final answer
\(S_{37} = 2146\)
Worked example 10: Finding \(n\) given the sum of an arithmetic sequence
The sum of the second and third terms of an arithmetic sequence is equal to zero and the sum of the first
\(\text{36}\) terms of the series is equal to \(\text{1 152}\). Find the first three terms in the
series.
Write down the given information
\begin{align*}
T_{2} + T_{3} &= 0 \\
\text{So} \quad (a + d) + (a + 2d) &= 0 \\
\therefore 2a + 3d &= 0 \ldots \ldots (1)
\end{align*}\begin{align*}
S_{n} &= \frac{n}{2}(2a + (n-1)d ) \\
S_{36} &= \frac{36}{2}(2a + (36-1)d ) \\
1152 &= 18(2a + 35d ) \\
\therefore 64 &= 2a + 35d \ldots \ldots (2)
\end{align*}
Solve the two equations simultaneously
\begin{align*}
2a + 3d &= 0 \ldots \ldots (1) \\
2a + 35d &= 64 \ldots \ldots (2) \\
\text{Eqn } (2) - (1): \quad 32d &= 64 \\
\therefore d &= 2 \\
\text{And } 2a + 3(2) &= 0 \\
2a &= -6 \\
\therefore a &= -3
\end{align*}
Write the final answer
The first three terms of the series are:
\begin{align*}
T_{1} &= a = -3 \\
T_{2} &= a + d = -3 + 2 = -1 \\
T_{3} &= a + 2d = -3 + 2(2) = 1
\end{align*}\[-3 -1 + 1\]
Calculating the value of a term given the sum of \(n\) terms:
If the first term in a series is \(T_{1}\), then \(S_{1} = T_{1}\).
We also know the sum of the first two terms \(S_{2} = T_{1} + T_{2}\), which we rearrange to make \(T_{2}\) the
subject of the equation:
\begin{align*}
T_{2} &= S_{2} - T_{1} \\
\text{Substitute } S_{1} &= T_{1} \\
\therefore T_{2} &= S_{2} - S_{1}
\end{align*}
Similarly, we could determine the third and fourth term in a series:
\begin{align*}
T_{3} &= S_{3} - S_{2} \\
\text{And } T_{4} &= S_{4} - S_{3}
\end{align*}
\(T_{n} = S_{n} - S_{n-1}, \text{ for } n \in \{2;3;4; \ldots \}\) and \(T_{1} = S_{1}\)
Sum of an arithmetic series
Textbook Exercise 1.8
Determine the value of \(k\): \[\sum _{n=1}^{k} \left( -2n \right) = - 20\]
\begin{align*}
\left( -2(1) \right) + \left( -2(2) \right) + \left( -2(3) \right) + \ldots + \left( -2(k) \right) &=
-20 \\
-2 - 4 - 6 + \ldots -2k & = -20
\end{align*}
This is an arithmetic series with \(a = -2\) and \(d = -2\):
\begin{align*}
S_{n} &= \frac{n}{2}[2a + (n-1)d] \\
-20 &= \frac{n}{2}[2(-2) + (n-1)(-2)] \\
-40 &= n [-4 + -2n + 2] \\
-40 &= n [ -2n - 2] \\
-40 &= -2n^{2} - 2n \\
2n^{2} + 2n - 40 &= 0 \\
n^{2} + n - 20 &= 0 \\
(n + 5)(n - 4) &= 0 \\
\therefore n = -5 &\text{ or } n = 4 \\
\therefore S_{4} &= -20 \\
\therefore k &= 4
\end{align*}
How many terms of the series must be added to give a sum of \(\text{425}\)?
\begin{align*}
{S}_{n} &= \frac{n}{2} \left(7n+15\right) \\
\therefore 425 &= \frac{n}{2} \left(7n+15\right) \\
850 &= n \left(7n+15\right) \\
&= 7n^{2} + 15n \\
0 &= 7n^{2} + 15n - 850 \\
&= (7n + 85)(n - 10) \\
\therefore n = -\frac{85}{7} &\text{ or } n = 10
\end{align*}
but \(n\) must be a positive integer, therefore \(n=10\).
Determine the sixth term of the series.
\begin{align*}
{S}_{n} &= \frac{n}{2} \left(7n+15\right) \\
{S}_{1} &= {T}_{1} = a \\
{S}_{1} &= \frac{n}{2} \left(7n+15\right) \\
&= \frac{1}{2} \left(7(1)+15\right) \\
\therefore a &= 11 \\
{S}_{2} &= \frac{2}{2} \left(7(2)+15\right) \\
&= 29 \\
\therefore T_{1} + T_{2} &= 29 \\
\therefore T_{2} &=29 - 11 \\
\text{And } d &= T_{2} - T_{1} \\
&= 18 - 11 \\
&= 7 \\
\therefore T_{n} &= a + (n-1)d \\
&= 11 + (n-1)(7) \\
&= 11 + 7n - 7 \\
&= 7n + 4 \\
\therefore T_{6} &= 7(6) + 4 \\
&= 46
\end{align*}
The common difference of an arithmetic series is \(\text{3}\). Calculate the values of \(n\) for
which the \(n\)\(^{\text{th}}\) term of the series is \(\text{93}\), and the sum of the first \(n\)
terms is \(\text{975}\).
\begin{align*}
d &= 3 \\
{T}_{n} &= a + (n - 1)d \\
93 &= a + 3(n-1) \\
&= a + 3n - 3 \\
96 &= a + 3n \\
\therefore a &= 96 - 3n \\
{S}_{n} &= \frac{n}{2}[2a + (n-1)d] \\
\therefore 975 &= \frac{n}{2}[2(96 - 3n) + 3(n-1)] \\
1950 &= n[192 - 6n + 3n - 3] \\
1950 &= 189n - 3n^{2} \\
0 &= - 3n^{2} + 189n - 1950 \\
0 &= n^{2} - 63n + 650 \\\
0 &= (n - 13)(n - 50) \\
\therefore n = 13 &\text{ or } n = 50
\end{align*}
Explain why there are two possible answers.
There are two series that satisfy the given parameters:
\begin{align*}
d &= 3 \\
a &= 96 - 3n \\
\text{If } n &= 13 \\
a &= 96 - 3(13) \\
&= 57 \\
\therefore 57 + 60 + 63 + & \ldots + T_{13} = 975 \\
\text{If } n &= 50 \\
a &= 96 - 3(50) \\
&= -54 \\
\therefore (-54) + (-51) + (-48) + & \ldots + T_{50} = 975
\end{align*}
The third term of an arithmetic sequence is \(-\text{7}\) and the seventh term is \(\text{9}\).
Determine the sum of the first \(\text{51}\) terms of the sequence.
\begin{align*}
{T}_{3} &= -7 = a + 2d \ldots \ldots (1) \\
{T}_{7} &= 9 = a + 6d \ldots \ldots (2) \\
\therefore \text{Subtract eqns: } (1) - (2) \quad -7 - (9) &= a + 2d -(a + 6d) \\
-16 &= - 4d \\
\therefore 4 &= d \\
\text{Substitute back into eqn. } (1) \quad a &= -7 -2 (4) \\
\therefore a &= -15 \\
S_{n} &= \frac{n}{2}[2a + (n-1)d] \\
S_{51} &= \frac{51}{2}[2(-15) + (51-1)(4)] \\
&= \frac{51}{2}[-30 + 200] \\
&= (51)(85) \\
\therefore S_{51}&= 4335
\end{align*}
Calculate the sum of the arithmetic series \(4+7+10+\cdots +901\).
\begin{align*}
a &= 4 \\
l &= 901 \\
d &= T_{2} - T_{1} \\
&= 7 - 4 \\
&= 3 \\
\text{And } T_{n} &= a +(n-1)d \\
&= 4 +(n-1)(3) \\
\therefore 901 &= 4 + 3n - 3 \\
900 &= 3n \\
\therefore 300 &= n \\
S_{n} &= \frac{n}{2}[a + l] \\
S_{300} &= \frac{300}{2}[4 + 901 ] \\
&= (150)(905) \\
\therefore S_{300} &= 135750
\end{align*}
Evaluate without using a calculator: \(\dfrac{4 + 8 + 12 + \cdots + 100}{3 + 10 + 17 + \cdots + 101}\)
Consider the numerator: \(4 + 8 + 12 + \ldots + 100\)
\begin{align*}
a &= 4 \\
l &= 100 \\
d &= T_{2} - T_{1} \\
&= 8 - 4 \\
&= 4 \\
\text{And } T_{n} &= a +(n-1)d \\
100 &= 4 +(n-1)(4) \\
100 &= 4n \\
\therefore 25 &= n \\
S_{n} &= \frac{n}{2}[a + l] \\
S_{25} &= \frac{25}{2}[4 + 100 ] \\
S_{25} &= (25)(52)
\end{align*}
Consider the denominator: \(3 + 10 + 17 + \ldots + 101\)
\begin{align*}
a &= 3 \\
l &= 101 \\
d &= T_{2} - T_{1} \\
&= 10 - 3 \\
&= 7 \\
\text{And } T_{n} &= a +(n-1)d \\
101 &= 3 +(n-1)(7) \\
101 &= 3 + 7n - 7 \\
105 &= 7n \\
\therefore 15 &= n \\
S_{n} &= \frac{n}{2}[a + l] \\
S_{15} &= \frac{15}{2}[3 + 101 ] \\
S_{15} &= (15)(52)
\end{align*}
Now consider the quotient of the two series:
\begin{align*}
\frac{ S_{25} }{ S_{15} } &= \frac{25 \times 52}{15 \times 52} \\
&= \frac{25}{15} \\
&= \frac{5}{3}
\end{align*}
Find the first term and the common difference.
\begin{align*}
T_{n} &= a + (n-1)d \\
T_{2} &= a + d \\
-4 &= a + d \ldots \ldots (1) \\
S_{n} &= \frac{n}{2}[2a + (n-1)d] \\
S_{6} &= \frac{6}{2}[2a + (6-1)d] \\
21 &= 3[2a + 5d] \\
\therefore 7 &= 2a + 5d \ldots \ldots (2) \\
\text{Eqn. } (1) \times 2: \quad -8 &= 2a + 2d \\
\text{Eqn. } (2) - 2(1): \quad 7 - (-8) &= (2a + 5d) - (2a + 2d) \\
15 &= 3d \\
\therefore 5 &= d \\
\text{And } a &= -4 - 5 \\
&= -9
\end{align*}
Hence determine \(T_{100}\).
[IEB, Nov 2004]
\begin{align*}
T_{n} &= a + (n-1)d \\
T_{100} &= -9 + (100-1)(5) \\
&= -9 + 495 \\
&=486
\end{align*}
\[\sum _{w=0}^{8}{(7w+8)}\]
Arithmetic series: \(8 + 15 + 22 + \ldots + 64\)
\begin{align*}
S_{n} &= \frac{n}{2}[2a + (n-1)d]\\
a &= 8 \\
d &= 15 - 8 = 7 \\
\therefore S_{9} &= \frac{9}{2}[2(8) + (9-1)(7)]\\
&= \frac{9}{2}[16 + 56]\\
&= \frac{9}{2}[72]\\
&= (9)(36) \\
&= 324
\end{align*}
\[\sum _{j=1}^{8}{7j+8}\]
Arithmetic series: \(7 + 14 + 21 + \ldots + 56\)
\begin{align*}
S_{n} &= \frac{n}{2}[2a + (n-1)d]\\
a &= 7 \\
d &= 14 - 7 = 7 \\
\therefore S_{8} &= \frac{8}{2}[2(7) + (8-1)(7)]\\
&= 4[14 + 49]\\
&= 4 (63) \\
&= 252 \\
\therefore S_{8} + 8 &= 260
\end{align*}
\begin{align*}
\text{Or } S_{n} &= \frac{n}{2}[a + l]\\
a &= 7 \\
l &= 56 \\
\therefore S_{8} &= \frac{8}{2}[7 + 56]\\
&= 4 (63) \\
&= 252 \\
\therefore S_{8} + 8 &= 260
\end{align*}
Determine the value of \(n\).
\[\sum _{c=1}^{n}{(2 - 3c)} = -330\]
Series: \(-1 - 4 - 7 \ldots + (2 - 3n)\)
\begin{align*}
a &= -1 \\
d &= T_{2} - T_{1} = -4 - (-1) = -3 \\
d &= T_{3} - T_{2} = -7 - (-4) = -3 \\
\therefore &\text{ this is an arithmetic series} \\
\therefore S_{n} &= \frac{n}{2}[2a + (n-1)d]\\
-330 &= \frac{n}{2}[2(-1) + (n-1)(-3)]\\
- 660 &= n[-2 - 3n + 3 ] \\
- 660 &= n - 3n^{2} \\
\therefore 0 &= -3n^{2} + n + 660 \\
0 &= 3n^{2} - n - 660 \\
0 &= (3n + 44)(n - 15) \\
\therefore n = -\frac{44}{3} &\text{ or } n = 15
\end{align*}
but \(n\) must be a positive integer, therefore \(n = 15\).
Alternative method:
\begin{align*}
a &= -1 \\
l &= 2 - 3n\\
\therefore S_{n} &= \frac{n}{2}[a + 1] \\
-330 &= \frac{n}{2}[-1 + 2 -3n] \\
- 660 &= n(1 -3n) \\
- 660 &= n - 3n^{2} \\
\therefore 0 &= -3n^{2} + n + 660 \\
0 &= 3n^{2} - n - 660 \\
0 &= (3n + 44)(n - 15) \\
\therefore n = -\frac{44}{3} &\text{ or } n = 15
\end{align*}
but \(n\) must be a positive integer, therefore \(n = 15\).
The sum of \(n\) terms of an arithmetic series is \(5{n}^{2}-11n\) for all values of \(n\). Determine
the common difference.
\begin{align*}
{S}_{n} &= 5n^{2} - 11n \\
\therefore S_{1} &= 5(1)^{2} - 11(1) \\
&= -6 \\
\text{And } S_{2} &= 5(2)^{2} - 11(2) \\
&= 20 - 22 \\
&= -2 \\
&= T_{1} + T_{2} \\
\therefore T _{2} &= S_{2} - S_{1} \\
&= -2 - (-6) \\
&= 4 \\
\therefore d &= T_{2} - T_{1} \\
&= 4 - (-6) \\
&= 10
\end{align*}
The sum of an arithmetic series is \(\text{100}\) times its first term, while the last term is
\(\text{9}\) times the first term. Calculate the number of terms in the series if the first term is not
equal to zero.
\begin{align*}
{S}_{n} &= 100a \\
l &= 9a \\
{S}_{n} &= \frac{n}{2}[a + l] \\
100a &= \frac{n}{2}[a + 9a] \\
100a &= \frac{n}{2}[10a] \\
100a &= 5a(n) \\
\frac{100a}{5a} &= n \\
\therefore 20 &= n
\end{align*}
