We think you are located in United States. Is this correct?

# 1.4 Finite arithmetic series

## 1.4 Finite arithmetic series (EMCDX)

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 ($$d$$):

${T}_{n}={a} + \left(n-1\right)d$

where

• $${T}_{n}$$ is the $$n$$$$^{\text{th}}$$ term of the sequence;

• $${a}$$ is the first term;

• $$d$$ is the common difference.

When we sum a finite number of terms in an arithmetic sequence, we get a finite arithmetic series.

The sum of the first one hundred integers

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:

• Write the numbers in ascending order.
• Write the numbers in descending order.
• Add the corresponding pairs of terms together.
• Simplify the equation by making $$S_{n}$$ the subject of the equation.
\begin{align*} S_{100} &= \text{1} \enspace + \enspace \text{2} \enspace + \enspace \text{3} + \cdots + \text{98} + \text{99} + \text{100} \\ + \quad \underline{S_{100 }} &= \underline{ \text{100} + \text{99} +\text{98} + \cdots + \text{3} \enspace + \enspace \text{2} \enspace + \enspace \text{1}} \\ \therefore 2S_{100} &= \text{101} + \text{101} +\text{101} + \cdots + \text{101} +\text{101} +\text{101} \\ \therefore 2S_{\text{100}} &= \text{101} \times \text{100} \\ &= \text{10 100} \\ \therefore S_{\text{100}} &= \frac{\text{10 100}}{2} \\ &= \text{5 050} \end{align*}

### 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*}

$$S_{30} = 3105$$

temp text

## 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*}

$$S_{65} = 3835$$

## 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.

$$S_{37} = 2146$$

temp text

## 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*}

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*}$-3 -1 + 1$
temp text

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*}

The sum to $$n$$ terms of an arithmetic series is $${S}_{n}=\frac{n}{2} \left(7n+15\right)$$.

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*}

The second term of an arithmetic sequence is $$-\text{4}$$ and the sum of the first six terms of the series is $$\text{21}$$.

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*}

Determine the value of the following:

$\sum _{w=0}^{8}{(7w+8)}$

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*}