General Formula for a Finite Arithmetic Series
General Formula for a Finite Arithmetic Series
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} & = \cfrac{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} & = \cfrac{n}{2}(a + [a + (n-1)d ]) \\ \therefore S_{n} &= \cfrac{n}{2}[2a + (n-1)d ]\end{align*}
The general formula for determining the sum of an arithmetic series is given by:
\[{S}_{n}= \cfrac{n}{2} [ 2{a} + (n-1) d ]\]or
\[{S}_{n}= \cfrac{n}{2} (a + l)\]For example, we can calculate the sum \({S}_{20}\) for the arithmetic sequence \({T}_{n}=3+7 (n-1)\) by summing all the individual terms:
\begin{align*}{S}_{20} & = \sum _{n=1}^{20} [3+7 (n-1)] \\& = 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} & = \cfrac{n}{2}(2a + (n-1)d ) \\{S}_{20} & = \cfrac{20}{2} [2(3) + 7 (20-1)] \\& = 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.
Example
Question
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} &= \cfrac{n}{2}[2a + (n-1)d ] \\ S_{30} &= \cfrac{30}{2}[2(2) + (30-1)(7) ] \\ &= 15(4 + 203) \\ &= 15 (207) \\ &= 3105 \end{align*}Write the final answer
\(S_{30} = 3105\)
Example
Question
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} &= \cfrac{n}{2}(a + l ) \\ S_{65} &= \cfrac{65}{2}(-5 + 123) \\ &= \cfrac{65}{2}(118) \\ &= 3835 \end{align*}Write the final answer
\(S_{65} = 3835\)
Example
Question
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} &= \cfrac{n}{2}(2a + (n-1)d ) \\ 2146 &= \cfrac{n}{2}(2(4) + (n-1)(3) ) \\ 4292 &= n(8 + 3n - 3) \\ \therefore 0 &= 3n^2 + 5n -4292 \\ &= (3n + 116)(n - 37 ) \\ \therefore n = -\cfrac{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
\(S_{37} = 2146\)
Example
Question
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} &= \cfrac{n}{2}(2a + (n-1)d ) \\ S_{36} &= \cfrac{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 \(T_{1} = S_{1}\)
This lesson is part of:
Sequences and Series