An Arithmetic Series is the sum of all the terms in an arithmetic sequence up to a specific term.
They can be expressed algebraically using a couple of different formulas:
The first formula is used when the difference, \(d\), is known whereas the second formula is used when you know the final series term.
Determine the following of the series \(S_{25} = \Sigma(5i-2)\):
i. We can find the first \(4\) terms by substituting \(i\) with the series position.
We can determine the first term as such:
\(\textcolor{magenta}{t_1} = 5(1)-2\)
\(\textcolor{magenta}{t_1} = 5-2\)
\(\textcolor{magenta}{t_1 = 3}\)
We can determine the second term as such:
\(\textcolor{magenta}{t_2} = 5(2)-2\)
\(\textcolor{magenta}{t_2} = 10-2\)
\(\textcolor{magenta}{t_2 = 8}\)
We can determine the third term as such:
\(\textcolor{magenta}{t_3} = 5(3)-2\)
\(\textcolor{magenta}{t_3} = 15-2\)
\(\textcolor{magenta}{t_3 = 13}\)
We can determine the fourth term as such:
\(\textcolor{magenta}{t_4} = 5(4)-2\)
\(\textcolor{magenta}{t_4} = 20-2\)
\(\textcolor{magenta}{t_4 = 18}\)
Therefore, we can determine the first 4 terms in the series as \(\boldsymbol{3, 8, 13, 18}\).
ii. We can find the \(25\)th term by performing the same process we did in Step i:
\(\textcolor{magenta}{t_{25}} = 5(25)-2\)
\(\textcolor{magenta}{t_{25}} = 125-2\)
\(\textcolor{magenta}{t_{25} = 123}\)
Therefore, we can determine the \(25\)th term in the series as \(\boldsymbol{123}\).
iii. Since we know the last series term, we can find the sum of all \(25\) terms by using the second Sum formula:
\(\textcolor{olive}{S_n} = \cfrac{\textcolor{blue}{n}}{2}[\textcolor{teal}{t_1} + \textcolor{magenta}{t_n}]\)
\(\textcolor{olive}{S_n} = \cfrac{\textcolor{blue}{25}}{2}[\textcolor{teal}{3} + \textcolor{magenta}{123}]\)
\(\textcolor{olive}{S_n} = \cfrac{\textcolor{blue}{25}}{2}[126]\)
\(\textcolor{olive}{S_n} = \cfrac{3150}{2}\)
\(\textcolor{olive}{S_n = 1575}\)
Therefore, we can determine that the sum of all \(25\) terms in this series is \(\boldsymbol{1575}\).
First, we can determine the explicit series equations for the respective terms:
\(t_n = \textcolor{red}{a} + \textcolor{green}{d}(\textcolor{blue}{n}-1)\)
We can determine the first equation by plugging in the appropriate values and simplifying:
\(\text{E1}: \textcolor{magenta}{t₅} = \textcolor{red}{a} + \textcolor{green}{d}(\textcolor{blue}{5}-1)\)
\(\text{E1}: \textcolor{magenta}{-8} = \textcolor{red}{a} + 4\textcolor{green}{d}\)
We can determine the second equation by plugging in the appropriate values and simplifying:
\(\text{E2}: \textcolor{magenta}{t₉} = \textcolor{red}{a} + \textcolor{green}{d}(\textcolor{blue}{9}-1)\)
\(\text{E2}: \textcolor{magenta}{4} = \textcolor{red}{a} + 8\textcolor{green}{d}\)
We can then subtract one equation from the other in order to determine the difference. In this case, we will subtract Equation 1 from Equation 2:
\(\textcolor{magenta}{4} - (\textcolor{magenta}{-8}) = \cancel{(\textcolor{red}{a}} + 8\textcolor{green}{d}) - (\cancel{\textcolor{red}{a}} + 4\textcolor{green}{d})\)
\(12 = 4\textcolor{green}{d}\)
\(\cfrac{\cancel{4}\textcolor{green}{d}}{\cancel{4}} = \cfrac{12}{4}\)
\(\textcolor{green}{d = 3}\)
We can now plug the difference into one of the equations in order to get the initial value, \(a\). In this case, we will use the first equation:
\(\textcolor{magenta}{-8} = \textcolor{red}{a} + 4(\textcolor{green}{3})\)
\(\textcolor{red}{a} = \textcolor{magenta}{-8} - 12\)
\(\textcolor{red}{a = -20}\)
Finally, as we have identified the difference, we can now plug all pertinent values into the first Sum formula:
\(\textcolor{olive}{S_n} = \cfrac{\textcolor{blue}{n}}{2}[2\textcolor{red}{a} + \textcolor{green}{d}(\textcolor{blue}{n}-1)]\)
\(\textcolor{olive}{S_{20}} = \cfrac{\textcolor{blue}{20}}{2}[2(\textcolor{red}{-20}) + \textcolor{green}{3}(\textcolor{blue}{20}-1)]\)
\(\textcolor{olive}{S_{20}} = 10[-40 + 3(19)]\)
\(\textcolor{olive}{S_{20}} = 10[-40 + 57]\)
\(\textcolor{olive}{S_{20}} = 10[17]\)
\(\textcolor{olive}{S_{20} = 170}\)
Therefore, we can determine that \(\boldsymbol{S_{20} = 170}\).
The third term of an arithmetic series is \(9\) and the sum of the first \(10\) terms is \(-60\). Find the partial sum of the first \(20\) terms.
We can use a system of equations in order to find the first term. To find our first equation, we can plug the pertinent values into the explicit series formula:
\(\textcolor{magenta}{t_n} = \textcolor{red}{a} + \textcolor{green}{d}(\textcolor{blue}{n}-1)\)
\(\textcolor{magenta}{t_3} = \textcolor{red}{a} + \textcolor{green}{d}(\textcolor{blue}{3}-1)\)
\(\textcolor{magenta}{9} = \textcolor{red}{a} + 2\textcolor{green}{d}\)
In order to find the second equation, we can plug the pertinent values into the first Sum equation:
\(\textcolor{olive}{S_n} = \cfrac{\textcolor{blue}{n}}{2}[2\textcolor{red}{a} + \textcolor{green}{d}(\textcolor{blue}{n}-1)]\)
\(\textcolor{olive}{S_{10}} = \cfrac{\textcolor{blue}{10}}{2}[2\textcolor{red}{a} + \textcolor{green}{d}(\textcolor{blue}{10}-1)]\)
\(\textcolor{olive}{-60} = 5[2\textcolor{red}{a} + 9\textcolor{green}{d}]\)
\(\textcolor{olive}{-60} = 10\textcolor{red}{a} + 45\textcolor{green}{d}\)
In order to cancel out \(a\) terms, we can multiply the first equation by a factor of \(10\):
\(10(\textcolor{magenta}{9} = \textcolor{red}{a} + 2\textcolor{green}{d})\)
\(\textcolor{olive}{90} = 10\textcolor{red}{a} + 20\textcolor{green}{d}\)
We can now subtract one equation from the other in order to determine the difference. In this case, we will subtract Equation 1 from Equation 2:
\(\textcolor{olive}{-60} - \textcolor{olive}{90} = (10\textcolor{red}{a} + 45\textcolor{green}{d}) - (10\textcolor{red}{a} + 20\textcolor{green}{d})\)
\(-150 = 25\textcolor{green}{d}\)
\( \cfrac{\cancel{25}\textcolor{green}{d}}{\cancel{25}}= \cfrac{-150}{25}\)
\(\textcolor{green}{d = -6}\)
Next, we can plug the difference into one of the equations in order to get the initial value, \(a\). In this case, we will use the first equation:
\(\textcolor{magenta}{9} = \textcolor{red}{a} + 2(\textcolor{green}{-6})\)
\(\textcolor{magenta}{9} = \textcolor{red}{a} - 12\)
\(\textcolor{red}{a} = 9 + 12\)
\(\textcolor{red}{a = 21}\)
Finally, we can plug all pertinent values into the first Sum formula to determine \(S_{20}\):
\(\textcolor{olive}{S_n} = \cfrac{\textcolor{blue}{n}}{2}[2\textcolor{red}{a} + \textcolor{green}{d}(\textcolor{blue}{n}-1)]\)
\(\textcolor{olive}{S_{20}} = \cfrac{\textcolor{blue}{20}}{2}[2(\textcolor{red}{21}) + (\textcolor{green}{-6})(\textcolor{blue}{20}-1)]\)
\(\textcolor{olive}{S_{20}} = 10[42 + (\textcolor{green}{-6})(19)]\)
\(\textcolor{olive}{S_{20}} = 10[42 + -114]\)
\(\textcolor{olive}{S_{20}} = 10[-72]\)
\(\textcolor{olive}{S_{20} = -720}\)
Therefore, we can determine that \(\boldsymbol{S₂₀ = -720}\).
As we can identify the first (\(175\)) and final series terms (\(71\)) but not the difference, we can use second sum equation and work backwards to determine the # of rows:
\(\textcolor{olive}{Sₙ} = \cfrac{\textcolor{blue}{n}}{2}[\textcolor{teal}{t₁} + \textcolor{magenta}{tₙ}]\)
\(\textcolor{olive}{6519} = \cfrac{\textcolor{blue}{n}}{2}[\textcolor{teal}{175} + \textcolor{magenta}{71}]\)
\(\textcolor{olive}{6519} = \cfrac{246\textcolor{blue}{n}}{2}\)
\(6519 = 123\textcolor{blue}{n}\)
\( \cfrac{\cancel{123}\textcolor{blue}{n}}{\cancel{123}} = \cfrac{6519}{123}\)
\(\textcolor{blue}{n = 53}\)
Therefore, we can determine that Colin was able to complete \(\boldsymbol{53}\) rows of his pyramid.