Guided Learning Extension
In this example of an arithemtic series, we'll be working backwards to find the number of terms whose sum is a given value.
Given \(a_{12}=51\) and \(a_{25}=12\), how many total terms of the sequences must be added together to get a sum of \(945\)?
Using the summation formula for arithmetic series: \(S_n=\frac{n}{2}\left(a_1+a_n\right)\), we observe that the only variable we can substitute for is \(S_n=945\). In this case, we need to find the explicit formula for the nth term before we proceed.
Using the given term values, write out two equations.
\(51=a_1+d\left(12-1\right)\)
\(51=a_1+11d\)
\(12=a_1+d\left(25-1\right)\)
\(12=a_1+24d\)
Use elimination to find the values of the first term and common difference.
Common Difference:
\(51=a_1+11d\)
\(-(12=a_1+24d)\)
\(39=-13d\)
\(-3=d\)
Substitute the common difference (\(d=-3\)) in to solve for \(a_1\).
\(51=a_1+11\left(-3\right)\)
\(51=a_1-33\)
\(84=a_1\)
Therefore, \(a_n=84-3\left(n-1\right)\), or, in simplified form, \(a_n=87-3n\).
We are now ready to use the summation formula for arithmetic series.
\(945=\frac{n}{2}\left(84+87-3n\right)\)
\(1890=n\left(171-3n\right)\)
\(1890=171n-3n^2\)
\(3n^2-171n+1890=0\)
When we solve this quadratic by graphing (or any other method) we get two solutions, \(15\) and \(42\).
Extension Challenge: Explain how the first \(15\) terms add to be \(945\) and how the first \(42\) terms of this sequence also add to be \(945\).
Guided Learning
In this example of an arithemtic series, we'll be working backwards to find the number of terms whose sum is a given value.
Given \(a_{12}=51\) and \(a_{25}=12\), how many total terms of the sequences must be added together to get a sum of \(945\)?
Using the summation formula for arithmetic series: \(S_n=\frac{n}{2}\left(a_1+a_n\right)\), we observe that the only variable we can substitute for is \(S_n=945\). In this case, we need to find the explicit formula for the nth term before we proceed.
Using the given term values, write out two equations.
\(51=a_1+d\left(12-1\right)\)
\(51=a_1+11d\)
\(12=a_1+d\left(25-1\right)\)
\(12=a_1+24d\)
Use elimination to find the values of the first term and common difference.
Common Difference:
\(51=a_1+11d\)
\(-(12=a_1+24d)\)
\(39=-13d\)
\(-3=d\)
Substitute the common difference (\(d=-3\)) in to solve for \(a_1\).
\(51=a_1+11\left(-3\right)\)
\(51=a_1-33\)
\(84=a_1\)
Therefore, \(a_n=84-3\left(n-1\right)\), or, in simplified form, \(a_n=87-3n\).
We are now ready to use the summation formula for arithmetic series.
\(945=\frac{n}{2}\left(84+87-3n\right)\)
\(1890=n\left(171-3n\right)\)
\(1890=171n-3n^2\)
\(3n^2-171n+1890=0\)
When we solve this quadratic by graphing (or any other method) we get two solutions, \(15\) and \(42\).
Extension Challenge: Explain how the first \(15\) terms add to be \(945\) and how the first \(42\) terms of this sequence also add to be \(945\).
Guided Learning