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Write the sum \(\displaystyle -48-40-32 \ldots +248+256\) in sigma notation and then find the sum.

The common difference is given by \(\displaystyle a_2-a_1=-40-(-48)=8\). Using the first term gives the sequene \(\displaystyle a_n= -48+(n-1)(8)\). Setting the general term equal to the last term and solving for \(\displaystyle n\) gives \(\displaystyle -48+(n-1)(8)=256 \implies n = 39 \). Writing in sigma notation gives \(\displaystyle \displaystyle \sum_{n=1}^{39} \left(8 n - 56\right)\). Using the formula for a finite arithmetic sum gives \(\displaystyle \frac{ 39(-48+256) }{2}=4056\).

Download \(\LaTeX\) 

\begin{question}Write the sum $-48-40-32 \ldots +248+256$ in sigma notation and then find the sum.
    \soln{9cm}{The common difference is given by $a_2-a_1=-40-(-48)=8$. Using the first term gives the sequene $a_n= -48+(n-1)(8)$. Setting the general term equal to the last term and solving for $n$ gives $-48+(n-1)(8)=256 \implies n = 39 $. Writing in sigma notation gives $\displaystyle \sum_{n=1}^{39} \left(8 n - 56\right)$. Using the formula for a finite arithmetic sum gives $\frac{ 39(-48+256) }{2}=4056$. }

\end{question}

Download Question and Solution Environment\(\LaTeX\)
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HTML for Canvas 
<p> <p>Write the sum  <img class="equation_image" title=" \displaystyle -48-40-32 \ldots +248+256 " src="/equation_images/%20%5Cdisplaystyle%20-48-40-32%20%5Cldots%20%2B248%2B256%20" alt="LaTeX:  \displaystyle -48-40-32 \ldots +248+256 " data-equation-content=" \displaystyle -48-40-32 \ldots +248+256 " />  in sigma notation and then find the sum.</p> </p>
HTML for Canvas
<p> <p>The common difference is given by  <img class="equation_image" title=" \displaystyle a_2-a_1=-40-(-48)=8 " src="/equation_images/%20%5Cdisplaystyle%20a_2-a_1%3D-40-%28-48%29%3D8%20" alt="LaTeX:  \displaystyle a_2-a_1=-40-(-48)=8 " data-equation-content=" \displaystyle a_2-a_1=-40-(-48)=8 " /> . Using the first term gives the sequene  <img class="equation_image" title=" \displaystyle a_n= -48+(n-1)(8) " src="/equation_images/%20%5Cdisplaystyle%20a_n%3D%20-48%2B%28n-1%29%288%29%20" alt="LaTeX:  \displaystyle a_n= -48+(n-1)(8) " data-equation-content=" \displaystyle a_n= -48+(n-1)(8) " /> . Setting the general term equal to the last term and solving for  <img class="equation_image" title=" \displaystyle n " src="/equation_images/%20%5Cdisplaystyle%20n%20" alt="LaTeX:  \displaystyle n " data-equation-content=" \displaystyle n " />  gives  <img class="equation_image" title=" \displaystyle -48+(n-1)(8)=256 \implies n = 39  " src="/equation_images/%20%5Cdisplaystyle%20-48%2B%28n-1%29%288%29%3D256%20%5Cimplies%20n%20%3D%2039%20%20" alt="LaTeX:  \displaystyle -48+(n-1)(8)=256 \implies n = 39  " data-equation-content=" \displaystyle -48+(n-1)(8)=256 \implies n = 39  " /> . Writing in sigma notation gives  <img class="equation_image" title=" \displaystyle \displaystyle \sum_{n=1}^{39} \left(8 n - 56\right) " src="/equation_images/%20%5Cdisplaystyle%20%5Cdisplaystyle%20%5Csum_%7Bn%3D1%7D%5E%7B39%7D%20%5Cleft%288%20n%20-%2056%5Cright%29%20" alt="LaTeX:  \displaystyle \displaystyle \sum_{n=1}^{39} \left(8 n - 56\right) " data-equation-content=" \displaystyle \displaystyle \sum_{n=1}^{39} \left(8 n - 56\right) " /> . Using the formula for a finite arithmetic sum gives  <img class="equation_image" title=" \displaystyle \frac{ 39(-48+256) }{2}=4056 " src="/equation_images/%20%5Cdisplaystyle%20%5Cfrac%7B%2039%28-48%2B256%29%20%7D%7B2%7D%3D4056%20" alt="LaTeX:  \displaystyle \frac{ 39(-48+256) }{2}=4056 " data-equation-content=" \displaystyle \frac{ 39(-48+256) }{2}=4056 " /> . </p> </p>