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Write the sum \(\displaystyle 5+0-5 \ldots -120-125\) in sigma notation and then find the sum.

The common difference is given by \(\displaystyle a_2-a_1=0-(5)=-5\). Using the first term gives the sequene \(\displaystyle a_n= 5+(n-1)(-5)\). Setting the general term equal to the last term and solving for \(\displaystyle n\) gives \(\displaystyle 5+(n-1)(-5)=-125 \implies n = 27 \). Writing in sigma notation gives \(\displaystyle \displaystyle \sum_{n=1}^{27} \left(10 - 5 n\right)\). Using the formula for a finite arithmetic sum gives \(\displaystyle \frac{ 27(5-125) }{2}=-1620\).

Download \(\LaTeX\) 

\begin{question}Write the sum $5+0-5 \ldots -120-125$ in sigma notation and then find the sum.
    \soln{9cm}{The common difference is given by $a_2-a_1=0-(5)=-5$. Using the first term gives the sequene $a_n= 5+(n-1)(-5)$. Setting the general term equal to the last term and solving for $n$ gives $5+(n-1)(-5)=-125 \implies n = 27 $. Writing in sigma notation gives $\displaystyle \sum_{n=1}^{27} \left(10 - 5 n\right)$. Using the formula for a finite arithmetic sum gives $\frac{ 27(5-125) }{2}=-1620$. }

\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 5+0-5 \ldots -120-125 " src="/equation_images/%20%5Cdisplaystyle%205%2B0-5%20%5Cldots%20-120-125%20" alt="LaTeX:  \displaystyle 5+0-5 \ldots -120-125 " data-equation-content=" \displaystyle 5+0-5 \ldots -120-125 " />  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=0-(5)=-5 " src="/equation_images/%20%5Cdisplaystyle%20a_2-a_1%3D0-%285%29%3D-5%20" alt="LaTeX:  \displaystyle a_2-a_1=0-(5)=-5 " data-equation-content=" \displaystyle a_2-a_1=0-(5)=-5 " /> . Using the first term gives the sequene  <img class="equation_image" title=" \displaystyle a_n= 5+(n-1)(-5) " src="/equation_images/%20%5Cdisplaystyle%20a_n%3D%205%2B%28n-1%29%28-5%29%20" alt="LaTeX:  \displaystyle a_n= 5+(n-1)(-5) " data-equation-content=" \displaystyle a_n= 5+(n-1)(-5) " /> . 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 5+(n-1)(-5)=-125 \implies n = 27  " src="/equation_images/%20%5Cdisplaystyle%205%2B%28n-1%29%28-5%29%3D-125%20%5Cimplies%20n%20%3D%2027%20%20" alt="LaTeX:  \displaystyle 5+(n-1)(-5)=-125 \implies n = 27  " data-equation-content=" \displaystyle 5+(n-1)(-5)=-125 \implies n = 27  " /> . Writing in sigma notation gives  <img class="equation_image" title=" \displaystyle \displaystyle \sum_{n=1}^{27} \left(10 - 5 n\right) " src="/equation_images/%20%5Cdisplaystyle%20%5Cdisplaystyle%20%5Csum_%7Bn%3D1%7D%5E%7B27%7D%20%5Cleft%2810%20-%205%20n%5Cright%29%20" alt="LaTeX:  \displaystyle \displaystyle \sum_{n=1}^{27} \left(10 - 5 n\right) " data-equation-content=" \displaystyle \displaystyle \sum_{n=1}^{27} \left(10 - 5 n\right) " /> . Using the formula for a finite arithmetic sum gives  <img class="equation_image" title=" \displaystyle \frac{ 27(5-125) }{2}=-1620 " src="/equation_images/%20%5Cdisplaystyle%20%5Cfrac%7B%2027%285-125%29%20%7D%7B2%7D%3D-1620%20" alt="LaTeX:  \displaystyle \frac{ 27(5-125) }{2}=-1620 " data-equation-content=" \displaystyle \frac{ 27(5-125) }{2}=-1620 " /> . </p> </p>