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Write the sum \(\displaystyle -11-8-5 \ldots +97+100\) in sigma notation and then find the sum.

The common difference is given by \(\displaystyle a_2-a_1=-8-(-11)=3\). Using the first term gives the sequene \(\displaystyle a_n= -11+(n-1)(3)\). Setting the general term equal to the last term and solving for \(\displaystyle n\) gives \(\displaystyle -11+(n-1)(3)=100 \implies n = 38 \). Writing in sigma notation gives \(\displaystyle \displaystyle \sum_{n=1}^{38} \left(3 n - 14\right)\). Using the formula for a finite arithmetic sum gives \(\displaystyle \frac{ 38(-11+100) }{2}=1691\).

Download \(\LaTeX\) 

\begin{question}Write the sum $-11-8-5 \ldots +97+100$ in sigma notation and then find the sum.
    \soln{9cm}{The common difference is given by $a_2-a_1=-8-(-11)=3$. Using the first term gives the sequene $a_n= -11+(n-1)(3)$. Setting the general term equal to the last term and solving for $n$ gives $-11+(n-1)(3)=100 \implies n = 38 $. Writing in sigma notation gives $\displaystyle \sum_{n=1}^{38} \left(3 n - 14\right)$. Using the formula for a finite arithmetic sum gives $\frac{ 38(-11+100) }{2}=1691$. }

\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 -11-8-5 \ldots +97+100 " src="/equation_images/%20%5Cdisplaystyle%20-11-8-5%20%5Cldots%20%2B97%2B100%20" alt="LaTeX:  \displaystyle -11-8-5 \ldots +97+100 " data-equation-content=" \displaystyle -11-8-5 \ldots +97+100 " />  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=-8-(-11)=3 " src="/equation_images/%20%5Cdisplaystyle%20a_2-a_1%3D-8-%28-11%29%3D3%20" alt="LaTeX:  \displaystyle a_2-a_1=-8-(-11)=3 " data-equation-content=" \displaystyle a_2-a_1=-8-(-11)=3 " /> . Using the first term gives the sequene  <img class="equation_image" title=" \displaystyle a_n= -11+(n-1)(3) " src="/equation_images/%20%5Cdisplaystyle%20a_n%3D%20-11%2B%28n-1%29%283%29%20" alt="LaTeX:  \displaystyle a_n= -11+(n-1)(3) " data-equation-content=" \displaystyle a_n= -11+(n-1)(3) " /> . 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 -11+(n-1)(3)=100 \implies n = 38  " src="/equation_images/%20%5Cdisplaystyle%20-11%2B%28n-1%29%283%29%3D100%20%5Cimplies%20n%20%3D%2038%20%20" alt="LaTeX:  \displaystyle -11+(n-1)(3)=100 \implies n = 38  " data-equation-content=" \displaystyle -11+(n-1)(3)=100 \implies n = 38  " /> . Writing in sigma notation gives  <img class="equation_image" title=" \displaystyle \displaystyle \sum_{n=1}^{38} \left(3 n - 14\right) " src="/equation_images/%20%5Cdisplaystyle%20%5Cdisplaystyle%20%5Csum_%7Bn%3D1%7D%5E%7B38%7D%20%5Cleft%283%20n%20-%2014%5Cright%29%20" alt="LaTeX:  \displaystyle \displaystyle \sum_{n=1}^{38} \left(3 n - 14\right) " data-equation-content=" \displaystyle \displaystyle \sum_{n=1}^{38} \left(3 n - 14\right) " /> . Using the formula for a finite arithmetic sum gives  <img class="equation_image" title=" \displaystyle \frac{ 38(-11+100) }{2}=1691 " src="/equation_images/%20%5Cdisplaystyle%20%5Cfrac%7B%2038%28-11%2B100%29%20%7D%7B2%7D%3D1691%20" alt="LaTeX:  \displaystyle \frac{ 38(-11+100) }{2}=1691 " data-equation-content=" \displaystyle \frac{ 38(-11+100) }{2}=1691 " /> . </p> </p>