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Write the sum \(\displaystyle -16-24-32 \ldots -184-192\) in sigma notation and then find the sum.

The common difference is given by \(\displaystyle a_2-a_1=-24-(-16)=-8\). Using the first term gives the sequene \(\displaystyle a_n= -16+(n-1)(-8)\). Setting the general term equal to the last term and solving for \(\displaystyle n\) gives \(\displaystyle -16+(n-1)(-8)=-192 \implies n = 23 \). Writing in sigma notation gives \(\displaystyle \displaystyle \sum_{n=1}^{23} \left(- 8 n - 8\right)\). Using the formula for a finite arithmetic sum gives \(\displaystyle \frac{ 23(-16-192) }{2}=-2392\).

Download \(\LaTeX\) 

\begin{question}Write the sum $-16-24-32 \ldots -184-192$ in sigma notation and then find the sum.
    \soln{9cm}{The common difference is given by $a_2-a_1=-24-(-16)=-8$. Using the first term gives the sequene $a_n= -16+(n-1)(-8)$. Setting the general term equal to the last term and solving for $n$ gives $-16+(n-1)(-8)=-192 \implies n = 23 $. Writing in sigma notation gives $\displaystyle \sum_{n=1}^{23} \left(- 8 n - 8\right)$. Using the formula for a finite arithmetic sum gives $\frac{ 23(-16-192) }{2}=-2392$. }

\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 -16-24-32 \ldots -184-192 " src="/equation_images/%20%5Cdisplaystyle%20-16-24-32%20%5Cldots%20-184-192%20" alt="LaTeX:  \displaystyle -16-24-32 \ldots -184-192 " data-equation-content=" \displaystyle -16-24-32 \ldots -184-192 " />  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=-24-(-16)=-8 " src="/equation_images/%20%5Cdisplaystyle%20a_2-a_1%3D-24-%28-16%29%3D-8%20" alt="LaTeX:  \displaystyle a_2-a_1=-24-(-16)=-8 " data-equation-content=" \displaystyle a_2-a_1=-24-(-16)=-8 " /> . Using the first term gives the sequene  <img class="equation_image" title=" \displaystyle a_n= -16+(n-1)(-8) " src="/equation_images/%20%5Cdisplaystyle%20a_n%3D%20-16%2B%28n-1%29%28-8%29%20" alt="LaTeX:  \displaystyle a_n= -16+(n-1)(-8) " data-equation-content=" \displaystyle a_n= -16+(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 -16+(n-1)(-8)=-192 \implies n = 23  " src="/equation_images/%20%5Cdisplaystyle%20-16%2B%28n-1%29%28-8%29%3D-192%20%5Cimplies%20n%20%3D%2023%20%20" alt="LaTeX:  \displaystyle -16+(n-1)(-8)=-192 \implies n = 23  " data-equation-content=" \displaystyle -16+(n-1)(-8)=-192 \implies n = 23  " /> . Writing in sigma notation gives  <img class="equation_image" title=" \displaystyle \displaystyle \sum_{n=1}^{23} \left(- 8 n - 8\right) " src="/equation_images/%20%5Cdisplaystyle%20%5Cdisplaystyle%20%5Csum_%7Bn%3D1%7D%5E%7B23%7D%20%5Cleft%28-%208%20n%20-%208%5Cright%29%20" alt="LaTeX:  \displaystyle \displaystyle \sum_{n=1}^{23} \left(- 8 n - 8\right) " data-equation-content=" \displaystyle \displaystyle \sum_{n=1}^{23} \left(- 8 n - 8\right) " /> . Using the formula for a finite arithmetic sum gives  <img class="equation_image" title=" \displaystyle \frac{ 23(-16-192) }{2}=-2392 " src="/equation_images/%20%5Cdisplaystyle%20%5Cfrac%7B%2023%28-16-192%29%20%7D%7B2%7D%3D-2392%20" alt="LaTeX:  \displaystyle \frac{ 23(-16-192) }{2}=-2392 " data-equation-content=" \displaystyle \frac{ 23(-16-192) }{2}=-2392 " /> . </p> </p>