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Write the sum \(\displaystyle -12-21-30 \ldots -354-363\) in sigma notation and then find the sum.

The common difference is given by \(\displaystyle a_2-a_1=-21-(-12)=-9\). Using the first term gives the sequene \(\displaystyle a_n= -12+(n-1)(-9)\). Setting the general term equal to the last term and solving for \(\displaystyle n\) gives \(\displaystyle -12+(n-1)(-9)=-363 \implies n = 40 \). Writing in sigma notation gives \(\displaystyle \displaystyle \sum_{n=1}^{40} \left(- 9 n - 3\right)\). Using the formula for a finite arithmetic sum gives \(\displaystyle \frac{ 40(-12-363) }{2}=-7500\).

Download \(\LaTeX\) 

\begin{question}Write the sum $-12-21-30 \ldots -354-363$ in sigma notation and then find the sum.
    \soln{9cm}{The common difference is given by $a_2-a_1=-21-(-12)=-9$. Using the first term gives the sequene $a_n= -12+(n-1)(-9)$. Setting the general term equal to the last term and solving for $n$ gives $-12+(n-1)(-9)=-363 \implies n = 40 $. Writing in sigma notation gives $\displaystyle \sum_{n=1}^{40} \left(- 9 n - 3\right)$. Using the formula for a finite arithmetic sum gives $\frac{ 40(-12-363) }{2}=-7500$. }

\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 -12-21-30 \ldots -354-363 " src="/equation_images/%20%5Cdisplaystyle%20-12-21-30%20%5Cldots%20-354-363%20" alt="LaTeX:  \displaystyle -12-21-30 \ldots -354-363 " data-equation-content=" \displaystyle -12-21-30 \ldots -354-363 " />  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=-21-(-12)=-9 " src="/equation_images/%20%5Cdisplaystyle%20a_2-a_1%3D-21-%28-12%29%3D-9%20" alt="LaTeX:  \displaystyle a_2-a_1=-21-(-12)=-9 " data-equation-content=" \displaystyle a_2-a_1=-21-(-12)=-9 " /> . Using the first term gives the sequene  <img class="equation_image" title=" \displaystyle a_n= -12+(n-1)(-9) " src="/equation_images/%20%5Cdisplaystyle%20a_n%3D%20-12%2B%28n-1%29%28-9%29%20" alt="LaTeX:  \displaystyle a_n= -12+(n-1)(-9) " data-equation-content=" \displaystyle a_n= -12+(n-1)(-9) " /> . 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 -12+(n-1)(-9)=-363 \implies n = 40  " src="/equation_images/%20%5Cdisplaystyle%20-12%2B%28n-1%29%28-9%29%3D-363%20%5Cimplies%20n%20%3D%2040%20%20" alt="LaTeX:  \displaystyle -12+(n-1)(-9)=-363 \implies n = 40  " data-equation-content=" \displaystyle -12+(n-1)(-9)=-363 \implies n = 40  " /> . Writing in sigma notation gives  <img class="equation_image" title=" \displaystyle \displaystyle \sum_{n=1}^{40} \left(- 9 n - 3\right) " src="/equation_images/%20%5Cdisplaystyle%20%5Cdisplaystyle%20%5Csum_%7Bn%3D1%7D%5E%7B40%7D%20%5Cleft%28-%209%20n%20-%203%5Cright%29%20" alt="LaTeX:  \displaystyle \displaystyle \sum_{n=1}^{40} \left(- 9 n - 3\right) " data-equation-content=" \displaystyle \displaystyle \sum_{n=1}^{40} \left(- 9 n - 3\right) " /> . Using the formula for a finite arithmetic sum gives  <img class="equation_image" title=" \displaystyle \frac{ 40(-12-363) }{2}=-7500 " src="/equation_images/%20%5Cdisplaystyle%20%5Cfrac%7B%2040%28-12-363%29%20%7D%7B2%7D%3D-7500%20" alt="LaTeX:  \displaystyle \frac{ 40(-12-363) }{2}=-7500 " data-equation-content=" \displaystyle \frac{ 40(-12-363) }{2}=-7500 " /> . </p> </p>