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Write the sum \(\displaystyle 15+14+13 \ldots -11-12\) in sigma notation and then find the sum.

The common difference is given by \(\displaystyle a_2-a_1=14-(15)=-1\). Using the first term gives the sequene \(\displaystyle a_n= 15+(n-1)(-1)\). Setting the general term equal to the last term and solving for \(\displaystyle n\) gives \(\displaystyle 15+(n-1)(-1)=-12 \implies n = 28 \). Writing in sigma notation gives \(\displaystyle \displaystyle \sum_{n=1}^{28} \left(16 - n\right)\). Using the formula for a finite arithmetic sum gives \(\displaystyle \frac{ 28(15-12) }{2}=42\).

Download \(\LaTeX\) 

\begin{question}Write the sum $15+14+13 \ldots -11-12$ in sigma notation and then find the sum.
    \soln{9cm}{The common difference is given by $a_2-a_1=14-(15)=-1$. Using the first term gives the sequene $a_n= 15+(n-1)(-1)$. Setting the general term equal to the last term and solving for $n$ gives $15+(n-1)(-1)=-12 \implies n = 28 $. Writing in sigma notation gives $\displaystyle \sum_{n=1}^{28} \left(16 - n\right)$. Using the formula for a finite arithmetic sum gives $\frac{ 28(15-12) }{2}=42$. }

\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 15+14+13 \ldots -11-12 " src="/equation_images/%20%5Cdisplaystyle%2015%2B14%2B13%20%5Cldots%20-11-12%20" alt="LaTeX:  \displaystyle 15+14+13 \ldots -11-12 " data-equation-content=" \displaystyle 15+14+13 \ldots -11-12 " />  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=14-(15)=-1 " src="/equation_images/%20%5Cdisplaystyle%20a_2-a_1%3D14-%2815%29%3D-1%20" alt="LaTeX:  \displaystyle a_2-a_1=14-(15)=-1 " data-equation-content=" \displaystyle a_2-a_1=14-(15)=-1 " /> . Using the first term gives the sequene  <img class="equation_image" title=" \displaystyle a_n= 15+(n-1)(-1) " src="/equation_images/%20%5Cdisplaystyle%20a_n%3D%2015%2B%28n-1%29%28-1%29%20" alt="LaTeX:  \displaystyle a_n= 15+(n-1)(-1) " data-equation-content=" \displaystyle a_n= 15+(n-1)(-1) " /> . 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 15+(n-1)(-1)=-12 \implies n = 28  " src="/equation_images/%20%5Cdisplaystyle%2015%2B%28n-1%29%28-1%29%3D-12%20%5Cimplies%20n%20%3D%2028%20%20" alt="LaTeX:  \displaystyle 15+(n-1)(-1)=-12 \implies n = 28  " data-equation-content=" \displaystyle 15+(n-1)(-1)=-12 \implies n = 28  " /> . Writing in sigma notation gives  <img class="equation_image" title=" \displaystyle \displaystyle \sum_{n=1}^{28} \left(16 - n\right) " src="/equation_images/%20%5Cdisplaystyle%20%5Cdisplaystyle%20%5Csum_%7Bn%3D1%7D%5E%7B28%7D%20%5Cleft%2816%20-%20n%5Cright%29%20" alt="LaTeX:  \displaystyle \displaystyle \sum_{n=1}^{28} \left(16 - n\right) " data-equation-content=" \displaystyle \displaystyle \sum_{n=1}^{28} \left(16 - n\right) " /> . Using the formula for a finite arithmetic sum gives  <img class="equation_image" title=" \displaystyle \frac{ 28(15-12) }{2}=42 " src="/equation_images/%20%5Cdisplaystyle%20%5Cfrac%7B%2028%2815-12%29%20%7D%7B2%7D%3D42%20" alt="LaTeX:  \displaystyle \frac{ 28(15-12) }{2}=42 " data-equation-content=" \displaystyle \frac{ 28(15-12) }{2}=42 " /> . </p> </p>