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Write the sum \(\displaystyle 40+46+52 \ldots +328+334\) in sigma notation and then find the sum.

The common difference is given by \(\displaystyle a_2-a_1=46-(40)=6\). Using the first term gives the sequene \(\displaystyle a_n= 40+(n-1)(6)\). Setting the general term equal to the last term and solving for \(\displaystyle n\) gives \(\displaystyle 40+(n-1)(6)=334 \implies n = 50 \). Writing in sigma notation gives \(\displaystyle \displaystyle \sum_{n=1}^{50} \left(6 n + 34\right)\). Using the formula for a finite arithmetic sum gives \(\displaystyle \frac{ 50(40+334) }{2}=9350\).

Download \(\LaTeX\) 

\begin{question}Write the sum $40+46+52 \ldots +328+334$ in sigma notation and then find the sum.
    \soln{9cm}{The common difference is given by $a_2-a_1=46-(40)=6$. Using the first term gives the sequene $a_n= 40+(n-1)(6)$. Setting the general term equal to the last term and solving for $n$ gives $40+(n-1)(6)=334 \implies n = 50 $. Writing in sigma notation gives $\displaystyle \sum_{n=1}^{50} \left(6 n + 34\right)$. Using the formula for a finite arithmetic sum gives $\frac{ 50(40+334) }{2}=9350$. }

\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 40+46+52 \ldots +328+334 " src="/equation_images/%20%5Cdisplaystyle%2040%2B46%2B52%20%5Cldots%20%2B328%2B334%20" alt="LaTeX:  \displaystyle 40+46+52 \ldots +328+334 " data-equation-content=" \displaystyle 40+46+52 \ldots +328+334 " />  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=46-(40)=6 " src="/equation_images/%20%5Cdisplaystyle%20a_2-a_1%3D46-%2840%29%3D6%20" alt="LaTeX:  \displaystyle a_2-a_1=46-(40)=6 " data-equation-content=" \displaystyle a_2-a_1=46-(40)=6 " /> . Using the first term gives the sequene  <img class="equation_image" title=" \displaystyle a_n= 40+(n-1)(6) " src="/equation_images/%20%5Cdisplaystyle%20a_n%3D%2040%2B%28n-1%29%286%29%20" alt="LaTeX:  \displaystyle a_n= 40+(n-1)(6) " data-equation-content=" \displaystyle a_n= 40+(n-1)(6) " /> . 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 40+(n-1)(6)=334 \implies n = 50  " src="/equation_images/%20%5Cdisplaystyle%2040%2B%28n-1%29%286%29%3D334%20%5Cimplies%20n%20%3D%2050%20%20" alt="LaTeX:  \displaystyle 40+(n-1)(6)=334 \implies n = 50  " data-equation-content=" \displaystyle 40+(n-1)(6)=334 \implies n = 50  " /> . Writing in sigma notation gives  <img class="equation_image" title=" \displaystyle \displaystyle \sum_{n=1}^{50} \left(6 n + 34\right) " src="/equation_images/%20%5Cdisplaystyle%20%5Cdisplaystyle%20%5Csum_%7Bn%3D1%7D%5E%7B50%7D%20%5Cleft%286%20n%20%2B%2034%5Cright%29%20" alt="LaTeX:  \displaystyle \displaystyle \sum_{n=1}^{50} \left(6 n + 34\right) " data-equation-content=" \displaystyle \displaystyle \sum_{n=1}^{50} \left(6 n + 34\right) " /> . Using the formula for a finite arithmetic sum gives  <img class="equation_image" title=" \displaystyle \frac{ 50(40+334) }{2}=9350 " src="/equation_images/%20%5Cdisplaystyle%20%5Cfrac%7B%2050%2840%2B334%29%20%7D%7B2%7D%3D9350%20" alt="LaTeX:  \displaystyle \frac{ 50(40+334) }{2}=9350 " data-equation-content=" \displaystyle \frac{ 50(40+334) }{2}=9350 " /> . </p> </p>