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Questions: Algebra BusinessCalculus
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Write the sum \(\displaystyle 24+21+18 \ldots -117-120\) in sigma notation and then find the sum.
The common difference is given by \(\displaystyle a_2-a_1=21-(24)=-3\). Using the first term gives the sequene \(\displaystyle a_n= 24+(n-1)(-3)\). Setting the general term equal to the last term and solving for \(\displaystyle n\) gives \(\displaystyle 24+(n-1)(-3)=-120 \implies n = 49 \). Writing in sigma notation gives \(\displaystyle \displaystyle \sum_{n=1}^{49} \left(27 - 3 n\right)\). Using the formula for a finite arithmetic sum gives \(\displaystyle \frac{ 49(24-120) }{2}=-2352\).
\begin{question}Write the sum $24+21+18 \ldots -117-120$ in sigma notation and then find the sum.
\soln{9cm}{The common difference is given by $a_2-a_1=21-(24)=-3$. Using the first term gives the sequene $a_n= 24+(n-1)(-3)$. Setting the general term equal to the last term and solving for $n$ gives $24+(n-1)(-3)=-120 \implies n = 49 $. Writing in sigma notation gives $\displaystyle \sum_{n=1}^{49} \left(27 - 3 n\right)$. Using the formula for a finite arithmetic sum gives $\frac{ 49(24-120) }{2}=-2352$. }
\end{question}
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\begin{document}\begin{question}(10pts) The question goes here!
\soln{9cm}{The solution goes here.}
\end{question}\end{document}<p> <p>Write the sum <img class="equation_image" title=" \displaystyle 24+21+18 \ldots -117-120 " src="/equation_images/%20%5Cdisplaystyle%2024%2B21%2B18%20%5Cldots%20-117-120%20" alt="LaTeX: \displaystyle 24+21+18 \ldots -117-120 " data-equation-content=" \displaystyle 24+21+18 \ldots -117-120 " /> in sigma notation and then find the sum.</p> </p>
<p> <p>The common difference is given by <img class="equation_image" title=" \displaystyle a_2-a_1=21-(24)=-3 " src="/equation_images/%20%5Cdisplaystyle%20a_2-a_1%3D21-%2824%29%3D-3%20" alt="LaTeX: \displaystyle a_2-a_1=21-(24)=-3 " data-equation-content=" \displaystyle a_2-a_1=21-(24)=-3 " /> . Using the first term gives the sequene <img class="equation_image" title=" \displaystyle a_n= 24+(n-1)(-3) " src="/equation_images/%20%5Cdisplaystyle%20a_n%3D%2024%2B%28n-1%29%28-3%29%20" alt="LaTeX: \displaystyle a_n= 24+(n-1)(-3) " data-equation-content=" \displaystyle a_n= 24+(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 24+(n-1)(-3)=-120 \implies n = 49 " src="/equation_images/%20%5Cdisplaystyle%2024%2B%28n-1%29%28-3%29%3D-120%20%5Cimplies%20n%20%3D%2049%20%20" alt="LaTeX: \displaystyle 24+(n-1)(-3)=-120 \implies n = 49 " data-equation-content=" \displaystyle 24+(n-1)(-3)=-120 \implies n = 49 " /> . Writing in sigma notation gives <img class="equation_image" title=" \displaystyle \displaystyle \sum_{n=1}^{49} \left(27 - 3 n\right) " src="/equation_images/%20%5Cdisplaystyle%20%5Cdisplaystyle%20%5Csum_%7Bn%3D1%7D%5E%7B49%7D%20%5Cleft%2827%20-%203%20n%5Cright%29%20" alt="LaTeX: \displaystyle \displaystyle \sum_{n=1}^{49} \left(27 - 3 n\right) " data-equation-content=" \displaystyle \displaystyle \sum_{n=1}^{49} \left(27 - 3 n\right) " /> . Using the formula for a finite arithmetic sum gives <img class="equation_image" title=" \displaystyle \frac{ 49(24-120) }{2}=-2352 " src="/equation_images/%20%5Cdisplaystyle%20%5Cfrac%7B%2049%2824-120%29%20%7D%7B2%7D%3D-2352%20" alt="LaTeX: \displaystyle \frac{ 49(24-120) }{2}=-2352 " data-equation-content=" \displaystyle \frac{ 49(24-120) }{2}=-2352 " /> . </p> </p>