Difference between revisions of "1994 AHSME Problems/Problem 14"

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===Brief Introduction===
 
===Brief Introduction===
  
For those that do not know the formula, the sum of an arithmetic series with first term <math>a_1</math>, last term <math>a_n</math> as <math>n</math> terms, is <cmath>n=\frac{a_1+a_n}{2}.</cmath> We can prove this as follows:
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For those that do not know the formula, the sum of an arithmetic series with first term <math>a_1</math>, last term <math>a_n</math> as <math>n</math> terms, is <cmath>S = \frac{n(a_1+a_n)}{2}.</cmath> We can prove this as follows:
  
Let <math>d</math> be the common difference between terms of our series and let <math>n</math> be the number of terms in our series. Let <math>a_1</math> be the first term. Our series is <cmath>a_1,~a_1+d,~a_1+2d,\dots,a_1+(n-1)d.</cmath> Note that we have <math>n-1</math> in the last term because <math>a_1</math> is a term. Let <math>S</math> be our sum such that <cmath>S=a_1+(a_1+d)+\dots+(a_1+(n-d)d).</cmath> We can rewrite our sums as <cmath>S=(a_1+(n-1)d)+(a_1+(n-2)d)+\dots+(a_1+d)+a_1.</cmath> Adding these two sums together essentially creates <math>n</math> pairs of <math>a_1+(a_1+(n-1)d)</math> as shown below: <cmath>2S=n(a_1+(a_1+(n-1)d))\implies S=n\left(\frac{a_1+a_n}{2}\right).</cmath> We use <math>a_n</math> in place of <math>a_1+(n-1)d</math> to represent the last term.
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Let <math>d</math> be the common difference between terms of our series and let <math>n</math> be the number of terms in our series. Let <math>a_1</math> be the first term. Our series is <cmath>a_1,~a_1+d,~a_1+2d,\dots,a_1+(n-1)d.</cmath> Note that we have <math>n-1</math> in the last term because <math>a_1</math> is a term. Let <math>S</math> be our sum such that <cmath>S=a_1+(a_1+d)+\dots+(a_1+(n-1)d).</cmath> We can rewrite our sums as <cmath>S=(a_1+(n-1)d)+(a_1+(n-2)d)+\dots+(a_1+d)+a_1.</cmath> Adding these two sums together essentially creates <math>n</math> pairs of <math>a_1+(a_1+(n-1)d)</math> as shown below: <cmath>2S=n(a_1+(a_1+(n-1)d))\implies S=\frac{n(a_1+a_n)}{2}</cmath> We use <math>a_n</math> in place of <math>a_1+(n-1)d</math> to represent the last term.
  
  

Latest revision as of 02:45, 28 May 2021

Problem

Find the sum of the arithmetic series \[20+20\frac{1}{5}+20\frac{2}{5}+\cdots+40\] $\textbf{(A)}\ 3000 \qquad\textbf{(B)}\ 3030 \qquad\textbf{(C)}\ 3150 \qquad\textbf{(D)}\ 4100 \qquad\textbf{(E)}\ 6000$

Solution

Brief Introduction

For those that do not know the formula, the sum of an arithmetic series with first term $a_1$, last term $a_n$ as $n$ terms, is \[S = \frac{n(a_1+a_n)}{2}.\] We can prove this as follows:

Let $d$ be the common difference between terms of our series and let $n$ be the number of terms in our series. Let $a_1$ be the first term. Our series is \[a_1,~a_1+d,~a_1+2d,\dots,a_1+(n-1)d.\] Note that we have $n-1$ in the last term because $a_1$ is a term. Let $S$ be our sum such that \[S=a_1+(a_1+d)+\dots+(a_1+(n-1)d).\] We can rewrite our sums as \[S=(a_1+(n-1)d)+(a_1+(n-2)d)+\dots+(a_1+d)+a_1.\] Adding these two sums together essentially creates $n$ pairs of $a_1+(a_1+(n-1)d)$ as shown below: \[2S=n(a_1+(a_1+(n-1)d))\implies S=\frac{n(a_1+a_n)}{2}\] We use $a_n$ in place of $a_1+(n-1)d$ to represent the last term.



Solving

Our first term is $20$ and our last term is $40$. To find the number of terms, $n$, we note that the common difference between each term is $\frac{1}{5}$. So we have \[20+\frac{1}{5}(n-1)=40\implies n-1=100\implies n=101.\] Using our formula, our sum is \[101\left(\frac{20+40}{2}\right)=101\times 30=\boxed{\textbf{(B) }3030.}\]


--Solution by TheMaskedMagician

See Also

1994 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
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