Difference between revisions of "1994 AHSME Problems/Problem 14"
m (→Solution) |
Logsobolev (talk | contribs) m (→Brief Introduction) |
||
Line 9: | Line 9: | ||
===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> | + | 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- | + | 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
Solution
Brief Introduction
For those that do not know the formula, the sum of an arithmetic series with first term , last term as terms, is We can prove this as follows:
Let be the common difference between terms of our series and let be the number of terms in our series. Let be the first term. Our series is Note that we have in the last term because is a term. Let be our sum such that We can rewrite our sums as Adding these two sums together essentially creates pairs of as shown below: We use in place of to represent the last term.
Solving
Our first term is and our last term is . To find the number of terms, , we note that the common difference between each term is . So we have Using our formula, our sum is
--Solution by TheMaskedMagician
See Also
1994 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.