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Difference between revisions of "1980 AHSME Problems/Problem 11"
(New solution using means of arithmetic sequences.)
 
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<math>\text{(A)} \ 90 \qquad \text{(B)} \ -90 \qquad \text{(C)} \ 110 \qquad \text{(D)} \ -110 \qquad \text{(E)} \ -100</math>
 
<math>\text{(A)} \ 90 \qquad \text{(B)} \ -90 \qquad \text{(C)} \ 110 \qquad \text{(D)} \ -110 \qquad \text{(E)} \ -100</math>
  
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== Solution ==
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== Solution 1 ==
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Let <math>a</math> be the first term of the sequence and let <math>d</math> be the common difference of the sequence.
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Let <math>a</math> and <math>d</math> be the first term and common difference of the sequence, respectively.
  
 
The sum of the first <math>10</math> terms is <math>\frac{10}{2}(2a+9d)=100</math>. This equation can be simplified to <math>2a+9d=20</math>.
 
The sum of the first <math>10</math> terms is <math>\frac{10}{2}(2a+9d)=100</math>. This equation can be simplified to <math>2a+9d=20</math>.
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Solving the system of these two equations yields <math>(a, d) = \left(\frac{1099}{100}, -\frac{11}{50}\right)</math>.  Therefore, the sum of the first <math>110</math> terms is <math>\frac{110}{2}(2a+109d)=55\left(2\cdot\frac{1099}{100} + 109 \cdot -\frac{11}{50}\right) = 55 \cdot -2 = \boxed{(\textbf{D}) -110}</math>.
 
Solving the system of these two equations yields <math>(a, d) = \left(\frac{1099}{100}, -\frac{11}{50}\right)</math>.  Therefore, the sum of the first <math>110</math> terms is <math>\frac{110}{2}(2a+109d)=55\left(2\cdot\frac{1099}{100} + 109 \cdot -\frac{11}{50}\right) = 55 \cdot -2 = \boxed{(\textbf{D}) -110}</math>.
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== Solution 2 ==
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Let <math>a_1</math>, <math>a_2</math>, ..., <math>a_n</math> be the terms of this arithmetic sequence, and let <math>d</math> be its common difference.  Recall that in an arithmetic sequence with an even number of terms, the mean of the middle two terms equals the mean of all terms in the sequence. 
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The mean of the first <math>10</math> terms is <math>\frac{a_5 + a_6}{2} = \frac{100}{10} = 10</math>, while the mean of the first <math>100</math> terms is <math>\frac{a_{50}+a_{51}}{2} = \frac{10}{100} = \frac{1}{10}</math>. 
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Then <math>\frac{1}{10} = \frac{a_{50}+a_{51}}{2} = \frac{a_5 + 45d + a_6 + 45d}{2} = \frac{a_5+a_6}{2} + 45d = 10 + 45d</math>, so <math>d = -\frac{11}{50}</math>.  So the mean of the first <math>110</math> terms is <math>\frac{a_{55}+a_{56}}{2} = \frac{a_{50}+5d+a_{51}+5d}{2} = \frac{a_{50}+a_{51}}{2}+5d=\frac{1}{10}+5\cdot-\frac{11}{50}=-1</math>.  Therefore, the sum of the first <math>110</math> terms is <math>110 \cdot -1 =  \boxed{(\textbf{D}) -110}</math>.
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-j314andrews
  
 
== See also ==
 
== See also ==

Latest revision as of 21:08, 25 July 2025

Problem

If the sum of the first $10$ terms and the sum of the first $100$ terms of a given arithmetic progression are $100$ and $10$, respectively, then the sum of first $110$ terms is:

$\text{(A)} \ 90 \qquad \text{(B)} \ -90 \qquad \text{(C)} \ 110 \qquad \text{(D)} \ -110 \qquad \text{(E)} \ -100$

Solution 1

Let $a$ and $d$ be the first term and common difference of the sequence, respectively.

The sum of the first $10$ terms is $\frac{10}{2}(2a+9d)=100$. This equation can be simplified to $2a+9d=20$.

The sum of the first $100$ terms is $\frac{100}{2}(2a+99d)=10$. This equation can be simplified to $2a+99d=\frac{1}{5}$.

Solving the system of these two equations yields $(a, d) = \left(\frac{1099}{100}, -\frac{11}{50}\right)$. Therefore, the sum of the first $110$ terms is $\frac{110}{2}(2a+109d)=55\left(2\cdot\frac{1099}{100} + 109 \cdot -\frac{11}{50}\right) = 55 \cdot -2 = \boxed{(\textbf{D}) -110}$.

Solution 2

Let $a_1$, $a_2$, ..., $a_n$ be the terms of this arithmetic sequence, and let $d$ be its common difference. Recall that in an arithmetic sequence with an even number of terms, the mean of the middle two terms equals the mean of all terms in the sequence.

The mean of the first $10$ terms is $\frac{a_5 + a_6}{2} = \frac{100}{10} = 10$, while the mean of the first $100$ terms is $\frac{a_{50}+a_{51}}{2} = \frac{10}{100} = \frac{1}{10}$.

Then $\frac{1}{10} = \frac{a_{50}+a_{51}}{2} = \frac{a_5 + 45d + a_6 + 45d}{2} = \frac{a_5+a_6}{2} + 45d = 10 + 45d$, so $d = -\frac{11}{50}$. So the mean of the first $110$ terms is $\frac{a_{55}+a_{56}}{2} = \frac{a_{50}+5d+a_{51}+5d}{2} = \frac{a_{50}+a_{51}}{2}+5d=\frac{1}{10}+5\cdot-\frac{11}{50}=-1$. Therefore, the sum of the first $110$ terms is $110 \cdot -1 = \boxed{(\textbf{D}) -110}$.

-j314andrews

See also

1980 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 10 		Followed by
Problem 12
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

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