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Difference between revisions of "1980 AHSME Problems/Problem 11"

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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.
 
Let <math>a</math> be the first term of the sequence and let <math>d</math> be the common difference of the sequence.
  
−
Sum of the first 10 terms: <math>\frac{10}{2}(2a+9d)=100 \Longleftrightarrow 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>.
−
Sum of the first 100 terms: <math>\frac{100}{2}(2a+99d)=10 \Longleftrightarrow 2a+99d=\frac{1}{5}</math>
 
  
−
Solving the system, we get <math>d=-\frac{11}{50}</math>, <math>a=\frac{1099}{100}</math>. The sum of the first 110 terms is <math>\frac{110}{2}(2a+109d)=55(-2)=-110</math>
+
The sum of the first <math>100</math> terms is <math>\frac{100}{2}(2a+99d)=10</math>. This equation can be simplified to <math>2a+99d=\frac{1}{5}</math>.
  
−
Therefore, <math>\boxed{D}</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>.
  
 
== See also ==
 
== See also ==

Revision as of 20:48, 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

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

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 $\frac{110}{2}(2a+109d)=55\left(2\cdot\frac{1099}{100} + 109 \cdot -\frac{11}{50}\right) = 55 \cdot -2 = \boxed{(\textbf{D}) -110}$.

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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