Difference between revisions of "1999 AHSME Problems/Problem 20"
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<math>\textrm{(A)} \ 29 \qquad \textrm{(B)} \ 59 \qquad \textrm{(C)} \ 79 \qquad \textrm{(D)} \ 99 \qquad \textrm{(E)} \ 179</math> | <math>\textrm{(A)} \ 29 \qquad \textrm{(B)} \ 59 \qquad \textrm{(C)} \ 79 \qquad \textrm{(D)} \ 99 \qquad \textrm{(E)} \ 179</math> | ||
− | == Solution == | + | == Solution 1== |
Let <math>m</math> be the arithmetic mean of <math>a_1</math> and <math>a_2</math>. We can then write <math>a_1=m-x</math> and <math>a_2=m+x</math> for some <math>x</math>. | Let <math>m</math> be the arithmetic mean of <math>a_1</math> and <math>a_2</math>. We can then write <math>a_1=m-x</math> and <math>a_2=m+x</math> for some <math>x</math>. |
Revision as of 09:15, 17 April 2022
Problem
The sequence satisfies , and, for all , is the arithmetic mean of the first terms. Find .
Solution 1
Let be the arithmetic mean of and . We can then write and for some .
By definition, .
Next, is the mean of , and , which is again .
Realizing this, one can easily prove by induction that .
It follows that . From we get that . And thus .
See also
1999 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 19 |
Followed by Problem 21 | |
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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