Difference between revisions of "2002 AMC 12B Problems/Problem 9"
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==Solution== | ==Solution== | ||
=== Solution 1 === | === Solution 1 === | ||
− | We can let a=1, b=2, c=3, and d=4. <math>\frac{a}{d}=\boxed{\frac{1}{4}} \Longrightarrow \mathrm{(C)}</math> | + | We can let <math>a=1</math>, <math>b=2</math>, <math>c=3</math>, and <math>d=4</math>. <math>\frac{a}{d}=\boxed{\frac{1}{4}} \Longrightarrow \mathrm{(C)}</math> |
=== Solution 2 === | === Solution 2 === |
Latest revision as of 21:42, 11 December 2017
Problem
If are positive real numbers such that
form an increasing arithmetic sequence and
form a geometric sequence, then
is
Solution
Solution 1
We can let ,
,
, and
.
Solution 2
As is a geometric sequence, let
and
for some
.
Now, is an arithmetic sequence. Its difference is
. Thus
.
Comparing the two expressions for we get
. The positive solution is
, and
.
Solution 3
Letting be the common difference of the arithmetic progression, we have
,
,
. We are given that
=
, or
Cross-multiplying, we get
So
.
See also
2002 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 8 |
Followed by Problem 10 |
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 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.