Difference between revisions of "1972 AHSME Problems/Problem 16"
Lopkiloinm (talk | contribs) (Created page with "== Problem 16 == There are two positive numbers that may be inserted between <math>3</math> and <math>9</math> such that the first three are in geometric progression while...") |
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\textbf{(E) }9\frac{1}{2}</math> | \textbf{(E) }9\frac{1}{2}</math> | ||
| − | == Solution == | + | == Solution 1 == |
Let <math>a</math> be first and <math>b</math> be second. We can then get equations based on our knowledge: <math>b-a = 9-b</math> and <math>b/a = a/3</math>. We then get <math>b = (9+a)/2</math> from our first equation and we substitute that into the second to get <math>\frac{9+a}{2a} = a/3</math> which simplifies to <math>2a^2-3a-27=0</math> which be <math>(2a-9)(a+3) = 0</math> and <math>a=9/2</math>. Then <math>b=27/4</math>. Their sums be <math>\boxed{\textbf{(B) }11\frac{1}{4}}</math>. ~lopkiloinm | Let <math>a</math> be first and <math>b</math> be second. We can then get equations based on our knowledge: <math>b-a = 9-b</math> and <math>b/a = a/3</math>. We then get <math>b = (9+a)/2</math> from our first equation and we substitute that into the second to get <math>\frac{9+a}{2a} = a/3</math> which simplifies to <math>2a^2-3a-27=0</math> which be <math>(2a-9)(a+3) = 0</math> and <math>a=9/2</math>. Then <math>b=27/4</math>. Their sums be <math>\boxed{\textbf{(B) }11\frac{1}{4}}</math>. ~lopkiloinm | ||
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| + | == Solution 2 == | ||
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| + | Like in Solution 1, let <math>a</math> be the first number and <math>b</math> be the second. From the problem, we have that <math>a = 3r</math> and <math>b = 3r^2</math> for some common ratio r. Also, <math>9 = a + 2d</math> for some common difference d. We use the first and second listed equations to get that <math>3r + d = 3r^2</math>, and we use the first and third equations to find that <math>9 = 3r + 2d</math>. Solving the latter equation for d and substituting into the former, we find that <math>3r + \frac{9 - 3r}{2} = 3r^2</math> or <math>6r + (9 - 3r) = 6r^2</math>. Moving every term to one side and factoring, we get that <math>3(2r - 3)(r + 1) = 0</math>. Clearly, the common ratio cannot be <math>-1</math>, so we ascertain that the common ratio is <math>3/2</math>. From here, we find that <math>a = 3 \cdot \frac{3}{2} = \frac{9}{2}</math>, and <math>b = \frac{9}{2} \cdot \frac{3}{2} = 27/4</math>. Then their sum is <math>\boxed{\textbf{(B) }11\frac{1}{4}}</math>. | ||
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| + | ~ cxsmi | ||
Revision as of 00:30, 20 January 2024
Problem 16
There are two positive numbers that may be inserted between 
and 
such that the first three are in geometric progression
while the last three are in arithmetic progression. The sum of those two positive numbers is
Solution 1
Let 
be first and 
be second. We can then get equations based on our knowledge: 
and 
. We then get 
from our first equation and we substitute that into the second to get 
which simplifies to 
which be 
and 
. Then 
. Their sums be 
. ~lopkiloinm
Solution 2
Like in Solution 1, let be the first number and be the second. From the problem, we have that 
and 
for some common ratio r. Also, 
for some common difference d. We use the first and second listed equations to get that 
, and we use the first and third equations to find that 
. Solving the latter equation for d and substituting into the former, we find that 
or 
. Moving every term to one side and factoring, we get that 
. Clearly, the common ratio cannot be 
, so we ascertain that the common ratio is 
. From here, we find that 
, and 
. Then their sum is .
~ cxsmi
