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In a set of four numbers, the first three are in a geometric sequence and the last three are in an arithmetic sequence with a common difference of 6. If the first number is the same as the fourth, find the four numbers.

 Jun 11, 2022

Best Answer 

 #1
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Let the common ratio of the first three numbers is \(r\) and the numbers be \(x_1,x_2,x_3,\) and \(x_4,\) respectively. Note that we can replace the value of the first number anytime with the fourth.

 

From the fact that the last three numbers are in an arithmetic sequence, we can rewrite the set as 

 

\(x_2+12,x_2,x_2+6,x_2+12\)

 

We can then get the pair of equations

 

\(r(x_2+12)r=x_2\)

\(rx_2=x_2+6\)

 

Solving, we find that \(x_2=-4\) and \(r=-1/2\). Therefore, the set of four numbers is

 

\(\boxed{8,-4,2,8}\)

 Jun 11, 2022
 #1
avatar+68 
+1
Best Answer

Let the common ratio of the first three numbers is \(r\) and the numbers be \(x_1,x_2,x_3,\) and \(x_4,\) respectively. Note that we can replace the value of the first number anytime with the fourth.

 

From the fact that the last three numbers are in an arithmetic sequence, we can rewrite the set as 

 

\(x_2+12,x_2,x_2+6,x_2+12\)

 

We can then get the pair of equations

 

\(r(x_2+12)r=x_2\)

\(rx_2=x_2+6\)

 

Solving, we find that \(x_2=-4\) and \(r=-1/2\). Therefore, the set of four numbers is

 

\(\boxed{8,-4,2,8}\)

RainbowSquirrel53 Jun 11, 2022

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