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Two arithmetic sequences A and B both begin with 30 and have common differences of absolute value 10, with sequence A increasing and sequence B decreasing. What is the absolute value of the difference between the 51st term of sequence A and the 51st term of sequence B?

Lightning  Aug 26, 2018

Best Answer 

 #1
avatar+343 
+2

Arithmetic sequence \(A\), starts at 30, and goes up 10 each step.

 

Arithmetic sequence \(B\), starts at 30, and goes down 10 each step.

 

To find the \(n\)th term of a sequenece, we have a formula: \(a + (n-1)d\), where \(a\) is the first number of the sequence and \(d\) is the difference between each term. Let's use this formula to find the \(51\)st term of sequence \(A\).

 

We have \(30 + (51-1)10= 30 + 50\times10 = 30+500=530\).

 

For sequence \(B\), we have \(30 + (51-1)(-10) = 30 + 50(-10) = 30 + -500 = -470\)

 

So, the difference would be \(530 - (-470) = 530 + 470 = 1000\).

 

- Daisy

dierdurst  Aug 26, 2018
 #1
avatar+343 
+2
Best Answer

Arithmetic sequence \(A\), starts at 30, and goes up 10 each step.

 

Arithmetic sequence \(B\), starts at 30, and goes down 10 each step.

 

To find the \(n\)th term of a sequenece, we have a formula: \(a + (n-1)d\), where \(a\) is the first number of the sequence and \(d\) is the difference between each term. Let's use this formula to find the \(51\)st term of sequence \(A\).

 

We have \(30 + (51-1)10= 30 + 50\times10 = 30+500=530\).

 

For sequence \(B\), we have \(30 + (51-1)(-10) = 30 + 50(-10) = 30 + -500 = -470\)

 

So, the difference would be \(530 - (-470) = 530 + 470 = 1000\).

 

- Daisy

dierdurst  Aug 26, 2018
 #3
avatar+701 
+1

Thanks! Your right!

Lightning  Aug 26, 2018
 #2
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0

deleted.

Guest Aug 26, 2018
edited by Guest  Aug 26, 2018

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