web2.0calc
 
+0  
 

 is the textbook wrong?

0
727
3
avatar+251 

Solve for x,

 

\(438 = \frac{n}{2}[40+(n-1)4]\)

 

the text book seems to think this simplifty's down to:

 

\(438 = 20n+2n-2\)

 

but I believe it works like this?

 

\(438=20n+2n^2-2n\)

 

any assistance is much appreciated.

 Dec 31, 2016

Best Answer 

 #3
avatar+129852 
+10

438  = [n / 2] [ 40 + ( n − 1 ) 4 ]

 

876   =   [n] [40  + ( n − 1 ) 4 ]

 

876  = n [ 40 + 4n − 4 ]

 

876  =  n [ 36 + 4n]

 

876  = 36n + 4n^2

 

219  = 9n + n^2

 

n^2 + 9n  −  219  = 0

 

n  =  [  −9 ± √ 957 ] / 2

 

 

cool cool cool

 Dec 31, 2016

3+0 Answers

 #1
avatar+37146 
+5

I believe  YOU have it correct vest4R....do you need any more help?

 Dec 31, 2016
 #2
avatar+251 
+5

Thanks for the feedback.

 

I'm good from here thankyou.

 

It's frustrating when this happens because this is the onlybexample on the sum of terms of an arithmetic sequence

 Dec 31, 2016
 #3
avatar+129852 
+10
Best Answer

438  = [n / 2] [ 40 + ( n − 1 ) 4 ]

 

876   =   [n] [40  + ( n − 1 ) 4 ]

 

876  = n [ 40 + 4n − 4 ]

 

876  =  n [ 36 + 4n]

 

876  = 36n + 4n^2

 

219  = 9n + n^2

 

n^2 + 9n  −  219  = 0

 

n  =  [  −9 ± √ 957 ] / 2

 

 

cool cool cool

CPhill Dec 31, 2016

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