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 Oct 22, 2018
 #1
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We  have this  arithmetic  series

 

10  +  12  + 14 + ......+  ???  =  220

 

The sum of an   arithmetic series is given  by :

 

S  =  (n / 2)  [ 2a  +  (n - 1) * d ]    

 

Where a  is the first term, n is the number of terms  [ and the number of rows in this case ] and  d  is the difference between terms

 

So we have

 

(n / 2)  [ 2(10)  + ( n - 1) * 2 ]   =  220        multiply both sides by 2

 

n [ 20 + 2n - 2 ]   =  440

 

n [ 18 + 2n ]  = 440

 

2n^2 + 18n - 440  =  0      divide through by 2

 

n^2 + 9n - 220  =  0       factor

 

(n + 20) ( n - 11)  =  0

 

Set eacn  factor to 0  and solve for n  and we get that

 

n = -20    [ reject ]    or

 

n  = 11    [ accept ]

 

So....11 rows

 

 

cool cool cool

 Oct 22, 2018

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