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Given that $a_1, a_2 , a_3, \cdots, a_{21}$ are in an arithmetic progression and the value of
\[\sum_{i = 1}^{21} a_i = 693\]
Find the value of
\[\sum_{r = 0}^{10} a_{2r + 1}\]

 Jan 28, 2021
 #1
avatar+120023 
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In th first, we have that the sum of an arithemetic series  is given  by

 

(first term +  last term)  (n /2)

 

So

 

(first term +  last term)  (21 /2  = 693

 

first term +  last term  =   693 ( 2)  / 21 =    66

 

And  in the  second  summation.....we  are summing  these terms  of the  first series  :

 

Term #   +      Term #

    1        +        21            =   66

    3        +        19            =  66

    5        +        17            =  66

    7        +        15            =  66

    9         +       13            =  66

   11                                 =  66/2   = 33

 

 

So   the sum  =   66(5) +  33    =   363                       

 

 

cool cool cool

 Jan 28, 2021

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