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Let a_1, a_2, ..., a_{21} be an arithmetic sequence such that

\(\displaystyle \sum_{i = 1}^{21} a_i = 693\)

 

Compute \(\displaystyle \sum_{r = 0}^{10} a_{2r + 1}\)

 May 4, 2020
 #1
avatar+111326 
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The first  summation implies  that

 

[a1 + a21] [21/2]  = 693

 

[ a1 + a21] [10.5]  = 693       divide both sides  by  10.5

 

a1  + a21  =  66

 

So   we have 10 pairs of terms  each summing to  66

 

So....the 11th term must be   693 - 10(66)  =  693 - 660 = 33

 

And the  second summation  sums  the   odd tems  of  the series

 

So....

 

(a1 + a21)  =  66

(a3 + a19) =   66

(a5 +  a17) =  66

(a7 + a15) =   66

(a9 + a13) =   66

  a11         =   33

 

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

 

 

cool cool cool

 May 4, 2020

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