Question 1209780
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We can subtract 21 from each term to go from this arithmetic sequence
21,27,33,...,6021
to this arithmetic sequence
0,6,12,...,6000
which are the multiples of 6 in the form 6k, where k ranges from k = 0 to k = 1000
There are 1001 integers in the set {0,1,2,...,1000}


The original arithmetic sequence starts at a<sub>1</sub> = 21 and has common difference d = 6.
There are n = 1001 terms being added.


S<sub>n</sub> = sum of first n terms of arithmetic sequence
S<sub>n</sub> = (n/2)*(2*a<sub>1</sub> + d*(n-1))
S<sub>1001</sub> = (1001/2)*(2*21 + 6*(1001-1))
S<sub>1001</sub> = <font color=red>3024021</font>


Or,
S<sub>n</sub> = (n/2)*(a<sub>1</sub> + a<sub>n</sub>)
S<sub>1001</sub> = (1001/2)*(a<sub>1</sub> + a<sub>1001</sub>)
S<sub>1001</sub> = (1001/2)*(21+6021)
S<sub>1001</sub> = <font color=red>3024021</font>


Therefore,
21+27+33+...+6021 = <font color=red>3024021</font>


Verification using <a href="https://www.wolframalpha.com/input?i=21%2B27%2B33%2B...%2B6021">WolframAlpha</a>
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