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Mathematics 13 Online
OpenStudy (anonymous):

Find the sum for the 1st 20 terms in the sequence. 4, 7,10,13

OpenStudy (anonymous):

First we begin by find the 20th terms, \[\huge a_{20}=a_{1}+(n-1)d\]

OpenStudy (anonymous):

\[\huge a_{1}=4,\; n=20,\; d=3\]

OpenStudy (anonymous):

Now we plug in our numbers:

OpenStudy (anonymous):

I am sill lost

OpenStudy (anonymous):

\[\huge a_{20}=61\]

OpenStudy (anonymous):

Well, ill let this overzealous user explain it to you

OpenStudy (anonymous):

Thank you for your help even though when I put in that answer it was wrong :(

OpenStudy (mr.math):

\[\sum_{n=0}^{20}4+3n=4(21)+\frac{(3\times20)(3\times 21)}{2}=84+630=714\]

OpenStudy (mr.math):

Notice here that \(\sum_{n=0}^k n=\frac{k(k+1)}{2}\).

OpenStudy (anonymous):

Thank you [=

OpenStudy (mr.math):

Or you can just add \(4+7+10+13+..+61\) :P

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