Question 995915
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These integers are  12, 18, 24, . . . , 60.


It is arithmetic progression with the first term  12  and the difference  6;  the last term is  60. The number of the terms is {{{(60-12)/6 + 1}}} = {{{48/6 + 1}}} = 8 + 1 = 9.


Use the formula for the sum of arithmetic progression


S = {{{((a[1]+a[n])/2)*n}}}


(see the lesson &nbsp;<A HREF=http://www.algebra.com/algebra/homework/Sequences-and-series/Arithmetic-progressions.lesson>Arithmetic progressions</A>&nbsp; in this site). &nbsp;It gives


S = {{{(12 + 60)/2*9}}} = {{{72/2*9}}} = 36*9 = 324.