Question 127233
The first term is a, the second term is a + d, the third, a + 2d, etc.


So the sum of the first 6 terms is {{{6a+15d}}}.  The 15 comes from {{{1+2+3+4+5=15}}}


So our first equation is {{{6a+15d=72}}}


The second term is {{{a+d}}} and the fifth term is {{{a+4d}}}, so the second equation is:


{{{a+d=7(a+4d)}}}, which simplifies to {{{-6a-27d=0}}}


Add the two equations, since the coefficients on the a terms are already additive inverses


{{{0a-12d=72}}}
{{{d=-6}}}


Substitute this value for d into the first equation:


{{{6a+15(-6)=72}}}
{{{6a=72+90}}}
{{{a=27}}}


Check:


{{{27+21+15+9+3-3=72}}}


{{{21=7(3)}}}


Answer checks.