Question 1071095
{{{S[n]=(a[1]+a[n])*n/2}}}
is my favorite way of calculating the sum.
Substituting,
{{{5235=(15+a[30])*30/2}}}
{{{10470=(15+a[30])*30}}}
{{{349=15+a[30]}}}
{{{349-15=a[30]}}}
{{{highlight(a[30]=334)}}}
The formula for {{{a[n]}}} 
based on {{{a[1]}}} and {{{d}}} is
{{{a[n]=a[1]+(n-1)*d}}}
so, substituting we get
{{{334=15+(30-1)d}}}
{{{334=15+29d}}}
{{{334-15=29d}}}
{{{319=29d}}}
{{{319/29=d}}}
{{{highlight(d=11)}}}