Question 1039756
{{{a[3]=a[0]+(3-1)*d=-12}}}
1.{{{a[0]+2d=-12}}}
.
.
{{{a[7]=a[0]+(7-1)*d=8}}}
2.{{{a[0]+6d=8}}}
Subtract 1 from 2,
{{{a[0]+6d-a[0]-2d=8-(-12)}}}
{{{4d=20}}}
{{{d=5}}}
and
{{{a[0]+6(5)=8}}}
{{{a[0]=8-30}}}
{{{a[0]=-22}}}
So then,
{{{a[10]=-22+(10-1)5}}}
{{{a[10]=-22+45}}}
{{{a[10]=23}}}
So,
{{{S[10]=(10/2)(-22+23)}}}
{{{S[10]=5}}}