Question 1021671
{{{a[4]=3a[1]}}}
{{{a[4]=a[1]+3d}}}
So then,
{{{3a[1]=a[1]+3d}}}
{{{2a[1]=3d}}}
{{{d=(2/3)a[1]}}}
You also know,
{{{a[1]+a[2]+a[3]+a[4]=120}}}
{{{a[1]+(a[1]+d)+(a[1]+2d)+(a[1]+3d)=120}}}
{{{4a[1]+6d=120}}}
Substituting,
{{{4a[1]+6(2/3)a[1]=120}}}
{{{4a[1]+4a[1]=120}}}
{{{8a[1]=120}}}
{{{a[1]=120/8}}}
{{{a[1]=15}}}
{{{a[4]=3(15)=45}}}