a 1+(a 1+4d)= a 1+5d
a 1+4d=5d
a 1=d
a2+a4=a6
(a 1+d)+(a 1+3d)= a 1+5d
an = a 1+(n- 1)d = d+(n- 1)d = nd
When d=a 1, arithmetic progression satisfies:
am+an=a(m+n)